chain_1.miz
    
    begin
    
    reserve X,x,y,z for
    set;
    
    reserve n,m,k,k9,d9 for
    Nat;
    
    theorem :: 
    
    CHAIN_1:1
    
    
    
    
    
    Th1: for x,y be 
    Real st x 
    < y holds ex z be 
    Element of 
    REAL st x 
    < z & z 
    < y 
    
    proof
    
      let x,y be
    Real;
    
      assume x
    < y; 
    
      then
    
      consider z be
    Real such that 
    
      
    
    A1: x 
    < z and 
    
      
    
    A2: z 
    < y by 
    XREAL_1: 5;
    
      reconsider z as
    Element of 
    REAL by 
    XREAL_0:def 1;
    
      take z;
    
      thus thesis by
    A1,
    A2;
    
    end;
    
    theorem :: 
    
    CHAIN_1:2
    
    
    
    
    
    Th2: for x,y be 
    Real holds ex z be 
    Element of 
    REAL st x 
    < z & y 
    < z 
    
    proof
    
      let x,y be
    Real;
    
      reconsider x, y as
    Real;
    
      reconsider z = ((
    max (x,y)) 
    + 1) as 
    Element of 
    REAL by 
    XREAL_0:def 1;
    
      take z;
    
      
    
      
    
    A1: (x 
    +  
    0 ) 
    < z by 
    XREAL_1: 8,
    XXREAL_0: 25;
    
      (y
    +  
    0 ) 
    < z by 
    XREAL_1: 8,
    XXREAL_0: 25;
    
      hence thesis by
    A1;
    
    end;
    
    scheme :: 
    
    CHAIN_1:sch1
    
    FrSet12 { D() -> non
    empty  
    set , A() -> non 
    empty  
    set , P[ 
    set, 
    set], F(
    set, 
    set) ->
    Element of D() } : 
    
{ F(x,y) where x,y be 
    Element of A() : P[x, y] } 
    c= D(); 
    
      let z be
    object;
    
      assume z
    in { F(x,y) where x,y be 
    Element of A() : P[x, y] }; 
    
      then ex x,y be
    Element of A() st z 
    = F(x,y) & P[x, y]; 
    
      hence thesis;
    
    end;
    
    definition
    
      let B be
    set;
    
      let A be
    Subset of B; 
    
      :: original:
    bool
    
      redefine
    
      func
    
    bool A -> 
    Subset-Family of B ; 
    
      coherence by
    ZFMISC_1: 67;
    
    end
    
    definition
    
      let d be
    real  
    Element of 
    NAT ; 
    
      :: original:
    zero
    
      redefine
    
      :: 
    
    CHAIN_1:def1
    
      attr d is
    
    zero means not d 
    >  
    0 ; 
    
      compatibility ;
    
    end
    
    definition
    
      let d be
    Nat;
    
      :: original:
    zero
    
      redefine
    
      :: 
    
    CHAIN_1:def2
    
      attr d is
    
    zero means 
    
      :
    
    Def2: not d 
    >= 1; 
    
      compatibility
    
      proof
    
        d is non
    zero iff d 
    >= (1 
    +  
    0 ) by 
    NAT_1: 13;
    
        hence thesis;
    
      end;
    
    end
    
    reserve d for non
    zero  
    Nat;
    
    reserve i,i0,i1 for
    Element of ( 
    Seg d); 
    
    theorem :: 
    
    CHAIN_1:3
    
    
    
    
    
    Th3: for x,y be 
    object holds 
    {x, y} is
    trivial iff x 
    = y 
    
    proof
    
      let x,y be
    object;
    
      hereby
    
        
    
        
    
    A1: x 
    in  
    {x, y} by
    TARSKI:def 2;
    
        y
    in  
    {x, y} by
    TARSKI:def 2;
    
        hence
    {x, y} is
    trivial implies x 
    = y by 
    A1;
    
      end;
    
      
    {x, x}
    =  
    {x} by
    ENUMSET1: 29;
    
      hence thesis;
    
    end;
    
    registration
    
      cluster non 
    trivial
    finite for 
    set;
    
      existence
    
      proof
    
        take
    {
    0 , 1}; 
    
        thus thesis by
    Th3;
    
      end;
    
    end
    
    registration
    
      let X be non
    trivial  
    set;
    
      let Y be
    set;
    
      cluster (X 
    \/ Y) -> non 
    trivial;
    
      coherence
    
      proof
    
        consider x,y be
    object such that 
    
        
    
    A1: x 
    in X and 
    
        
    
    A2: y 
    in X and 
    
        
    
    A3: x 
    <> y by 
    ZFMISC_1:def 10;
    
        take x, y;
    
        thus thesis by
    A1,
    A2,
    A3,
    XBOOLE_0:def 3;
    
      end;
    
      cluster (Y 
    \/ X) -> non 
    trivial;
    
      coherence ;
    
    end
    
    registration
    
      let X be non
    trivial  
    set;
    
      cluster non 
    trivial
    finite for 
    Subset of X; 
    
      existence
    
      proof
    
        consider x,y be
    object such that 
    
        
    
    A1: x 
    in X and 
    
        
    
    A2: y 
    in X and 
    
        
    
    A3: x 
    <> y by 
    ZFMISC_1:def 10;
    
        take
    {x, y};
    
        thus thesis by
    A1,
    A2,
    A3,
    Th3,
    ZFMISC_1: 32;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:4
    
    
    
    
    
    Th4: X is 
    trivial & (X 
    \/  
    {y}) is non
    trivial implies ex x be 
    object st X 
    =  
    {x}
    
    proof
    
      assume that
    
      
    
    A1: X is 
    trivial and 
    
      
    
    A2: (X 
    \/  
    {y}) is non
    trivial;
    
      X is
    empty or ex x be 
    object st X 
    =  
    {x} by
    A1,
    ZFMISC_1: 131;
    
      hence thesis by
    A2;
    
    end;
    
    scheme :: 
    
    CHAIN_1:sch2
    
    NonEmptyFinite { X() -> non
    empty  
    set , A() -> non 
    empty
    finite  
    Subset of X() , P[ 
    set] } :
    
P[A()]
    
      provided
    
      
    
    A1: for x be 
    Element of X() st x 
    in A() holds P[ 
    {x}]
    
       and 
    
      
    
    A2: for x be 
    Element of X(), B be non 
    empty
    finite  
    Subset of X() st x 
    in A() & B 
    c= A() & not x 
    in B & P[B] holds P[(B 
    \/  
    {x})];
    
      defpred
    
    Q[
    set] means $1 is
    empty or P[$1]; 
    
      
    
      
    
    A3: A() is 
    finite;
    
      
    
      
    
    A4: 
    Q[
    {} ]; 
    
      
    
      
    
    A5: for x,B be 
    set st x 
    in A() & B 
    c= A() & 
    Q[B] holds
    Q[(B
    \/  
    {x})]
    
      proof
    
        let x,B be
    set;
    
        assume that
    
        
    
    A6: x 
    in A() and 
    
        
    
    A7: B 
    c= A() and 
    
        
    
    A8: 
    Q[B];
    
        reconsider B as
    Subset of X() by 
    A7,
    XBOOLE_1: 1;
    
        per cases ;
    
          suppose x
    in B; 
    
          then
    {x}
    c= B by 
    ZFMISC_1: 31;
    
          hence thesis by
    A8,
    XBOOLE_1: 12;
    
        end;
    
          suppose B is non
    empty & not x 
    in B; 
    
          hence thesis by
    A2,
    A6,
    A7,
    A8;
    
        end;
    
          suppose B is
    empty;
    
          hence thesis by
    A1,
    A6;
    
        end;
    
      end;
    
      
    Q[A()] from
    FINSET_1:sch 2(
    A3,
    A4,
    A5);
    
      hence thesis;
    
    end;
    
    scheme :: 
    
    CHAIN_1:sch3
    
    NonTrivialFinite { X() -> non
    trivial  
    set , A() -> non 
    trivial
    finite  
    Subset of X() , P[ 
    set] } :
    
P[A()]
    
      provided
    
      
    
    A1: for x,y be 
    Element of X() st x 
    in A() & y 
    in A() & x 
    <> y holds P[ 
    {x, y}]
    
       and 
    
      
    
    A2: for x be 
    Element of X(), B be non 
    trivial
    finite  
    Subset of X() st x 
    in A() & B 
    c= A() & not x 
    in B & P[B] holds P[(B 
    \/  
    {x})];
    
      defpred
    
    Q[
    set] means $1 is
    trivial or P[$1]; 
    
      
    
      
    
    A3: A() is 
    finite;
    
      
    
      
    
    A4: 
    Q[
    {} ]; 
    
      
    
      
    
    A5: for x,B be 
    set st x 
    in A() & B 
    c= A() & 
    Q[B] holds
    Q[(B
    \/  
    {x})]
    
      proof
    
        let x,B be
    set;
    
        assume that
    
        
    
    A6: x 
    in A() and 
    
        
    
    A7: B 
    c= A() and 
    
        
    
    A8: 
    Q[B];
    
        reconsider B as
    Subset of X() by 
    A7,
    XBOOLE_1: 1;
    
        per cases ;
    
          suppose (B
    \/  
    {x}) is
    trivial;
    
          hence thesis;
    
        end;
    
          suppose x
    in B; 
    
          then
    {x}
    c= B by 
    ZFMISC_1: 31;
    
          hence thesis by
    A8,
    XBOOLE_1: 12;
    
        end;
    
          suppose B is non
    trivial & not x 
    in B; 
    
          hence thesis by
    A2,
    A6,
    A7,
    A8;
    
        end;
    
          suppose
    
          
    
    A9: B is 
    trivial & (B 
    \/  
    {x}) is non
    trivial;
    
          then
    
          consider y be
    object such that 
    
          
    
    A10: B 
    =  
    {y} by
    Th4;
    
          
    
          
    
    A11: x 
    <> y by 
    A9,
    A10;
    
          
    
          
    
    A12: (B 
    \/  
    {x})
    =  
    {x, y} by
    A10,
    ENUMSET1: 1;
    
          y
    in B by 
    A10,
    TARSKI:def 1;
    
          hence thesis by
    A1,
    A6,
    A7,
    A11,
    A12;
    
        end;
    
      end;
    
      
    Q[A()] from
    FINSET_1:sch 2(
    A3,
    A4,
    A5);
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:5
    
    
    
    
    
    Th5: ( 
    card X) 
    = 2 iff ex x, y st x 
    in X & y 
    in X & x 
    <> y & for z st z 
    in X holds z 
    = x or z 
    = y 
    
    proof
    
      hereby
    
        assume
    
        
    
    A1: ( 
    card X) 
    = 2; 
    
        then
    
        reconsider X9 = X as
    finite  
    set;
    
        consider x,y be
    object such that 
    
        
    
    A2: x 
    <> y and 
    
        
    
    A3: X9 
    =  
    {x, y} by
    A1,
    CARD_2: 60;
    
        reconsider x, y as
    set by 
    TARSKI: 1;
    
        take x, y;
    
        thus x
    in X & y 
    in X & x 
    <> y & for z st z 
    in X holds z 
    = x or z 
    = y by 
    A2,
    A3,
    TARSKI:def 2;
    
      end;
    
      given x, y such that
    
      
    
    A4: x 
    in X and 
    
      
    
    A5: y 
    in X and 
    
      
    
    A6: x 
    <> y and 
    
      
    
    A7: for z st z 
    in X holds z 
    = x or z 
    = y; 
    
      for z be
    object holds z 
    in X iff z 
    = x or z 
    = y by 
    A4,
    A5,
    A7;
    
      then X
    =  
    {x, y} by
    TARSKI:def 2;
    
      hence thesis by
    A6,
    CARD_2: 57;
    
    end;
    
    ::$Canceled
    
    theorem :: 
    
    CHAIN_1:7
    
    
    
    
    
    Th6: for X,Y be 
    finite  
    set st X 
    misses Y holds (( 
    card X) is 
    even iff ( 
    card Y) is 
    even) iff (
    card (X 
    \/ Y)) is 
    even
    
    proof
    
      let X,Y be
    finite  
    set;
    
      assume X
    misses Y; 
    
      then (
    card (X 
    \/ Y)) 
    = (( 
    card X) 
    + ( 
    card Y)) by 
    CARD_2: 40;
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:8
    
    
    
    
    
    Th7: for X,Y be 
    finite  
    set holds (( 
    card X) is 
    even iff ( 
    card Y) is 
    even) iff (
    card (X 
    \+\ Y)) is 
    even
    
    proof
    
      let X,Y be
    finite  
    set;
    
      
    
      
    
    A1: (X 
    \ Y) 
    misses (X 
    /\ Y) by 
    XBOOLE_1: 89;
    
      
    
      
    
    A2: X 
    = ((X 
    \ Y) 
    \/ (X 
    /\ Y)) by 
    XBOOLE_1: 51;
    
      
    
      
    
    A3: (Y 
    \ X) 
    misses (X 
    /\ Y) by 
    XBOOLE_1: 89;
    
      
    
      
    
    A4: Y 
    = ((Y 
    \ X) 
    \/ (X 
    /\ Y)) by 
    XBOOLE_1: 51;
    
      
    
      
    
    A5: (X 
    \ Y) 
    misses (Y 
    \ X) by 
    XBOOLE_1: 82;
    
      
    
      
    
    A6: (X 
    \+\ Y) 
    = ((X 
    \ Y) 
    \/ (Y 
    \ X)) by 
    XBOOLE_0:def 6;
    
      
    
      
    
    A7: (( 
    card (X 
    \ Y)) is 
    even iff ( 
    card (X 
    /\ Y)) is 
    even) iff (
    card X) is 
    even by 
    A1,
    A2,
    Th6;
    
      ((
    card (Y 
    \ X)) is 
    even iff ( 
    card (X 
    /\ Y)) is 
    even) iff (
    card Y) is 
    even by 
    A3,
    A4,
    Th6;
    
      hence thesis by
    A5,
    A6,
    A7,
    Th6;
    
    end;
    
    definition
    
      let n;
    
      :: original:
    REAL
    
      redefine
    
      :: 
    
    CHAIN_1:def3
    
      func
    
    REAL n means 
    
      :
    
    Def3: for x be 
    object holds x 
    in it iff x is 
    Function of ( 
    Seg n), 
    REAL ; 
    
      compatibility
    
      proof
    
        
    
        
    
    A1: for x holds x 
    in ( 
    REAL n) iff x is 
    Function of ( 
    Seg n), 
    REAL  
    
        proof
    
          let x;
    
          hereby
    
            assume x
    in ( 
    REAL n); 
    
            then x
    in (n 
    -tuples_on  
    REAL ) by 
    EUCLID:def 1;
    
            then x
    in ( 
    Funcs (( 
    Seg n), 
    REAL )) by 
    FINSEQ_2: 93;
    
            hence x is
    Function of ( 
    Seg n), 
    REAL by 
    FUNCT_2: 66;
    
          end;
    
          assume x is
    Function of ( 
    Seg n), 
    REAL ; 
    
          then x
    in ( 
    Funcs (( 
    Seg n), 
    REAL )) by 
    FUNCT_2: 8;
    
          then x
    in (n 
    -tuples_on  
    REAL ) by 
    FINSEQ_2: 93;
    
          hence thesis by
    EUCLID:def 1;
    
        end;
    
        let X be
    FinSequenceSet of 
    REAL ; 
    
        thus X
    = ( 
    REAL n) implies for x be 
    object holds x 
    in X iff x is 
    Function of ( 
    Seg n), 
    REAL by 
    A1;
    
        assume
    
        
    
    A2: for x be 
    object holds x 
    in X iff x is 
    Function of ( 
    Seg n), 
    REAL ; 
    
        now
    
          let x be
    object;
    
          x
    in X iff x is 
    Function of ( 
    Seg n), 
    REAL by 
    A2;
    
          hence x
    in X iff x 
    in ( 
    REAL n) by 
    A1;
    
        end;
    
        hence thesis by
    TARSKI: 2;
    
      end;
    
    end
    
    reserve l,r,l9,r9,l99,r99,x,x9,l1,r1,l2,r2 for
    Element of ( 
    REAL d); 
    
    reserve Gi for non
    trivial
    finite  
    Subset of 
    REAL ; 
    
    reserve li,ri,li9,ri9,xi,xi9 for
    Real;
    
    begin
    
    definition
    
      let d;
    
      :: 
    
    CHAIN_1:def4
    
      mode
    
    Grating of d -> 
    Function of ( 
    Seg d), ( 
    bool  
    REAL ) means 
    
      :
    
    Def4: for i holds (it 
    . i) is non 
    trivial
    finite;
    
      existence
    
      proof
    
        defpred
    
    P[
    object, 
    object] means $2 is non
    trivial
    finite  
    Subset of 
    REAL ; 
    
        
    
        
    
    A1: for i be 
    object st i 
    in ( 
    Seg d) holds ex X be 
    object st 
    P[i, X]
    
        proof
    
          let i be
    object;
    
          assume i
    in ( 
    Seg d); 
    
          set X = the non
    trivial
    finite  
    Subset of 
    REAL ; 
    
          take X;
    
          thus thesis;
    
        end;
    
        consider G be
    Function such that 
    
        
    
    A2: ( 
    dom G) 
    = ( 
    Seg d) & for i be 
    object st i 
    in ( 
    Seg d) holds 
    P[i, (G
    . i)] from 
    CLASSES1:sch 1(
    A1);
    
        for i be
    object st i 
    in ( 
    Seg d) holds (G 
    . i) 
    in ( 
    bool  
    REAL ) 
    
        proof
    
          let i be
    object;
    
          assume i
    in ( 
    Seg d); 
    
          then (G
    . i) is 
    Subset of 
    REAL by 
    A2;
    
          hence thesis;
    
        end;
    
        then
    
        reconsider G as
    Function of ( 
    Seg d), ( 
    bool  
    REAL ) by 
    A2,
    FUNCT_2: 3;
    
        take G;
    
        thus thesis by
    A2;
    
      end;
    
    end
    
    reserve G for
    Grating of d; 
    
    registration
    
      let d;
    
      cluster -> 
    finite-yielding for 
    Grating of d; 
    
      coherence
    
      proof
    
        let G;
    
        let i be
    object;
    
        assume i
    in ( 
    Seg d); 
    
        hence (G
    . i) is 
    finite by 
    Def4;
    
      end;
    
    end
    
    definition
    
      let d, G, i;
    
      :: original:
    .
    
      redefine
    
      func G
    
    . i -> non 
    trivial
    finite  
    Subset of 
    REAL ; 
    
      coherence by
    Def4;
    
    end
    
    theorem :: 
    
    CHAIN_1:9
    
    
    
    
    
    Th8: x 
    in ( 
    product G) iff for i holds (x 
    . i) 
    in (G 
    . i) 
    
    proof
    
      x is
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
      then
    
      
    
    A1: ( 
    dom x) 
    = ( 
    Seg d) by 
    FUNCT_2:def 1;
    
      
    
      
    
    A2: ( 
    dom G) 
    = ( 
    Seg d) by 
    FUNCT_2:def 1;
    
      hence x
    in ( 
    product G) implies for i holds (x 
    . i) 
    in (G 
    . i) by 
    CARD_3: 9;
    
      assume for i holds (x
    . i) 
    in (G 
    . i); 
    
      then for i be
    object st i 
    in ( 
    Seg d) holds (x 
    . i) 
    in (G 
    . i); 
    
      hence thesis by
    A1,
    A2,
    CARD_3: 9;
    
    end;
    
    ::$Canceled
    
    theorem :: 
    
    CHAIN_1:11
    
    
    
    
    
    Th9: for X be non 
    empty
    finite  
    Subset of 
    REAL holds ex ri be 
    Element of 
    REAL st ri 
    in X & for xi st xi 
    in X holds ri 
    >= xi 
    
    proof
    
      defpred
    
    P[
    set] means ex ri be
    Element of 
    REAL st ri 
    in $1 & for xi st xi 
    in $1 holds ri 
    >= xi; 
    
      let X be non
    empty
    finite  
    Subset of 
    REAL ; 
    
      
    
      
    
    A1: for xi be 
    Element of 
    REAL st xi 
    in X holds 
    P[
    {xi}]
    
      proof
    
        let xi be
    Element of 
    REAL ; 
    
        assume xi
    in X; 
    
        take xi;
    
        thus thesis by
    TARSKI:def 1;
    
      end;
    
      
    
      
    
    A2: for x be 
    Element of 
    REAL , B be non 
    empty
    finite  
    Subset of 
    REAL st x 
    in X & B 
    c= X & not x 
    in B & 
    P[B] holds
    P[(B
    \/  
    {x})]
    
      proof
    
        let x be
    Element of 
    REAL ; 
    
        let B be non
    empty
    finite  
    Subset of 
    REAL ; 
    
        assume that x
    in X and B 
    c= X and not x 
    in B and 
    
        
    
    A3: 
    P[B];
    
        consider ri such that
    
        
    
    A4: ri 
    in B and 
    
        
    
    A5: for xi st xi 
    in B holds ri 
    >= xi by 
    A3;
    
        set B9 = (B
    \/  
    {x});
    
        
    
    A6: 
    
        now
    
          let xi;
    
          xi
    in  
    {x} iff xi
    = x by 
    TARSKI:def 1;
    
          hence xi
    in B9 iff xi 
    in B or xi 
    = x by 
    XBOOLE_0:def 3;
    
        end;
    
        per cases ;
    
          suppose
    
          
    
    A7: x 
    <= ri; 
    
          reconsider ri as
    Element of 
    REAL by 
    XREAL_0:def 1;
    
          take ri;
    
          thus ri
    in B9 by 
    A4,
    A6;
    
          let xi;
    
          assume xi
    in B9; 
    
          then xi
    in B or xi 
    = x by 
    A6;
    
          hence thesis by
    A5,
    A7;
    
        end;
    
          suppose
    
          
    
    A8: ri 
    < x; 
    
          take x;
    
          thus x
    in B9 by 
    A6;
    
          let xi;
    
          assume xi
    in B9; 
    
          then xi
    in B or xi 
    = x by 
    A6;
    
          then ri
    >= xi or xi 
    = x by 
    A5;
    
          hence thesis by
    A8,
    XXREAL_0: 2;
    
        end;
    
      end;
    
      thus
    P[X] from
    NonEmptyFinite(
    A1,
    A2);
    
    end;
    
    theorem :: 
    
    CHAIN_1:12
    
    
    
    
    
    Th10: for X be non 
    empty
    finite  
    Subset of 
    REAL holds ex li be 
    Element of 
    REAL st li 
    in X & for xi st xi 
    in X holds li 
    <= xi 
    
    proof
    
      defpred
    
    P[
    set] means ex li be
    Element of 
    REAL st li 
    in $1 & for xi st xi 
    in $1 holds li 
    <= xi; 
    
      let X be non
    empty
    finite  
    Subset of 
    REAL ; 
    
      
    
      
    
    A1: for xi be 
    Element of 
    REAL st xi 
    in X holds 
    P[
    {xi}]
    
      proof
    
        let xi be
    Element of 
    REAL ; 
    
        assume xi
    in X; 
    
        take xi;
    
        thus thesis by
    TARSKI:def 1;
    
      end;
    
      
    
      
    
    A2: for x be 
    Element of 
    REAL , B be non 
    empty
    finite  
    Subset of 
    REAL st x 
    in X & B 
    c= X & not x 
    in B & 
    P[B] holds
    P[(B
    \/  
    {x})]
    
      proof
    
        let x be
    Element of 
    REAL ; 
    
        let B be non
    empty
    finite  
    Subset of 
    REAL ; 
    
        assume that x
    in X and B 
    c= X and not x 
    in B and 
    
        
    
    A3: 
    P[B];
    
        consider li such that
    
        
    
    A4: li 
    in B and 
    
        
    
    A5: for xi st xi 
    in B holds li 
    <= xi by 
    A3;
    
        set B9 = (B
    \/  
    {x});
    
        
    
    A6: 
    
        now
    
          let xi;
    
          xi
    in  
    {x} iff xi
    = x by 
    TARSKI:def 1;
    
          hence xi
    in B9 iff xi 
    in B or xi 
    = x by 
    XBOOLE_0:def 3;
    
        end;
    
        per cases ;
    
          suppose
    
          
    
    A7: li 
    <= x; 
    
          reconsider li as
    Element of 
    REAL by 
    XREAL_0:def 1;
    
          take li;
    
          thus li
    in B9 by 
    A4,
    A6;
    
          let xi;
    
          assume xi
    in B9; 
    
          then xi
    in B or xi 
    = x by 
    A6;
    
          hence thesis by
    A5,
    A7;
    
        end;
    
          suppose
    
          
    
    A8: x 
    < li; 
    
          take x;
    
          thus x
    in B9 by 
    A6;
    
          let xi;
    
          assume xi
    in B9; 
    
          then xi
    in B or xi 
    = x by 
    A6;
    
          then li
    <= xi or xi 
    = x by 
    A5;
    
          hence thesis by
    A8,
    XXREAL_0: 2;
    
        end;
    
      end;
    
      thus
    P[X] from
    NonEmptyFinite(
    A1,
    A2);
    
    end;
    
    theorem :: 
    
    CHAIN_1:13
    
    
    
    
    
    Th11: ex li, ri st li 
    in Gi & ri 
    in Gi & li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri) 
    
    proof
    
      defpred
    
    P[
    set] means ex li, ri st li
    in $1 & ri 
    in $1 & li 
    < ri & for xi st xi 
    in $1 holds not (li 
    < xi & xi 
    < ri); 
    
      
    
    A1: 
    
      now
    
        let li,ri be
    Element of 
    REAL ; 
    
        assume that li
    in Gi and ri 
    in Gi and 
    
        
    
    A2: li 
    <> ri; 
    
        
    
    A3: 
    
        now
    
          let li, ri;
    
          assume
    
          
    
    A4: li 
    < ri; 
    
          thus
    P[
    {li, ri}]
    
          proof
    
            take li, ri;
    
            thus thesis by
    A4,
    TARSKI:def 2;
    
          end;
    
        end;
    
        li
    < ri or ri 
    < li by 
    A2,
    XXREAL_0: 1;
    
        hence
    P[
    {li, ri}] by
    A3;
    
      end;
    
      
    
      
    
    A5: for x be 
    Element of 
    REAL , B be non 
    trivial
    finite  
    Subset of 
    REAL st x 
    in Gi & B 
    c= Gi & not x 
    in B & 
    P[B] holds
    P[(B
    \/  
    {x})]
    
      proof
    
        let x be
    Element of 
    REAL ; 
    
        let B be non
    trivial
    finite  
    Subset of 
    REAL ; 
    
        assume that x
    in Gi and B 
    c= Gi and 
    
        
    
    A6: not x 
    in B and 
    
        
    
    A7: 
    P[B];
    
        consider li, ri such that
    
        
    
    A8: li 
    in B and 
    
        
    
    A9: ri 
    in B and 
    
        
    
    A10: li 
    < ri and 
    
        
    
    A11: for xi st xi 
    in B holds not (li 
    < xi & xi 
    < ri) by 
    A7;
    
        per cases by
    A6,
    A8,
    A9,
    XXREAL_0: 1;
    
          suppose
    
          
    
    A12: x 
    < li; 
    
          take li, ri;
    
          thus li
    in (B 
    \/  
    {x}) & ri
    in (B 
    \/  
    {x}) & li
    < ri by 
    A8,
    A9,
    A10,
    XBOOLE_0:def 3;
    
          let xi;
    
          assume xi
    in (B 
    \/  
    {x});
    
          then xi
    in B or xi 
    in  
    {x} by
    XBOOLE_0:def 3;
    
          hence thesis by
    A11,
    A12,
    TARSKI:def 1;
    
        end;
    
          suppose
    
          
    
    A13: li 
    < x & x 
    < ri; 
    
          take li, x;
    
          x
    in  
    {x} by
    TARSKI:def 1;
    
          hence li
    in (B 
    \/  
    {x}) & x
    in (B 
    \/  
    {x}) & li
    < x by 
    A8,
    A13,
    XBOOLE_0:def 3;
    
          let xi;
    
          assume xi
    in (B 
    \/  
    {x});
    
          then xi
    in B or xi 
    in  
    {x} by
    XBOOLE_0:def 3;
    
          then not (li
    < xi & xi 
    < ri) or xi 
    = x by 
    A11,
    TARSKI:def 1;
    
          hence thesis by
    A13,
    XXREAL_0: 2;
    
        end;
    
          suppose
    
          
    
    A14: ri 
    < x; 
    
          take li, ri;
    
          thus li
    in (B 
    \/  
    {x}) & ri
    in (B 
    \/  
    {x}) & li
    < ri by 
    A8,
    A9,
    A10,
    XBOOLE_0:def 3;
    
          let xi;
    
          assume xi
    in (B 
    \/  
    {x});
    
          then xi
    in B or xi 
    in  
    {x} by
    XBOOLE_0:def 3;
    
          hence thesis by
    A11,
    A14,
    TARSKI:def 1;
    
        end;
    
      end;
    
      thus
    P[Gi] from
    NonTrivialFinite(
    A1,
    A5);
    
    end;
    
    ::$Canceled
    
    theorem :: 
    
    CHAIN_1:15
    
    ex li, ri st li
    in Gi & ri 
    in Gi & ri 
    < li & for xi st xi 
    in Gi holds not (xi 
    < ri or li 
    < xi) 
    
    proof
    
      consider li be
    Element of 
    REAL such that 
    
      
    
    A1: li 
    in Gi and 
    
      
    
    A2: for xi st xi 
    in Gi holds li 
    >= xi by 
    Th9;
    
      consider ri be
    Element of 
    REAL such that 
    
      
    
    A3: ri 
    in Gi and 
    
      
    
    A4: for xi st xi 
    in Gi holds ri 
    <= xi by 
    Th10;
    
      take li, ri;
    
      
    
      
    
    A5: ri 
    <= li by 
    A2,
    A3;
    
      now
    
        assume
    
        
    
    A6: li 
    = ri; 
    
        consider x1,x2 be
    object such that 
    
        
    
    A7: x1 
    in Gi and 
    
        
    
    A8: x2 
    in Gi and 
    
        
    
    A9: x1 
    <> x2 by 
    ZFMISC_1:def 10;
    
        reconsider x1, x2 as
    Element of 
    REAL by 
    A7,
    A8;
    
        
    
        
    
    A10: ri 
    <= x1 by 
    A4,
    A7;
    
        
    
        
    
    A11: x1 
    <= li by 
    A2,
    A7;
    
        
    
        
    
    A12: ri 
    <= x2 by 
    A4,
    A8;
    
        
    
        
    
    A13: x2 
    <= li by 
    A2,
    A8;
    
        x1
    = li by 
    A6,
    A10,
    A11,
    XXREAL_0: 1;
    
        hence contradiction by
    A6,
    A9,
    A12,
    A13,
    XXREAL_0: 1;
    
      end;
    
      hence thesis by
    A1,
    A2,
    A3,
    A4,
    A5,
    XXREAL_0: 1;
    
    end;
    
    definition
    
      let Gi;
    
      :: 
    
    CHAIN_1:def5
    
      mode
    
    Gap of Gi -> 
    Element of 
    [:
    REAL , 
    REAL :] means 
    
      :
    
    Def5: ex li, ri st it 
    =  
    [li, ri] & li
    in Gi & ri 
    in Gi & ((li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri)) or (ri 
    < li & for xi st xi 
    in Gi holds not (li 
    < xi or xi 
    < ri))); 
    
      existence
    
      proof
    
        consider li, ri such that
    
        
    
    A1: li 
    in Gi and 
    
        
    
    A2: ri 
    in Gi and 
    
        
    
    A3: li 
    < ri and 
    
        
    
    A4: for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri) by 
    Th11;
    
        reconsider ri, li as
    Element of 
    REAL by 
    XREAL_0:def 1;
    
        take
    [li, ri], li, ri;
    
        thus thesis by
    A1,
    A2,
    A3,
    A4;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:16
    
    
    
    
    
    Th13: 
    [li, ri] is
    Gap of Gi iff li 
    in Gi & ri 
    in Gi & ((li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri)) or (ri 
    < li & for xi st xi 
    in Gi holds not (li 
    < xi or xi 
    < ri))) 
    
    proof
    
      thus
    [li, ri] is
    Gap of Gi implies li 
    in Gi & ri 
    in Gi & ((li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri)) or (ri 
    < li & for xi st xi 
    in Gi holds not (li 
    < xi or xi 
    < ri))) 
    
      proof
    
        assume
    [li, ri] is
    Gap of Gi; 
    
        then
    
        consider li9, ri9 such that
    
        
    
    A1: 
    [li, ri]
    =  
    [li9, ri9] and
    
        
    
    A2: li9 
    in Gi and 
    
        
    
    A3: ri9 
    in Gi and 
    
        
    
    A4: (li9 
    < ri9 & for xi st xi 
    in Gi holds not (li9 
    < xi & xi 
    < ri9)) or (ri9 
    < li9 & for xi st xi 
    in Gi holds not (li9 
    < xi or xi 
    < ri9)) by 
    Def5;
    
        
    
        
    
    A5: li9 
    = li by 
    A1,
    XTUPLE_0: 1;
    
        ri9
    = ri by 
    A1,
    XTUPLE_0: 1;
    
        hence thesis by
    A2,
    A3,
    A4,
    A5;
    
      end;
    
      li
    in  
    REAL & ri 
    in  
    REAL by 
    XREAL_0:def 1;
    
      then
    [li, ri]
    in  
    [:
    REAL , 
    REAL :] by 
    ZFMISC_1: 87;
    
      hence thesis by
    Def5;
    
    end;
    
    theorem :: 
    
    CHAIN_1:17
    
    Gi
    =  
    {li, ri} implies (
    [li9, ri9] is
    Gap of Gi iff li9 
    = li & ri9 
    = ri or li9 
    = ri & ri9 
    = li) 
    
    proof
    
      assume
    
      
    
    A1: Gi 
    =  
    {li, ri};
    
      hereby
    
        assume
    
        
    
    A2: 
    [li9, ri9] is
    Gap of Gi; 
    
        then
    
        
    
    A3: li9 
    in Gi by 
    Th13;
    
        
    
        
    
    A4: ri9 
    in Gi by 
    A2,
    Th13;
    
        
    
        
    
    A5: li9 
    = li or li9 
    = ri by 
    A1,
    A3,
    TARSKI:def 2;
    
        li9
    <> ri9 by 
    A2,
    Th13;
    
        hence li9
    = li & ri9 
    = ri or li9 
    = ri & ri9 
    = li by 
    A1,
    A4,
    A5,
    TARSKI:def 2;
    
      end;
    
      assume
    
      
    
    A6: li9 
    = li & ri9 
    = ri or li9 
    = ri & ri9 
    = li; 
    
      li9
    in  
    REAL & ri9 
    in  
    REAL by 
    XREAL_0:def 1;
    
      then
    [li9, ri9]
    in  
    [:
    REAL , 
    REAL :] by 
    ZFMISC_1: 87;
    
      then
    
      reconsider liri =
    [li9, ri9] as
    Element of 
    [:
    REAL , 
    REAL :]; 
    
      liri is
    Gap of Gi 
    
      proof
    
        take li9, ri9;
    
        li
    <> ri by 
    A1,
    Th3;
    
        hence thesis by
    A1,
    TARSKI:def 2,
    XXREAL_0: 1,
    A6;
    
      end;
    
      hence thesis;
    
    end;
    
    deffunc
    
    f(
    set) = $1;
    
    theorem :: 
    
    CHAIN_1:18
    
    
    
    
    
    Th15: xi 
    in Gi implies ex ri be 
    Element of 
    REAL st 
    [xi, ri] is
    Gap of Gi 
    
    proof
    
      assume
    
      
    
    A1: xi 
    in Gi; 
    
      defpred
    
    P[
    Element of 
    REAL ] means $1 
    > xi; 
    
      set Gi9 = {
    f(ri9) where ri9 be
    Element of 
    REAL : 
    f(ri9)
    in Gi & 
    P[ri9] };
    
      
    
      
    
    A2: Gi9 
    c= Gi from 
    FRAENKEL:sch 17;
    
      then
    
      reconsider Gi9 as
    finite  
    Subset of 
    REAL by 
    XBOOLE_1: 1;
    
      per cases ;
    
        suppose
    
        
    
    A3: Gi9 is 
    empty;
    
        
    
    A4: 
    
        now
    
          let xi9;
    
          assume that
    
          
    
    A5: xi9 
    in Gi and 
    
          
    
    A6: xi9 
    > xi; 
    
          xi9
    in Gi9 by 
    A5,
    A6;
    
          hence contradiction by
    A3;
    
        end;
    
        consider li be
    Element of 
    REAL such that 
    
        
    
    A7: li 
    in Gi and 
    
        
    
    A8: for xi9 st xi9 
    in Gi holds li 
    <= xi9 by 
    Th10;
    
        take li;
    
        
    
    A9: 
    
        now
    
          assume
    
          
    
    A10: li 
    = xi; 
    
          for xi9 be
    object holds xi9 
    in Gi iff xi9 
    = xi 
    
          proof
    
            let xi9 be
    object;
    
            hereby
    
              assume
    
              
    
    A11: xi9 
    in Gi; 
    
              then
    
              reconsider xi99 = xi9 as
    Element of 
    REAL ; 
    
              
    
              
    
    A12: li 
    <= xi99 by 
    A8,
    A11;
    
              xi99
    <= xi by 
    A4,
    A11;
    
              hence xi9
    = xi by 
    A10,
    A12,
    XXREAL_0: 1;
    
            end;
    
            thus thesis by
    A1;
    
          end;
    
          hence Gi
    =  
    {xi} by
    TARSKI:def 1;
    
          hence contradiction;
    
        end;
    
        li
    <= xi by 
    A1,
    A8;
    
        then
    
        
    
    A13: li 
    < xi by 
    A9,
    XXREAL_0: 1;
    
        for xi9 st xi9
    in Gi holds not (xi 
    < xi9 or xi9 
    < li) by 
    A4,
    A8;
    
        hence thesis by
    A1,
    A7,
    A13,
    Th13;
    
      end;
    
        suppose Gi9 is non
    empty;
    
        then
    
        reconsider Gi9 as non
    empty
    finite  
    Subset of 
    REAL ; 
    
        consider ri be
    Element of 
    REAL such that 
    
        
    
    A14: ri 
    in Gi9 and 
    
        
    
    A15: for ri9 st ri9 
    in Gi9 holds ri9 
    >= ri by 
    Th10;
    
        take ri;
    
        now
    
          thus xi
    in Gi by 
    A1;
    
          thus ri
    in Gi by 
    A2,
    A14;
    
          ex ri9 be
    Element of 
    REAL st ri9 
    = ri & ri9 
    in Gi & xi 
    < ri9 by 
    A14;
    
          hence xi
    < ri; 
    
          hereby
    
            let xi9;
    
            assume xi9
    in Gi; 
    
            then xi9
    <= xi or xi9 
    in Gi9; 
    
            hence not (xi
    < xi9 & xi9 
    < ri) by 
    A15;
    
          end;
    
        end;
    
        hence thesis by
    Th13;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:19
    
    
    
    
    
    Th16: xi 
    in Gi implies ex li be 
    Element of 
    REAL st 
    [li, xi] is
    Gap of Gi 
    
    proof
    
      assume
    
      
    
    A1: xi 
    in Gi; 
    
      defpred
    
    P[
    Element of 
    REAL ] means $1 
    < xi; 
    
      set Gi9 = {
    f(li9) where li9 be
    Element of 
    REAL : 
    f(li9)
    in Gi & 
    P[li9] };
    
      
    
      
    
    A2: Gi9 
    c= Gi from 
    FRAENKEL:sch 17;
    
      then
    
      reconsider Gi9 as
    finite  
    Subset of 
    REAL by 
    XBOOLE_1: 1;
    
      per cases ;
    
        suppose
    
        
    
    A3: Gi9 is 
    empty;
    
        
    
    A4: 
    
        now
    
          let xi9;
    
          assume that
    
          
    
    A5: xi9 
    in Gi and 
    
          
    
    A6: xi9 
    < xi; 
    
          xi9
    in Gi9 by 
    A5,
    A6;
    
          hence contradiction by
    A3;
    
        end;
    
        consider ri be
    Element of 
    REAL such that 
    
        
    
    A7: ri 
    in Gi and 
    
        
    
    A8: for xi9 st xi9 
    in Gi holds ri 
    >= xi9 by 
    Th9;
    
        take ri;
    
        
    
    A9: 
    
        now
    
          assume
    
          
    
    A10: ri 
    = xi; 
    
          for xi9 be
    object holds xi9 
    in Gi iff xi9 
    = xi 
    
          proof
    
            let xi9 be
    object;
    
            hereby
    
              assume
    
              
    
    A11: xi9 
    in Gi; 
    
              then
    
              reconsider xi99 = xi9 as
    Element of 
    REAL ; 
    
              
    
              
    
    A12: ri 
    >= xi99 by 
    A8,
    A11;
    
              xi99
    >= xi by 
    A4,
    A11;
    
              hence xi9
    = xi by 
    A10,
    A12,
    XXREAL_0: 1;
    
            end;
    
            thus thesis by
    A1;
    
          end;
    
          hence Gi
    =  
    {xi} by
    TARSKI:def 1;
    
          hence contradiction;
    
        end;
    
        ri
    >= xi by 
    A1,
    A8;
    
        then
    
        
    
    A13: ri 
    > xi by 
    A9,
    XXREAL_0: 1;
    
        for xi9 st xi9
    in Gi holds not (xi9 
    > ri or xi 
    > xi9) by 
    A4,
    A8;
    
        hence thesis by
    A1,
    A7,
    A13,
    Th13;
    
      end;
    
        suppose Gi9 is non
    empty;
    
        then
    
        reconsider Gi9 as non
    empty
    finite  
    Subset of 
    REAL ; 
    
        consider li be
    Element of 
    REAL such that 
    
        
    
    A14: li 
    in Gi9 and 
    
        
    
    A15: for li9 st li9 
    in Gi9 holds li9 
    <= li by 
    Th9;
    
        take li;
    
        now
    
          thus xi
    in Gi by 
    A1;
    
          thus li
    in Gi by 
    A2,
    A14;
    
          ex li9 be
    Element of 
    REAL st li9 
    = li & li9 
    in Gi & xi 
    > li9 by 
    A14;
    
          hence xi
    > li; 
    
          hereby
    
            let xi9;
    
            assume xi9
    in Gi; 
    
            then xi9
    >= xi or xi9 
    in Gi9; 
    
            hence not (xi9
    > li & xi 
    > xi9) by 
    A15;
    
          end;
    
        end;
    
        hence thesis by
    Th13;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:20
    
    
    
    
    
    Th17: 
    [li, ri] is
    Gap of Gi & 
    [li, ri9] is
    Gap of Gi implies ri 
    = ri9 
    
    proof
    
      
    
      
    
    A1: ri 
    <= ri9 & ri9 
    <= ri implies ri 
    = ri9 by 
    XXREAL_0: 1;
    
      assume that
    
      
    
    A2: 
    [li, ri] is
    Gap of Gi and 
    
      
    
    A3: 
    [li, ri9] is
    Gap of Gi; 
    
      
    
      
    
    A4: ri 
    in Gi by 
    A2,
    Th13;
    
      
    
      
    
    A5: ri9 
    in Gi by 
    A3,
    Th13;
    
      per cases by
    A2,
    Th13;
    
        suppose
    
        
    
    A6: li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri); 
    
        ri9
    <= li or li 
    < ri9 & ri9 
    < ri or ri 
    <= ri9; 
    
        hence thesis by
    A1,
    A3,
    A4,
    A5,
    A6,
    Th13;
    
      end;
    
        suppose
    
        
    
    A7: ri 
    < li & for xi st xi 
    in Gi holds not (li 
    < xi or xi 
    < ri); 
    
        ri9
    < ri or ri 
    <= ri9 & ri9 
    <= li or li 
    < ri9; 
    
        hence thesis by
    A1,
    A3,
    A4,
    A5,
    A7,
    Th13;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:21
    
    
    
    
    
    Th18: 
    [li, ri] is
    Gap of Gi & 
    [li9, ri] is
    Gap of Gi implies li 
    = li9 
    
    proof
    
      
    
      
    
    A1: li 
    <= li9 & li9 
    <= li implies li 
    = li9 by 
    XXREAL_0: 1;
    
      assume that
    
      
    
    A2: 
    [li, ri] is
    Gap of Gi and 
    
      
    
    A3: 
    [li9, ri] is
    Gap of Gi; 
    
      
    
      
    
    A4: li 
    in Gi by 
    A2,
    Th13;
    
      
    
      
    
    A5: li9 
    in Gi by 
    A3,
    Th13;
    
      per cases by
    A2,
    Th13;
    
        suppose
    
        
    
    A6: li 
    < ri & for xi st xi 
    in Gi holds not (li 
    < xi & xi 
    < ri); 
    
        li9
    <= li or li 
    < li9 & li9 
    < ri or ri 
    <= li9; 
    
        hence thesis by
    A1,
    A3,
    A4,
    A5,
    A6,
    Th13;
    
      end;
    
        suppose
    
        
    
    A7: ri 
    < li & for xi st xi 
    in Gi holds not (li 
    < xi or xi 
    < ri); 
    
        li9
    < ri or ri 
    <= li9 & li9 
    <= li or li 
    < li9; 
    
        hence thesis by
    A1,
    A3,
    A4,
    A5,
    A7,
    Th13;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:22
    
    
    
    
    
    Th19: ri 
    < li & 
    [li, ri] is
    Gap of Gi & ri9 
    < li9 & 
    [li9, ri9] is
    Gap of Gi implies li 
    = li9 & ri 
    = ri9 
    
    proof
    
      assume that
    
      
    
    A1: ri 
    < li and 
    
      
    
    A2: 
    [li, ri] is
    Gap of Gi and 
    
      
    
    A3: ri9 
    < li9 and 
    
      
    
    A4: 
    [li9, ri9] is
    Gap of Gi; 
    
      
    
      
    
    A5: li 
    in Gi by 
    A2,
    Th13;
    
      
    
      
    
    A6: ri 
    in Gi by 
    A2,
    Th13;
    
      
    
      
    
    A7: li9 
    in Gi by 
    A4,
    Th13;
    
      
    
      
    
    A8: ri9 
    in Gi by 
    A4,
    Th13;
    
      hereby
    
        assume li
    <> li9; 
    
        then li
    < li9 or li9 
    < li by 
    XXREAL_0: 1;
    
        hence contradiction by
    A1,
    A2,
    A3,
    A4,
    A5,
    A7,
    Th13;
    
      end;
    
      hereby
    
        assume ri
    <> ri9; 
    
        then ri
    < ri9 or ri9 
    < ri by 
    XXREAL_0: 1;
    
        hence contradiction by
    A1,
    A2,
    A3,
    A4,
    A6,
    A8,
    Th13;
    
      end;
    
    end;
    
    definition
    
      let d, l, r;
    
      :: 
    
    CHAIN_1:def6
    
      func
    
    cell (l,r) -> non 
    empty  
    Subset of ( 
    REAL d) equals { x : (for i holds (l 
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r 
    . i)) or (ex i st (r 
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i))) }; 
    
      coherence
    
      proof
    
        defpred
    
    P[
    Element of ( 
    REAL d)] means (for i holds (l 
    . i) 
    <= ($1 
    . i) & ($1 
    . i) 
    <= (r 
    . i)) or (ex i st (r 
    . i) 
    < (l 
    . i) & (($1 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= ($1 
    . i))); 
    
        set CELL = { x :
    P[x] };
    
        
    P[l];
    
        then
    
        
    
    A1: l 
    in CELL; 
    
        CELL
    c= ( 
    REAL d) from 
    FRAENKEL:sch 10;
    
        hence thesis by
    A1;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:23
    
    
    
    
    
    Th20: x 
    in ( 
    cell (l,r)) iff ((for i holds (l 
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r 
    . i)) or ex i st (r 
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i))) 
    
    proof
    
      defpred
    
    P[
    Element of ( 
    REAL d)] means (for i holds (l 
    . i) 
    <= ($1 
    . i) & ($1 
    . i) 
    <= (r 
    . i)) or (ex i st (r 
    . i) 
    < (l 
    . i) & (($1 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= ($1 
    . i))); 
    
      
    
      
    
    A1: ( 
    cell (l,r)) 
    = { x9 : 
    P[x9] };
    
      thus x
    in ( 
    cell (l,r)) iff 
    P[x] from
    LMOD_7:sch 7(
    A1);
    
    end;
    
    theorem :: 
    
    CHAIN_1:24
    
    
    
    
    
    Th21: (for i holds (l 
    . i) 
    <= (r 
    . i)) implies (x 
    in ( 
    cell (l,r)) iff for i holds (l 
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: for i holds (l 
    . i) 
    <= (r 
    . i); 
    
      hereby
    
        assume x
    in ( 
    cell (l,r)); 
    
        then (for i holds (l
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r 
    . i)) or ex i st (r 
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i)) by 
    Th20;
    
        hence for i holds (l
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r 
    . i) by 
    A1;
    
      end;
    
      thus thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:25
    
    
    
    
    
    Th22: (ex i st (r 
    . i) 
    < (l 
    . i)) implies (x 
    in ( 
    cell (l,r)) iff ex i st (r 
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i))) 
    
    proof
    
      given i0 such that
    
      
    
    A1: (r 
    . i0) 
    < (l 
    . i0); 
    
      (x
    . i0) 
    < (l 
    . i0) or (r 
    . i0) 
    < (x 
    . i0) by 
    A1,
    XXREAL_0: 2;
    
      hence x
    in ( 
    cell (l,r)) implies ex i st (r 
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i)) by 
    Th20;
    
      thus thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:26
    
    
    
    
    
    Th23: l 
    in ( 
    cell (l,r)) & r 
    in ( 
    cell (l,r)) 
    
    proof
    
      
    
      
    
    A1: (for i holds (l 
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i)) or ex i st (r 
    . i) 
    < (l 
    . i) & ((l 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (l 
    . i)); 
    
      (for i holds (l
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r 
    . i)) or ex i st (r 
    . i) 
    < (l 
    . i) & ((r 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (r 
    . i)); 
    
      hence thesis by
    A1;
    
    end;
    
    theorem :: 
    
    CHAIN_1:27
    
    
    
    
    
    Th24: ( 
    cell (x,x)) 
    =  
    {x}
    
    proof
    
      for x9 be
    object holds x9 
    in ( 
    cell (x,x)) iff x9 
    = x 
    
      proof
    
        let x9 be
    object;
    
        thus x9
    in ( 
    cell (x,x)) implies x9 
    = x 
    
        proof
    
          assume
    
          
    
    A1: x9 
    in ( 
    cell (x,x)); 
    
          then
    
          reconsider x, x9 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
          now
    
            let i;
    
            
    
            
    
    A2: for i holds (x 
    . i) 
    <= (x 
    . i); 
    
            then
    
            
    
    A3: (x 
    . i) 
    <= (x9 
    . i) by 
    A1,
    Th21;
    
            (x9
    . i) 
    <= (x 
    . i) by 
    A1,
    A2,
    Th21;
    
            hence (x9
    . i) 
    = (x 
    . i) by 
    A3,
    XXREAL_0: 1;
    
          end;
    
          hence thesis by
    FUNCT_2: 63;
    
        end;
    
        thus thesis by
    Th23;
    
      end;
    
      hence thesis by
    TARSKI:def 1;
    
    end;
    
    theorem :: 
    
    CHAIN_1:28
    
    
    
    
    
    Th25: (for i holds (l9 
    . i) 
    <= (r9 
    . i)) implies (( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) iff for i holds (l9 
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r9 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: for i holds (l9 
    . i) 
    <= (r9 
    . i); 
    
      thus (
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) implies for i holds (l9 
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r9 
    . i) 
    
      proof
    
        assume
    
        
    
    A2: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
        let i0;
    
        per cases ;
    
          suppose
    
          
    
    A3: (r 
    . i0) 
    < (l 
    . i0); 
    
          defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means $2 
    > (l 
    . $1) & $2 
    > (r9 
    . $1); 
    
          
    
          
    
    A4: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi] by
    Th2;
    
          consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
          
    
    A5: for i holds 
    P[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A4);
    
          reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
          ex i st (r
    . i) 
    < (l 
    . i) & ((x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i)) 
    
          proof
    
            take i0;
    
            thus thesis by
    A3,
    A5;
    
          end;
    
          then
    
          
    
    A6: x 
    in ( 
    cell (l,r)); 
    
          ex i st (x
    . i) 
    < (l9 
    . i) or (r9 
    . i) 
    < (x 
    . i) 
    
          proof
    
            take i0;
    
            thus thesis by
    A5;
    
          end;
    
          hence thesis by
    A1,
    A2,
    A6,
    Th21;
    
        end;
    
          suppose
    
          
    
    A7: (l 
    . i0) 
    <= (r 
    . i0); 
    
          
    
          
    
    A8: l 
    in ( 
    cell (l,r)) by 
    Th23;
    
          r
    in ( 
    cell (l,r)) by 
    Th23;
    
          hence thesis by
    A1,
    A2,
    A7,
    A8,
    Th21;
    
        end;
    
      end;
    
      assume
    
      
    
    A9: for i holds (l9 
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r9 
    . i); 
    
      let x be
    object;
    
      assume
    
      
    
    A10: x 
    in ( 
    cell (l,r)); 
    
      then
    
      reconsider x as
    Element of ( 
    REAL d); 
    
      now
    
        let i;
    
        
    
        
    
    A11: (l9 
    . i) 
    <= (l 
    . i) by 
    A9;
    
        
    
        
    
    A12: (l 
    . i) 
    <= (x 
    . i) by 
    A9,
    A10,
    Th21;
    
        
    
        
    
    A13: (x 
    . i) 
    <= (r 
    . i) by 
    A9,
    A10,
    Th21;
    
        (r
    . i) 
    <= (r9 
    . i) by 
    A9;
    
        hence (l9
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r9 
    . i) by 
    A11,
    A12,
    A13,
    XXREAL_0: 2;
    
        hence (l9
    . i) 
    <= (r9 
    . i) by 
    XXREAL_0: 2;
    
      end;
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:29
    
    
    
    
    
    Th26: (for i holds (r 
    . i) 
    < (l 
    . i)) implies (( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) iff for i holds (r 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    < (l9 
    . i) & (l9 
    . i) 
    <= (l 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: for i holds (r 
    . i) 
    < (l 
    . i); 
    
      thus (
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) implies for i holds (r 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    < (l9 
    . i) & (l9 
    . i) 
    <= (l 
    . i) 
    
      proof
    
        assume
    
        
    
    A2: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
        
    
        
    
    A3: for i holds (r9 
    . i) 
    < (l9 
    . i) 
    
        proof
    
          let i0;
    
          assume
    
          
    
    A4: (l9 
    . i0) 
    <= (r9 
    . i0); 
    
          defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    = i0 implies (l 
    . $1) 
    < $2 & (r9 
    . $1) 
    < $2) & ((r9 
    . $1) 
    < (l9 
    . $1) implies (r9 
    . $1) 
    < $2 & $2 
    < (l9 
    . $1)); 
    
          
    
          
    
    A5: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi]
    
          proof
    
            let i;
    
            per cases ;
    
              suppose i
    = i0 & (r9 
    . i) 
    < (l9 
    . i); 
    
              hence thesis by
    A4;
    
            end;
    
              suppose
    
              
    
    A6: i 
    <> i0; 
    
              (r9
    . i) 
    < (l9 
    . i) implies ex xi be 
    Element of 
    REAL st (r9 
    . i) 
    < xi & xi 
    < (l9 
    . i) by 
    Th1;
    
              hence thesis by
    A6;
    
            end;
    
              suppose
    
              
    
    A7: (l9 
    . i) 
    <= (r9 
    . i); 
    
              ex xi be
    Element of 
    REAL st (l 
    . i) 
    < xi & (r9 
    . i) 
    < xi by 
    Th2;
    
              hence thesis by
    A7;
    
            end;
    
          end;
    
          consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
          
    
    A8: for i holds 
    P[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A5);
    
          reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
          
    
          
    
    A9: (r 
    . i0) 
    < (l 
    . i0) by 
    A1;
    
          (x
    . i0) 
    <= (r 
    . i0) or (l 
    . i0) 
    <= (x 
    . i0) by 
    A8;
    
          then
    
          
    
    A10: x 
    in ( 
    cell (l,r)) by 
    A9;
    
          per cases by
    A2,
    A10,
    Th20;
    
            suppose for i holds (l9
    . i) 
    <= (x 
    . i) & (x 
    . i) 
    <= (r9 
    . i); 
    
            then (x
    . i0) 
    <= (r9 
    . i0); 
    
            hence contradiction by
    A8;
    
          end;
    
            suppose ex i st (r9
    . i) 
    < (l9 
    . i) & ((x 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i)); 
    
            hence contradiction by
    A8;
    
          end;
    
        end;
    
        let i0;
    
        hereby
    
          assume
    
          
    
    A11: (r9 
    . i0) 
    < (r 
    . i0); 
    
          defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means (r9 
    . $1) 
    < $2 & $2 
    < (l9 
    . $1) & ($1 
    = i0 implies $2 
    < (r 
    . $1)); 
    
          
    
          
    
    A12: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi]
    
          proof
    
            let i;
    
            per cases ;
    
              suppose
    
              
    
    A13: i 
    = i0 & (l9 
    . i) 
    <= (r 
    . i); 
    
              (r9
    . i) 
    < (l9 
    . i) by 
    A3;
    
              then
    
              consider xi be
    Element of 
    REAL such that 
    
              
    
    A14: (r9 
    . i) 
    < xi and 
    
              
    
    A15: xi 
    < (l9 
    . i) by 
    Th1;
    
              xi
    < (r 
    . i) by 
    A13,
    A15,
    XXREAL_0: 2;
    
              hence thesis by
    A14,
    A15;
    
            end;
    
              suppose
    
              
    
    A16: i 
    = i0 & (r 
    . i) 
    <= (l9 
    . i); 
    
              then
    
              consider xi be
    Element of 
    REAL such that 
    
              
    
    A17: (r9 
    . i) 
    < xi and 
    
              
    
    A18: xi 
    < (r 
    . i) by 
    A11,
    Th1;
    
              xi
    < (l9 
    . i) by 
    A16,
    A18,
    XXREAL_0: 2;
    
              hence thesis by
    A17,
    A18;
    
            end;
    
              suppose
    
              
    
    A19: i 
    <> i0; 
    
              (r9
    . i) 
    < (l9 
    . i) by 
    A3;
    
              then ex xi be
    Element of 
    REAL st ((r9 
    . i) 
    < xi) & (xi 
    < (l9 
    . i)) by 
    Th1;
    
              hence thesis by
    A19;
    
            end;
    
          end;
    
          consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
          
    
    A20: for i holds 
    P[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A12);
    
          reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
          
    
          
    
    A21: (r 
    . i0) 
    < (l 
    . i0) by 
    A1;
    
          (x
    . i0) 
    <= (r 
    . i0) or (l 
    . i0) 
    <= (x 
    . i0) by 
    A20;
    
          then
    
          
    
    A22: x 
    in ( 
    cell (l,r)) by 
    A21;
    
           not ((l9
    . i0) 
    <= (x 
    . i0) & (x 
    . i0) 
    <= (r9 
    . i0)) by 
    A3,
    XXREAL_0: 2;
    
          then ex i st (r9
    . i) 
    < (l9 
    . i) & ((x 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i)) by 
    A2,
    A22,
    Th20;
    
          hence contradiction by
    A20;
    
        end;
    
        thus (r9
    . i0) 
    < (l9 
    . i0) by 
    A3;
    
        hereby
    
          assume
    
          
    
    A23: (l9 
    . i0) 
    > (l 
    . i0); 
    
          defpred
    
    R[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means (l9 
    . $1) 
    > $2 & $2 
    > (r9 
    . $1) & ($1 
    = i0 implies $2 
    > (l 
    . $1)); 
    
          
    
          
    
    A24: for i holds ex xi be 
    Element of 
    REAL st 
    R[i, xi]
    
          proof
    
            let i;
    
            per cases ;
    
              suppose
    
              
    
    A25: i 
    = i0 & (r9 
    . i) 
    >= (l 
    . i); 
    
              (l9
    . i) 
    > (r9 
    . i) by 
    A3;
    
              then
    
              consider xi be
    Element of 
    REAL such that 
    
              
    
    A26: (r9 
    . i) 
    < xi and 
    
              
    
    A27: xi 
    < (l9 
    . i) by 
    Th1;
    
              xi
    > (l 
    . i) by 
    A25,
    A26,
    XXREAL_0: 2;
    
              hence thesis by
    A26,
    A27;
    
            end;
    
              suppose
    
              
    
    A28: i 
    = i0 & (l 
    . i) 
    >= (r9 
    . i); 
    
              then
    
              consider xi be
    Element of 
    REAL such that 
    
              
    
    A29: (l 
    . i) 
    < xi and 
    
              
    
    A30: xi 
    < (l9 
    . i) by 
    A23,
    Th1;
    
              xi
    > (r9 
    . i) by 
    A28,
    A29,
    XXREAL_0: 2;
    
              hence thesis by
    A29,
    A30;
    
            end;
    
              suppose
    
              
    
    A31: i 
    <> i0; 
    
              (l9
    . i) 
    > (r9 
    . i) by 
    A3;
    
              then ex xi be
    Element of 
    REAL st ((r9 
    . i) 
    < xi) & (xi 
    < (l9 
    . i)) by 
    Th1;
    
              hence thesis by
    A31;
    
            end;
    
          end;
    
          consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
          
    
    A32: for i holds 
    R[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A24);
    
          reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
          
    
          
    
    A33: (l 
    . i0) 
    > (r 
    . i0) by 
    A1;
    
          (x
    . i0) 
    >= (l 
    . i0) or (r 
    . i0) 
    >= (x 
    . i0) by 
    A32;
    
          then
    
          
    
    A34: x 
    in ( 
    cell (l,r)) by 
    A33;
    
           not ((r9
    . i0) 
    >= (x 
    . i0) & (x 
    . i0) 
    >= (l9 
    . i0)) by 
    A3,
    XXREAL_0: 2;
    
          then ex i st (l9
    . i) 
    > (r9 
    . i) & ((x 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i)) by 
    A2,
    A34,
    Th20;
    
          hence contradiction by
    A32;
    
        end;
    
      end;
    
      assume
    
      
    
    A35: for i holds (r 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    < (l9 
    . i) & (l9 
    . i) 
    <= (l 
    . i); 
    
      let x be
    object;
    
      assume
    
      
    
    A36: x 
    in ( 
    cell (l,r)); 
    
      then
    
      reconsider x as
    Element of ( 
    REAL d); 
    
      set i0 = the
    Element of ( 
    Seg d); 
    
      
    
      
    
    A37: (r 
    . i0) 
    <= (r9 
    . i0) by 
    A35;
    
      (r9
    . i0) 
    < (l9 
    . i0) by 
    A35;
    
      then
    
      
    
    A38: (r 
    . i0) 
    < (l9 
    . i0) by 
    A37,
    XXREAL_0: 2;
    
      (l9
    . i0) 
    <= (l 
    . i0) by 
    A35;
    
      then (r
    . i0) 
    < (l 
    . i0) by 
    A38,
    XXREAL_0: 2;
    
      then (x
    . i0) 
    < (l 
    . i0) or (r 
    . i0) 
    < (x 
    . i0) by 
    XXREAL_0: 2;
    
      then
    
      consider i such that (r
    . i) 
    < (l 
    . i) and 
    
      
    
    A39: (x 
    . i) 
    <= (r 
    . i) or (l 
    . i) 
    <= (x 
    . i) by 
    A36,
    Th20;
    
      
    
      
    
    A40: (r 
    . i) 
    <= (r9 
    . i) by 
    A35;
    
      
    
      
    
    A41: (l9 
    . i) 
    <= (l 
    . i) by 
    A35;
    
      
    
      
    
    A42: (r9 
    . i) 
    < (l9 
    . i) by 
    A35;
    
      (x
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i) by 
    A39,
    A40,
    A41,
    XXREAL_0: 2;
    
      hence thesis by
    A42;
    
    end;
    
    theorem :: 
    
    CHAIN_1:30
    
    
    
    
    
    Th27: (for i holds (l 
    . i) 
    <= (r 
    . i)) & (for i holds (r9 
    . i) 
    < (l9 
    . i)) implies (( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) iff ex i st (r 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (l 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: for i holds (l 
    . i) 
    <= (r 
    . i); 
    
      assume
    
      
    
    A2: for i holds (r9 
    . i) 
    < (l9 
    . i); 
    
      thus (
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) implies ex i st (r 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (l 
    . i) 
    
      proof
    
        assume
    
        
    
    A3: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
        assume
    
        
    
    A4: for i holds (r9 
    . i) 
    < (r 
    . i) & (l 
    . i) 
    < (l9 
    . i); 
    
        defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means (l 
    . $1) 
    <= $2 & $2 
    <= (r 
    . $1) & (r9 
    . $1) 
    < $2 & $2 
    < (l9 
    . $1); 
    
        
    
        
    
    A5: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi]
    
        proof
    
          let i;
    
          per cases ;
    
            suppose
    
            
    
    A6: (l 
    . i) 
    <= (r9 
    . i) & (l9 
    . i) 
    <= (r 
    . i); 
    
            (r9
    . i) 
    < (l9 
    . i) by 
    A2;
    
            then
    
            consider xi be
    Element of 
    REAL such that 
    
            
    
    A7: (r9 
    . i) 
    < xi and 
    
            
    
    A8: xi 
    < (l9 
    . i) by 
    Th1;
    
            take xi;
    
            thus thesis by
    A6,
    A7,
    A8,
    XXREAL_0: 2;
    
          end;
    
            suppose
    
            
    
    A9: (r9 
    . i) 
    < (l 
    . i) & (l9 
    . i) 
    <= (r 
    . i); 
    
            reconsider li = (l
    . i) as 
    Element of 
    REAL by 
    XREAL_0:def 1;
    
            take li;
    
            thus thesis by
    A1,
    A4,
    A9;
    
          end;
    
            suppose
    
            
    
    A10: (r 
    . i) 
    < (l9 
    . i); 
    
            reconsider ri = (r
    . i) as 
    Element of 
    REAL by 
    XREAL_0:def 1;
    
            take ri;
    
            thus thesis by
    A1,
    A4,
    A10;
    
          end;
    
        end;
    
        consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A11: for i holds 
    P[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A5);
    
        reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
        
    
        
    
    A12: x 
    in ( 
    cell (l,r)) by 
    A11;
    
        set i0 = the
    Element of ( 
    Seg d); 
    
        (r9
    . i0) 
    < (l9 
    . i0) by 
    A2;
    
        then ex i st (r9
    . i) 
    < (l9 
    . i) & ((x 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i)) by 
    A3,
    A12,
    Th22;
    
        hence contradiction by
    A11;
    
      end;
    
      given i0 such that
    
      
    
    A13: (r 
    . i0) 
    <= (r9 
    . i0) or (l9 
    . i0) 
    <= (l 
    . i0); 
    
      let x be
    object;
    
      assume
    
      
    
    A14: x 
    in ( 
    cell (l,r)); 
    
      then
    
      reconsider x as
    Element of ( 
    REAL d); 
    
      
    
      
    
    A15: (l 
    . i0) 
    <= (x 
    . i0) by 
    A1,
    A14,
    Th21;
    
      
    
      
    
    A16: (x 
    . i0) 
    <= (r 
    . i0) by 
    A1,
    A14,
    Th21;
    
      ex i st (r9
    . i) 
    < (l9 
    . i) & ((x 
    . i) 
    <= (r9 
    . i) or (l9 
    . i) 
    <= (x 
    . i)) 
    
      proof
    
        take i0;
    
        thus thesis by
    A2,
    A13,
    A15,
    A16,
    XXREAL_0: 2;
    
      end;
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:31
    
    
    
    
    
    Th28: (for i holds (l 
    . i) 
    <= (r 
    . i)) or (for i holds (l 
    . i) 
    > (r 
    . i)) implies (( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) iff l 
    = l9 & r 
    = r9) 
    
    proof
    
      assume
    
      
    
    A1: (for i holds (l 
    . i) 
    <= (r 
    . i)) or for i holds (l 
    . i) 
    > (r 
    . i); 
    
      thus (
    cell (l,r)) 
    = ( 
    cell (l9,r9)) implies l 
    = l9 & r 
    = r9 
    
      proof
    
        assume
    
        
    
    A2: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)); 
    
        per cases by
    A1;
    
          suppose
    
          
    
    A3: for i holds (l 
    . i) 
    <= (r 
    . i); 
    
          then
    
          
    
    A4: for i holds (l 
    . i) 
    <= (l9 
    . i) & (l9 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    <= (r 
    . i) by 
    A2,
    Th25;
    
          reconsider l, r, l9, r9 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
          
    
    A5: 
    
          now
    
            let i;
    
            
    
            
    
    A6: (l 
    . i) 
    <= (l9 
    . i) by 
    A2,
    A3,
    Th25;
    
            (l9
    . i) 
    <= (l 
    . i) by 
    A2,
    A4,
    Th25;
    
            hence (l
    . i) 
    = (l9 
    . i) by 
    A6,
    XXREAL_0: 1;
    
          end;
    
          now
    
            let i;
    
            
    
            
    
    A7: (r 
    . i) 
    <= (r9 
    . i) by 
    A2,
    A4,
    Th25;
    
            (r9
    . i) 
    <= (r 
    . i) by 
    A2,
    A3,
    Th25;
    
            hence (r
    . i) 
    = (r9 
    . i) by 
    A7,
    XXREAL_0: 1;
    
          end;
    
          hence thesis by
    A5,
    FUNCT_2: 63;
    
        end;
    
          suppose
    
          
    
    A8: for i holds (l 
    . i) 
    > (r 
    . i); 
    
          then
    
          
    
    A9: for i holds (r 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    < (l9 
    . i) & (l9 
    . i) 
    <= (l 
    . i) by 
    A2,
    Th26;
    
          reconsider l, r, l9, r9 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
          
    
    A10: 
    
          now
    
            let i;
    
            
    
            
    
    A11: (l 
    . i) 
    <= (l9 
    . i) by 
    A2,
    A9,
    Th26;
    
            (l9
    . i) 
    <= (l 
    . i) by 
    A2,
    A8,
    Th26;
    
            hence (l
    . i) 
    = (l9 
    . i) by 
    A11,
    XXREAL_0: 1;
    
          end;
    
          now
    
            let i;
    
            
    
            
    
    A12: (r 
    . i) 
    <= (r9 
    . i) by 
    A2,
    A8,
    Th26;
    
            (r9
    . i) 
    <= (r 
    . i) by 
    A2,
    A9,
    Th26;
    
            hence (r
    . i) 
    = (r9 
    . i) by 
    A12,
    XXREAL_0: 1;
    
          end;
    
          hence thesis by
    A10,
    FUNCT_2: 63;
    
        end;
    
      end;
    
      thus thesis;
    
    end;
    
    definition
    
      let d, G, k;
    
      assume
    
      
    
    A1: k 
    <= d; 
    
      :: 
    
    CHAIN_1:def7
    
      func
    
    cells (k,G) -> 
    finite non 
    empty  
    Subset-Family of ( 
    REAL d) equals 
    
      :
    
    Def7: { ( 
    cell (l,r)) : ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i))) }; 
    
      coherence
    
      proof
    
        defpred
    
    R[
    Element of ( 
    REAL d), 
    Element of ( 
    REAL d)] means ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & ($1 
    . i) 
    < ($2 
    . i) & 
    [($1
    . i), ($2 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & ($1 
    . i) 
    = ($2 
    . i) & ($1 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds ($2 
    . i) 
    < ($1 
    . i) & 
    [($1
    . i), ($2 
    . i)] is 
    Gap of (G 
    . i))); 
    
        deffunc
    
    f(
    Element of ( 
    REAL d), 
    Element of ( 
    REAL d)) = ( 
    cell ($1,$2)); 
    
        set CELLS = {
    f(l,r) :
    R[l, r] };
    
        reconsider X = (
    Seg k) as 
    Subset of ( 
    Seg d) by 
    A1,
    FINSEQ_1: 5;
    
        defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    [:
    REAL , 
    REAL :]] means ($1 
    in X & ex li, ri st $2 
    =  
    [li, ri] & li
    < ri & $2 is 
    Gap of (G 
    . $1)) or ( not $1 
    in X & ex li st $2 
    =  
    [li, li] & li
    in (G 
    . $1)); 
    
        
    
    A2: 
    
        now
    
          let i;
    
          thus ex lri be
    Element of 
    [:
    REAL , 
    REAL :] st 
    P[i, lri]
    
          proof
    
            per cases ;
    
              suppose
    
              
    
    A3: i 
    in X; 
    
              consider li, ri such that
    
              
    
    A4: li 
    in (G 
    . i) and 
    
              
    
    A5: ri 
    in (G 
    . i) and 
    
              
    
    A6: li 
    < ri and 
    
              
    
    A7: for xi st xi 
    in (G 
    . i) holds not (li 
    < xi & xi 
    < ri) by 
    Th11;
    
              reconsider li, ri as
    Element of 
    REAL by 
    XREAL_0:def 1;
    
              take
    [li, ri];
    
              
    [li, ri] is
    Gap of (G 
    . i) by 
    A4,
    A5,
    A6,
    A7,
    Def5;
    
              hence thesis by
    A3,
    A6;
    
            end;
    
              suppose
    
              
    
    A8: not i 
    in X; 
    
              set li = the
    Element of (G 
    . i); 
    
              reconsider li as
    Element of 
    REAL ; 
    
              reconsider lri =
    [li, li] as
    Element of 
    [:
    REAL , 
    REAL :]; 
    
              take lri;
    
              thus thesis by
    A8;
    
            end;
    
          end;
    
        end;
    
        consider lr be
    Function of ( 
    Seg d), 
    [:
    REAL , 
    REAL :] such that 
    
        
    
    A9: for i holds 
    P[i, (lr
    . i)] from 
    FUNCT_2:sch 3(
    A2);
    
        deffunc
    
    l(
    Element of ( 
    Seg d)) = ((lr 
    . $1) 
    `1 ); 
    
        consider l be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A10: for i holds (l 
    . i) 
    =  
    l(i) from
    FUNCT_2:sch 4;
    
        deffunc
    
    r(
    Element of ( 
    Seg d)) = ((lr 
    . $1) 
    `2 ); 
    
        consider r be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A11: for i holds (r 
    . i) 
    =  
    r(i) from
    FUNCT_2:sch 4;
    
        reconsider l, r as
    Element of ( 
    REAL d) by 
    Def3;
    
        
    
    A12: 
    
        now
    
          let i;
    
          
    
          
    
    A13: (l 
    . i) 
    = ((lr 
    . i) 
    `1 ) by 
    A10;
    
          (r
    . i) 
    = ((lr 
    . i) 
    `2 ) by 
    A11;
    
          hence (lr
    . i) 
    =  
    [(l
    . i), (r 
    . i)] by 
    A13,
    MCART_1: 21;
    
        end;
    
        now
    
          take A = (
    cell (l,r)); 
    
          thus A
    = ( 
    cell (l,r)); 
    
          now
    
            take X;
    
            thus (
    card X) 
    = k by 
    FINSEQ_1: 57;
    
            take l, r;
    
            thus A
    = ( 
    cell (l,r)); 
    
            let i;
    
            per cases ;
    
              suppose
    
              
    
    A14: i 
    in X; 
    
              then
    
              consider li, ri such that
    
              
    
    A15: (lr 
    . i) 
    =  
    [li, ri] and
    
              
    
    A16: li 
    < ri and 
    
              
    
    A17: (lr 
    . i) is 
    Gap of (G 
    . i) by 
    A9;
    
              
    
              
    
    A18: (lr 
    . i) 
    =  
    [(l
    . i), (r 
    . i)] by 
    A12;
    
              then li
    = (l 
    . i) by 
    A15,
    XTUPLE_0: 1;
    
              hence i
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A14,
    A15,
    A16,
    A17,
    A18,
    XTUPLE_0: 1;
    
            end;
    
              suppose
    
              
    
    A19: not i 
    in X; 
    
              then
    
              consider li such that
    
              
    
    A20: (lr 
    . i) 
    =  
    [li, li] and
    
              
    
    A21: li 
    in (G 
    . i) by 
    A9;
    
              
    
              
    
    A22: 
    [li, li]
    =  
    [(l
    . i), (r 
    . i)] by 
    A12,
    A20;
    
              then li
    = (l 
    . i) by 
    XTUPLE_0: 1;
    
              hence i
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A19,
    A21,
    A22,
    XTUPLE_0: 1;
    
            end;
    
          end;
    
          hence ex l, r st A
    = ( 
    cell (l,r)) & ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i))); 
    
        end;
    
        then
    
        
    
    A23: ( 
    cell (l,r)) 
    in CELLS; 
    
        defpred
    
    Q[
    object, 
    Element of ( 
    REAL d), 
    Element of ( 
    REAL d), 
    object] means $2
    in ( 
    product G) & $3 
    in ( 
    product G) & (($4 
    =  
    [
    0 , 
    [$2, $3]] & $1
    = ( 
    cell ($2,$3))) or ($4 
    =  
    [1,
    [$2, $3]] & $1
    = ( 
    cell ($2,$3)))); 
    
        defpred
    
    S[
    object, 
    object] means ex l, r st
    Q[$1, l, r, $2];
    
        
    
        
    
    A24: for A be 
    object st A 
    in CELLS holds ex lr be 
    object st 
    S[A, lr]
    
        proof
    
          let A be
    object;
    
          assume A
    in CELLS; 
    
          then
    
          consider l, r such that
    
          
    
    A25: A 
    = ( 
    cell (l,r)) and 
    
          
    
    A26: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)); 
    
          per cases by
    A26;
    
            suppose
    
            
    
    A27: ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            take
    [
    0 , 
    [l, r]], l, r;
    
            now
    
              let i;
    
              (l
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A27;
    
              hence (l
    . i) 
    in (G 
    . i) & (r 
    . i) 
    in (G 
    . i) by 
    Th13;
    
            end;
    
            hence l
    in ( 
    product G) & r 
    in ( 
    product G) by 
    Th8;
    
            thus thesis by
    A25;
    
          end;
    
            suppose
    
            
    
    A28: k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
            take
    [1,
    [l, r]], l, r;
    
            now
    
              let i;
    
              
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A28;
    
              hence (l
    . i) 
    in (G 
    . i) & (r 
    . i) 
    in (G 
    . i) by 
    Th13;
    
            end;
    
            hence l
    in ( 
    product G) & r 
    in ( 
    product G) by 
    Th8;
    
            thus thesis by
    A25;
    
          end;
    
        end;
    
        consider f be
    Function such that 
    
        
    
    A29: ( 
    dom f) 
    = CELLS & for A be 
    object st A 
    in CELLS holds 
    S[A, (f
    . A)] from 
    CLASSES1:sch 1(
    A24);
    
        
    
        
    
    A30: f is 
    one-to-one
    
        proof
    
          let A,A9 be
    object;
    
          assume that
    
          
    
    A31: A 
    in ( 
    dom f) and 
    
          
    
    A32: A9 
    in ( 
    dom f) and 
    
          
    
    A33: (f 
    . A) 
    = (f 
    . A9); 
    
          consider l, r such that
    
          
    
    A34: 
    Q[A, l, r, (f
    . A)] by 
    A29,
    A31;
    
          consider l9, r9 such that
    
          
    
    A35: 
    Q[A9, l9, r9, (f
    . A9)] by 
    A29,
    A32;
    
          per cases by
    A34;
    
            suppose
    
            
    
    A36: (f 
    . A) 
    =  
    [
    0 , 
    [l, r]] & A
    = ( 
    cell (l,r)); 
    
            then
    
            
    
    A37: 
    [l, r]
    =  
    [l9, r9] by
    A33,
    A35,
    XTUPLE_0: 1;
    
            then l
    = l9 by 
    XTUPLE_0: 1;
    
            hence thesis by
    A35,
    A36,
    A37,
    XTUPLE_0: 1;
    
          end;
    
            suppose
    
            
    
    A38: (f 
    . A) 
    =  
    [1,
    [l, r]] & A
    = ( 
    cell (l,r)); 
    
            then
    
            
    
    A39: 
    [l, r]
    =  
    [l9, r9] by
    A33,
    A35,
    XTUPLE_0: 1;
    
            then l
    = l9 by 
    XTUPLE_0: 1;
    
            hence thesis by
    A35,
    A38,
    A39,
    XTUPLE_0: 1;
    
          end;
    
        end;
    
        reconsider X = (
    product G) as 
    finite  
    set;
    
        
    
        
    
    A40: ( 
    rng f) 
    c=  
    [:
    {
    0 , 1}, 
    [:X, X:]:]
    
        proof
    
          let lr be
    object;
    
          assume lr
    in ( 
    rng f); 
    
          then
    
          consider A be
    object such that 
    
          
    
    A41: A 
    in ( 
    dom f) and 
    
          
    
    A42: lr 
    = (f 
    . A) by 
    FUNCT_1:def 3;
    
          consider l, r such that
    
          
    
    A43: 
    Q[A, l, r, (f
    . A)] by 
    A29,
    A41;
    
          
    
          
    
    A44: 
    0  
    in  
    {
    0 , 1} by 
    TARSKI:def 2;
    
          
    
          
    
    A45: 1 
    in  
    {
    0 , 1} by 
    TARSKI:def 2;
    
          
    [l, r]
    in  
    [:X, X:] by
    A43,
    ZFMISC_1: 87;
    
          hence thesis by
    A42,
    A43,
    A44,
    A45,
    ZFMISC_1: 87;
    
        end;
    
        CELLS
    c= ( 
    bool ( 
    REAL d)) from 
    FrSet12;
    
        hence thesis by
    A23,
    A29,
    A30,
    A40,
    CARD_1: 59;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:32
    
    
    
    
    
    Th29: k 
    <= d implies for A be 
    Subset of ( 
    REAL d) holds A 
    in ( 
    cells (k,G)) iff ex l, r st A 
    = ( 
    cell (l,r)) & ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i))) 
    
    proof
    
      assume k
    <= d; 
    
      then (
    cells (k,G)) 
    = { ( 
    cell (l,r)) : ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i))) } by 
    Def7;
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:33
    
    
    
    
    
    Th30: k 
    <= d implies (( 
    cell (l,r)) 
    in ( 
    cells (k,G)) iff ((ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)))) 
    
    proof
    
      assume
    
      
    
    A1: k 
    <= d; 
    
      hereby
    
        assume (
    cell (l,r)) 
    in ( 
    cells (k,G)); 
    
        then
    
        consider l9, r9 such that
    
        
    
    A2: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A3: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) by 
    A1,
    Th29;
    
        l
    = l9 & r 
    = r9 
    
        proof
    
          per cases by
    A3;
    
            suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i); 
    
            then for i holds (l9
    . i) 
    <= (r9 
    . i); 
    
            hence thesis by
    A2,
    Th28;
    
          end;
    
            suppose for i holds (r9
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis by
    A2,
    Th28;
    
          end;
    
        end;
    
        hence (ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    A3;
    
      end;
    
      thus thesis by
    A1,
    Th29;
    
    end;
    
    theorem :: 
    
    CHAIN_1:34
    
    
    
    
    
    Th31: k 
    <= d & ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) implies (for i holds ((l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ((l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) 
    
    proof
    
      assume that
    
      
    
    A1: k 
    <= d and 
    
      
    
    A2: ( 
    cell (l,r)) 
    in ( 
    cells (k,G)); 
    
      per cases by
    A1,
    A2,
    Th30;
    
        suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
        hence thesis;
    
      end;
    
        suppose k
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
        hence thesis;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:35
    
    
    
    
    
    Th32: k 
    <= d & ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) implies for i holds (l 
    . i) 
    in (G 
    . i) & (r 
    . i) 
    in (G 
    . i) 
    
    proof
    
      assume that
    
      
    
    A1: k 
    <= d and 
    
      
    
    A2: ( 
    cell (l,r)) 
    in ( 
    cells (k,G)); 
    
      let i;
    
      (l
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) or (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A1,
    A2,
    Th31;
    
      hence thesis by
    Th13;
    
    end;
    
    theorem :: 
    
    CHAIN_1:36
    
    k
    <= d & ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) implies (for i holds (l 
    . i) 
    <= (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i) by 
    Th31;
    
    theorem :: 
    
    CHAIN_1:37
    
    
    
    
    
    Th34: for A be 
    Subset of ( 
    REAL d) holds A 
    in ( 
    cells ( 
    0 ,G)) iff ex x st A 
    = ( 
    cell (x,x)) & for i holds (x 
    . i) 
    in (G 
    . i) 
    
    proof
    
      let A be
    Subset of ( 
    REAL d); 
    
      hereby
    
        assume A
    in ( 
    cells ( 
    0 ,G)); 
    
        then
    
        consider l, r such that
    
        
    
    A1: A 
    = ( 
    cell (l,r)) and 
    
        
    
    A2: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    =  
    0 & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or ( 
    0  
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    Th29;
    
        consider X be
    Subset of ( 
    Seg d) such that 
    
        
    
    A3: ( 
    card X) 
    =  
    0 and 
    
        
    
    A4: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A2;
    
        reconsider l9 = l, r9 = r as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
        X
    =  
    {} by 
    A3;
    
        then
    
        
    
    A5: for i holds (l9 
    . i) 
    = (r9 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A4;
    
        then l9
    = r9 by 
    FUNCT_2: 63;
    
        hence ex x st A
    = ( 
    cell (x,x)) & for i holds (x 
    . i) 
    in (G 
    . i) by 
    A1,
    A5;
    
      end;
    
      given x such that
    
      
    
    A6: A 
    = ( 
    cell (x,x)) and 
    
      
    
    A7: for i holds (x 
    . i) 
    in (G 
    . i); 
    
      ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    =  
    0 & for i holds i 
    in X & (x 
    . i) 
    < (x 
    . i) & 
    [(x
    . i), (x 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (x 
    . i) 
    = (x 
    . i) & (x 
    . i) 
    in (G 
    . i) 
    
      proof
    
        reconsider X =
    {} as 
    Subset of ( 
    Seg d) by 
    XBOOLE_1: 2;
    
        take X;
    
        thus thesis by
    A7;
    
      end;
    
      hence thesis by
    A6,
    Th29;
    
    end;
    
    theorem :: 
    
    CHAIN_1:38
    
    
    
    
    
    Th35: ( 
    cell (l,r)) 
    in ( 
    cells ( 
    0 ,G)) iff l 
    = r & for i holds (l 
    . i) 
    in (G 
    . i) 
    
    proof
    
      hereby
    
        assume (
    cell (l,r)) 
    in ( 
    cells ( 
    0 ,G)); 
    
        then
    
        consider x such that
    
        
    
    A1: ( 
    cell (l,r)) 
    = ( 
    cell (x,x)) and 
    
        
    
    A2: for i holds (x 
    . i) 
    in (G 
    . i) by 
    Th34;
    
        
    
        
    
    A3: for i holds (x 
    . i) 
    <= (x 
    . i); 
    
        then l
    = x by 
    A1,
    Th28;
    
        hence l
    = r & for i holds (l 
    . i) 
    in (G 
    . i) by 
    A1,
    A2,
    A3,
    Th28;
    
      end;
    
      thus thesis by
    Th34;
    
    end;
    
    theorem :: 
    
    CHAIN_1:39
    
    
    
    
    
    Th36: for A be 
    Subset of ( 
    REAL d) holds A 
    in ( 
    cells (d,G)) iff ex l, r st A 
    = ( 
    cell (l,r)) & (for i holds 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) & ((for i holds (l 
    . i) 
    < (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i)) 
    
    proof
    
      let A be
    Subset of ( 
    REAL d); 
    
      hereby
    
        assume A
    in ( 
    cells (d,G)); 
    
        then
    
        consider l, r such that
    
        
    
    A1: A 
    = ( 
    cell (l,r)) and 
    
        
    
    A2: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = d & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (d 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    Th29;
    
        thus ex l, r st A
    = ( 
    cell (l,r)) & (for i holds 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) & ((for i holds (l 
    . i) 
    < (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i)) 
    
        proof
    
          take l, r;
    
          per cases by
    A2;
    
            suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = d & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            then
    
            consider X be
    Subset of ( 
    Seg d) such that 
    
            
    
    A3: ( 
    card X) 
    = d and 
    
            
    
    A4: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            (
    card X) 
    = ( 
    card ( 
    Seg d)) by 
    A3,
    FINSEQ_1: 57;
    
            then not X
    c< ( 
    Seg d) by 
    CARD_2: 48;
    
            then X
    = ( 
    Seg d) by 
    XBOOLE_0:def 8;
    
            hence thesis by
    A1,
    A4;
    
          end;
    
            suppose for i holds (r
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis by
    A1;
    
          end;
    
        end;
    
      end;
    
      given l, r such that
    
      
    
    A5: A 
    = ( 
    cell (l,r)) and 
    
      
    
    A6: for i holds 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) and 
    
      
    
    A7: (for i holds (l 
    . i) 
    < (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i); 
    
      per cases by
    A7;
    
        suppose
    
        
    
    A8: for i holds (l 
    . i) 
    < (r 
    . i); 
    
        ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = d & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) 
    
        proof
    
          (
    Seg d) 
    c= ( 
    Seg d); 
    
          then
    
          reconsider X = (
    Seg d) as 
    Subset of ( 
    Seg d); 
    
          take X;
    
          thus (
    card X) 
    = d by 
    FINSEQ_1: 57;
    
          thus thesis by
    A6,
    A8;
    
        end;
    
        hence thesis by
    A5,
    Th29;
    
      end;
    
        suppose for i holds (r
    . i) 
    < (l 
    . i); 
    
        hence thesis by
    A5,
    A6,
    Th29;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:40
    
    
    
    
    
    Th37: ( 
    cell (l,r)) 
    in ( 
    cells (d,G)) iff (for i holds 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) & ((for i holds (l 
    . i) 
    < (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i)) 
    
    proof
    
      hereby
    
        assume (
    cell (l,r)) 
    in ( 
    cells (d,G)); 
    
        then
    
        consider l9, r9 such that
    
        
    
    A1: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A2: for i holds 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) and 
    
        
    
    A3: (for i holds (l9 
    . i) 
    < (r9 
    . i)) or for i holds (r9 
    . i) 
    < (l9 
    . i) by 
    Th36;
    
        
    
        
    
    A4: (for i holds (l9 
    . i) 
    <= (r9 
    . i)) or for i holds (r9 
    . i) 
    < (l9 
    . i) by 
    A3;
    
        then
    
        
    
    A5: l 
    = l9 by 
    A1,
    Th28;
    
        r
    = r9 by 
    A1,
    A4,
    Th28;
    
        hence (for i holds
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) & ((for i holds (l 
    . i) 
    < (r 
    . i)) or for i holds (r 
    . i) 
    < (l 
    . i)) by 
    A2,
    A3,
    A5;
    
      end;
    
      thus thesis by
    Th36;
    
    end;
    
    theorem :: 
    
    CHAIN_1:41
    
    
    
    
    
    Th38: d 
    = (d9 
    + 1) implies for A be 
    Subset of ( 
    REAL d) holds A 
    in ( 
    cells (d9,G)) iff ex l, r, i0 st A 
    = ( 
    cell (l,r)) & (l 
    . i0) 
    = (r 
    . i0) & (l 
    . i0) 
    in (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) 
    
    proof
    
      assume
    
      
    
    A1: d 
    = (d9 
    + 1); 
    
      then
    
      
    
    A2: d9 
    < d by 
    NAT_1: 13;
    
      let A be
    Subset of ( 
    REAL d); 
    
      hereby
    
        assume A
    in ( 
    cells (d9,G)); 
    
        then
    
        consider l, r such that
    
        
    
    A3: A 
    = ( 
    cell (l,r)) and 
    
        
    
    A4: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = d9 & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (d9 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    A2,
    Th29;
    
        take l, r;
    
        consider X be
    Subset of ( 
    Seg d) such that 
    
        
    
    A5: ( 
    card X) 
    = d9 and 
    
        
    
    A6: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A1,
    A4;
    
        (
    card (( 
    Seg d) 
    \ X)) 
    = (( 
    card ( 
    Seg d)) 
    - ( 
    card X)) by 
    CARD_2: 44
    
        .= (d
    - d9) by 
    A5,
    FINSEQ_1: 57
    
        .= 1 by
    A1;
    
        then
    
        consider i0 be
    object such that 
    
        
    
    A7: (( 
    Seg d) 
    \ X) 
    =  
    {i0} by
    CARD_2: 42;
    
        i0
    in (( 
    Seg d) 
    \ X) by 
    A7,
    TARSKI:def 1;
    
        then
    
        reconsider i0 as
    Element of ( 
    Seg d) by 
    XBOOLE_0:def 5;
    
        take i0;
    
        
    
    A8: 
    
        now
    
          let i;
    
          i
    in (( 
    Seg d) 
    \ X) iff i 
    = i0 by 
    A7,
    TARSKI:def 1;
    
          hence i
    in X iff i 
    <> i0 by 
    XBOOLE_0:def 5;
    
        end;
    
        thus A
    = ( 
    cell (l,r)) by 
    A3;
    
         not i0
    in X by 
    A8;
    
        hence (l
    . i0) 
    = (r 
    . i0) & (l 
    . i0) 
    in (G 
    . i0) by 
    A6;
    
        let i;
    
        assume i
    <> i0; 
    
        then i
    in X by 
    A8;
    
        hence (l
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A6;
    
      end;
    
      given l, r, i0 such that
    
      
    
    A9: A 
    = ( 
    cell (l,r)) and 
    
      
    
    A10: (l 
    . i0) 
    = (r 
    . i0) and 
    
      
    
    A11: (l 
    . i0) 
    in (G 
    . i0) and 
    
      
    
    A12: for i st i 
    <> i0 holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
      reconsider X = ((
    Seg d) 
    \  
    {i0}) as
    Subset of ( 
    Seg d) by 
    XBOOLE_1: 36;
    
      (
    card X) 
    = d9 & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) 
    
      proof
    
        
    
        thus (
    card X) 
    = (( 
    card ( 
    Seg d)) 
    - ( 
    card  
    {i0})) by
    CARD_2: 44
    
        .= (d
    - ( 
    card  
    {i0})) by
    FINSEQ_1: 57
    
        .= (d
    - 1) by 
    CARD_1: 30
    
        .= d9 by
    A1;
    
        let i;
    
        i
    in  
    {i0} iff i
    = i0 by 
    TARSKI:def 1;
    
        hence thesis by
    A10,
    A11,
    A12,
    XBOOLE_0:def 5;
    
      end;
    
      hence thesis by
    A2,
    A9,
    Th29;
    
    end;
    
    theorem :: 
    
    CHAIN_1:42
    
    d
    = (d9 
    + 1) implies (( 
    cell (l,r)) 
    in ( 
    cells (d9,G)) iff ex i0 st (l 
    . i0) 
    = (r 
    . i0) & (l 
    . i0) 
    in (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: d 
    = (d9 
    + 1); 
    
      hereby
    
        assume (
    cell (l,r)) 
    in ( 
    cells (d9,G)); 
    
        then
    
        consider l9, r9, i0 such that
    
        
    
    A2: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A3: (l9 
    . i0) 
    = (r9 
    . i0) and 
    
        
    
    A4: (l9 
    . i0) 
    in (G 
    . i0) and 
    
        
    
    A5: for i st i 
    <> i0 holds (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    A1,
    Th38;
    
        take i0;
    
        
    
    A6: 
    
        now
    
          let i;
    
          i
    = i0 or i 
    <> i0; 
    
          hence (l9
    . i) 
    <= (r9 
    . i) by 
    A3,
    A5;
    
        end;
    
        then
    
        
    
    A7: l 
    = l9 by 
    A2,
    Th28;
    
        r
    = r9 by 
    A2,
    A6,
    Th28;
    
        hence (l
    . i0) 
    = (r 
    . i0) & (l 
    . i0) 
    in (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A3,
    A4,
    A5,
    A7;
    
      end;
    
      thus thesis by
    A1,
    Th38;
    
    end;
    
    theorem :: 
    
    CHAIN_1:43
    
    
    
    
    
    Th40: for A be 
    Subset of ( 
    REAL d) holds A 
    in ( 
    cells (1,G)) iff ex l, r, i0 st A 
    = ( 
    cell (l,r)) & ((l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) 
    
    proof
    
      
    
      
    
    A1: d 
    >= 1 by 
    Def2;
    
      let A be
    Subset of ( 
    REAL d); 
    
      hereby
    
        assume A
    in ( 
    cells (1,G)); 
    
        then
    
        consider l, r such that
    
        
    
    A2: A 
    = ( 
    cell (l,r)) and 
    
        
    
    A3: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = 1 & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (1 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    A1,
    Th29;
    
        take l, r;
    
        thus ex i0 st A
    = ( 
    cell (l,r)) & ((l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) 
    
        proof
    
          per cases by
    A3;
    
            suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = 1 & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            then
    
            consider X be
    Subset of ( 
    Seg d) such that 
    
            
    
    A4: ( 
    card X) 
    = 1 and 
    
            
    
    A5: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            consider i0 be
    object such that 
    
            
    
    A6: X 
    =  
    {i0} by
    A4,
    CARD_2: 42;
    
            
    
            
    
    A7: i0 
    in X by 
    A6,
    TARSKI:def 1;
    
            then
    
            reconsider i0 as
    Element of ( 
    Seg d); 
    
            take i0;
    
            thus A
    = ( 
    cell (l,r)) & ((l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) by 
    A2,
    A5,
    A7;
    
            let i;
    
             not i
    in X iff i 
    <> i0 by 
    A6,
    TARSKI:def 1;
    
            hence thesis by
    A5;
    
          end;
    
            suppose
    
            
    
    A8: d 
    = 1 & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
            reconsider i0 = 1 as
    Element of ( 
    Seg d) by 
    A1,
    FINSEQ_1: 1;
    
            take i0;
    
            thus A
    = ( 
    cell (l,r)) & ((l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) by 
    A2,
    A8;
    
            let i;
    
            
    
            
    
    A9: 1 
    <= i by 
    FINSEQ_1: 1;
    
            i
    <= d by 
    FINSEQ_1: 1;
    
            hence thesis by
    A8,
    A9,
    XXREAL_0: 1;
    
          end;
    
        end;
    
      end;
    
      given l, r, i0 such that
    
      
    
    A10: A 
    = ( 
    cell (l,r)) and 
    
      
    
    A11: (l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0) and 
    
      
    
    A12: 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) and 
    
      
    
    A13: for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
      set X =
    {i0};
    
      per cases by
    A11;
    
        suppose
    
        
    
    A14: (l 
    . i0) 
    < (r 
    . i0); 
    
        
    
        
    
    A15: ( 
    card X) 
    = 1 by 
    CARD_1: 30;
    
        now
    
          let i;
    
          i
    in X iff i 
    = i0 by 
    TARSKI:def 1;
    
          hence i
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A12,
    A13,
    A14;
    
        end;
    
        hence thesis by
    A1,
    A10,
    A15,
    Th29;
    
      end;
    
        suppose
    
        
    
    A16: d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0); 
    
        now
    
          let i;
    
          
    
          
    
    A17: 1 
    <= i by 
    FINSEQ_1: 1;
    
          
    
          
    
    A18: i 
    <= d by 
    FINSEQ_1: 1;
    
          
    
          
    
    A19: 1 
    <= i0 by 
    FINSEQ_1: 1;
    
          
    
          
    
    A20: i0 
    <= d by 
    FINSEQ_1: 1;
    
          
    
          
    
    A21: i 
    = 1 by 
    A16,
    A17,
    A18,
    XXREAL_0: 1;
    
          i0
    = 1 by 
    A16,
    A19,
    A20,
    XXREAL_0: 1;
    
          hence (r
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A12,
    A16,
    A21;
    
        end;
    
        hence thesis by
    A10,
    A11,
    Th29;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:44
    
    (
    cell (l,r)) 
    in ( 
    cells (1,G)) iff ex i0 st ((l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) 
    
    proof
    
      hereby
    
        assume (
    cell (l,r)) 
    in ( 
    cells (1,G)); 
    
        then
    
        consider l9, r9, i0 such that
    
        
    
    A1: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A2: (l9 
    . i0) 
    < (r9 
    . i0) or d 
    = 1 & (r9 
    . i0) 
    < (l9 
    . i0) and 
    
        
    
    A3: 
    [(l9
    . i0), (r9 
    . i0)] is 
    Gap of (G 
    . i0) and 
    
        
    
    A4: for i st i 
    <> i0 holds (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i) by 
    Th40;
    
        
    
        
    
    A5: (for i holds (l9 
    . i) 
    <= (r9 
    . i)) or for i holds (r9 
    . i) 
    < (l9 
    . i) 
    
        proof
    
          per cases by
    A2;
    
            suppose
    
            
    
    A6: (l9 
    . i0) 
    < (r9 
    . i0); 
    
            now
    
              let i;
    
              i
    = i0 or i 
    <> i0; 
    
              hence (l9
    . i) 
    <= (r9 
    . i) by 
    A4,
    A6;
    
            end;
    
            hence thesis;
    
          end;
    
            suppose
    
            
    
    A7: d 
    = 1 & (r9 
    . i0) 
    < (l9 
    . i0); 
    
            now
    
              let i;
    
              
    
              
    
    A8: 1 
    <= i by 
    FINSEQ_1: 1;
    
              
    
              
    
    A9: i 
    <= d by 
    FINSEQ_1: 1;
    
              
    
              
    
    A10: 1 
    <= i0 by 
    FINSEQ_1: 1;
    
              
    
              
    
    A11: i0 
    <= d by 
    FINSEQ_1: 1;
    
              
    
              
    
    A12: i 
    = 1 by 
    A7,
    A8,
    A9,
    XXREAL_0: 1;
    
              i0
    = 1 by 
    A7,
    A10,
    A11,
    XXREAL_0: 1;
    
              hence (r9
    . i) 
    < (l9 
    . i) by 
    A7,
    A12;
    
            end;
    
            hence thesis;
    
          end;
    
        end;
    
        then
    
        
    
    A13: l 
    = l9 by 
    A1,
    Th28;
    
        r
    = r9 by 
    A1,
    A5,
    Th28;
    
        hence ex i0 st ((l
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0)) & 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) & for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A2,
    A3,
    A4,
    A13;
    
      end;
    
      thus thesis by
    Th40;
    
    end;
    
    theorem :: 
    
    CHAIN_1:45
    
    
    
    
    
    Th42: k 
    <= d & k9 
    <= d & ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) & ( 
    cell (l9,r9)) 
    in ( 
    cells (k9,G)) & ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) implies for i holds (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i) or (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) or (l 
    . i) 
    <= (r 
    . i) & (r9 
    . i) 
    < (l9 
    . i) & (r9 
    . i) 
    <= (l 
    . i) & (r 
    . i) 
    <= (l9 
    . i) 
    
    proof
    
      assume that
    
      
    
    A1: k 
    <= d and 
    
      
    
    A2: k9 
    <= d and 
    
      
    
    A3: ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) and 
    
      
    
    A4: ( 
    cell (l9,r9)) 
    in ( 
    cells (k9,G)); 
    
      assume
    
      
    
    A5: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
      let i;
    
      per cases by
    A2,
    A4,
    Th31;
    
        suppose
    
        
    
    A6: for i holds (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) or (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i); 
    
        then
    
        
    
    A7: for i holds (l9 
    . i) 
    <= (r9 
    . i); 
    
        then
    
        
    
    A8: (l9 
    . i) 
    <= (l 
    . i) by 
    A5,
    Th25;
    
        
    
        
    
    A9: (l 
    . i) 
    <= (r 
    . i) by 
    A5,
    A7,
    Th25;
    
        
    
        
    
    A10: (r 
    . i) 
    <= (r9 
    . i) by 
    A5,
    A7,
    Th25;
    
        
    
        
    
    A11: (l9 
    . i) 
    <= (r 
    . i) by 
    A8,
    A9,
    XXREAL_0: 2;
    
        
    
        
    
    A12: (l 
    . i) 
    <= (r9 
    . i) by 
    A9,
    A10,
    XXREAL_0: 2;
    
        thus thesis
    
        proof
    
          per cases by
    A6;
    
            suppose
    
            
    
    A13: 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i); 
    
            
    
    A14: 
    
            now
    
              assume that
    
              
    
    A15: (l9 
    . i) 
    <> (l 
    . i) and 
    
              
    
    A16: (l 
    . i) 
    <> (r9 
    . i); 
    
              
    
              
    
    A17: (l9 
    . i) 
    < (l 
    . i) by 
    A8,
    A15,
    XXREAL_0: 1;
    
              
    
              
    
    A18: (l 
    . i) 
    < (r9 
    . i) by 
    A12,
    A16,
    XXREAL_0: 1;
    
              (l
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
              hence contradiction by
    A13,
    A17,
    A18,
    Th13;
    
            end;
    
            now
    
              assume that
    
              
    
    A19: (l9 
    . i) 
    <> (r 
    . i) and 
    
              
    
    A20: (r 
    . i) 
    <> (r9 
    . i); 
    
              
    
              
    
    A21: (l9 
    . i) 
    < (r 
    . i) by 
    A11,
    A19,
    XXREAL_0: 1;
    
              
    
              
    
    A22: (r 
    . i) 
    < (r9 
    . i) by 
    A10,
    A20,
    XXREAL_0: 1;
    
              (r
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
              hence contradiction by
    A13,
    A21,
    A22,
    Th13;
    
            end;
    
            hence thesis by
    A9,
    A14,
    XXREAL_0: 1;
    
          end;
    
            suppose (l9
    . i) 
    = (r9 
    . i); 
    
            hence thesis by
    A8,
    A10,
    A11,
    A12,
    XXREAL_0: 1;
    
          end;
    
        end;
    
      end;
    
        suppose
    
        
    
    A23: for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i); 
    
        then
    
        
    
    A24: (r9 
    . i) 
    < (l9 
    . i); 
    
        
    
        
    
    A25: 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    A23;
    
        thus thesis
    
        proof
    
          per cases by
    A1,
    A3,
    Th31;
    
            suppose
    
            
    
    A26: for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
            then
    
            
    
    A27: (r 
    . i) 
    <= (r9 
    . i) by 
    A5,
    Th26;
    
            
    
            
    
    A28: (l9 
    . i) 
    <= (l 
    . i) by 
    A5,
    A26,
    Th26;
    
            
    
    A29: 
    
            now
    
              assume (l9
    . i) 
    <> (l 
    . i); 
    
              then
    
              
    
    A30: (l9 
    . i) 
    < (l 
    . i) by 
    A28,
    XXREAL_0: 1;
    
              (l
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
              hence contradiction by
    A24,
    A25,
    A30,
    Th13;
    
            end;
    
            now
    
              assume (r
    . i) 
    <> (r9 
    . i); 
    
              then
    
              
    
    A31: (r 
    . i) 
    < (r9 
    . i) by 
    A27,
    XXREAL_0: 1;
    
              (r
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
              hence contradiction by
    A24,
    A25,
    A31,
    Th13;
    
            end;
    
            hence thesis by
    A29;
    
          end;
    
            suppose
    
            
    
    A32: for i holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            
    
            
    
    A33: (l 
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
            (r
    . i) 
    in (G 
    . i) by 
    A1,
    A3,
    Th32;
    
            hence thesis by
    A24,
    A25,
    A32,
    A33,
    Th13;
    
          end;
    
        end;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:46
    
    
    
    
    
    Th43: k 
    < k9 & k9 
    <= d & ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) & ( 
    cell (l9,r9)) 
    in ( 
    cells (k9,G)) & ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) implies ex i st (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) 
    
    proof
    
      assume that
    
      
    
    A1: k 
    < k9 and 
    
      
    
    A2: k9 
    <= d and 
    
      
    
    A3: ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) and 
    
      
    
    A4: ( 
    cell (l9,r9)) 
    in ( 
    cells (k9,G)); 
    
      
    
      
    
    A5: (k 
    +  
    0 ) 
    < d by 
    A1,
    A2,
    XXREAL_0: 2;
    
      assume
    
      
    
    A6: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
      consider X be
    Subset of ( 
    Seg d) such that 
    
      
    
    A7: ( 
    card X) 
    = k and 
    
      
    
    A8: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A3,
    A5,
    Th30;
    
      
    
      
    
    A9: (d 
    - k) 
    >  
    0 by 
    A5,
    XREAL_1: 20;
    
      (
    card (( 
    Seg d) 
    \ X)) 
    = (( 
    card ( 
    Seg d)) 
    - ( 
    card X)) by 
    CARD_2: 44
    
      .= (d
    - k) by 
    A7,
    FINSEQ_1: 57;
    
      then
    
      consider i0 be
    object such that 
    
      
    
    A10: i0 
    in (( 
    Seg d) 
    \ X) by 
    A9,
    CARD_1: 27,
    XBOOLE_0:def 1;
    
      reconsider i0 as
    Element of ( 
    Seg d) by 
    A10,
    XBOOLE_0:def 5;
    
       not i0
    in X by 
    A10,
    XBOOLE_0:def 5;
    
      then
    
      
    
    A11: (l 
    . i0) 
    = (r 
    . i0) by 
    A8;
    
      per cases by
    A2,
    A3,
    A4,
    A5,
    A6,
    Th42;
    
        suppose (l
    . i0) 
    = (l9 
    . i0) & (r 
    . i0) 
    = (r9 
    . i0); 
    
        hence thesis by
    A11;
    
      end;
    
        suppose (l
    . i0) 
    = (l9 
    . i0) & (r 
    . i0) 
    = (l9 
    . i0); 
    
        hence thesis;
    
      end;
    
        suppose (l
    . i0) 
    = (r9 
    . i0) & (r 
    . i0) 
    = (r9 
    . i0); 
    
        hence thesis;
    
      end;
    
        suppose
    
        
    
    A12: (r9 
    . i0) 
    < (l9 
    . i0); 
    
        assume
    
        
    
    A13: for i holds ((l 
    . i) 
    <> (l9 
    . i) or (r 
    . i) 
    <> (l9 
    . i)) & ((l 
    . i) 
    <> (r9 
    . i) or (r 
    . i) 
    <> (r9 
    . i)); 
    
        defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means (l 
    . $1) 
    <= $2 & $2 
    <= (r 
    . $1) & (r9 
    . $1) 
    < $2 & $2 
    < (l9 
    . $1); 
    
        
    
        
    
    A14: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi]
    
        proof
    
          let i;
    
          
    
          
    
    A15: (l 
    . i) 
    in (G 
    . i) by 
    A3,
    A5,
    Th32;
    
          
    
          
    
    A16: (r 
    . i) 
    in (G 
    . i) by 
    A3,
    A5,
    Th32;
    
          
    
          
    
    A17: (r9 
    . i) 
    < (l9 
    . i) by 
    A2,
    A4,
    A12,
    Th31;
    
          
    
          
    
    A18: 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    A2,
    A4,
    A12,
    Th31;
    
          per cases ;
    
            suppose
    
            
    
    A19: (r9 
    . i) 
    < (l 
    . i) & (l 
    . i) 
    < (l9 
    . i); 
    
            reconsider li = (l
    . i) as 
    Element of 
    REAL by 
    XREAL_0:def 1;
    
            take li;
    
            thus thesis by
    A8,
    A19;
    
          end;
    
            suppose
    
            
    
    A20: (l 
    . i) 
    <= (r9 
    . i); 
    
            
    
            
    
    A21: (l 
    . i) 
    >= (r9 
    . i) by 
    A15,
    A17,
    A18,
    Th13;
    
            then
    
            
    
    A22: (l 
    . i) 
    = (r9 
    . i) by 
    A20,
    XXREAL_0: 1;
    
            then (r
    . i) 
    <> (r9 
    . i) by 
    A13;
    
            then (l
    . i) 
    < (r 
    . i) by 
    A8,
    A22;
    
            then
    
            consider xi be
    Element of 
    REAL such that 
    
            
    
    A23: (l 
    . i) 
    < xi and 
    
            
    
    A24: xi 
    < (r 
    . i) by 
    Th1;
    
            take xi;
    
            (r
    . i) 
    <= (l9 
    . i) by 
    A16,
    A17,
    A18,
    Th13;
    
            hence thesis by
    A21,
    A23,
    A24,
    XXREAL_0: 2;
    
          end;
    
            suppose
    
            
    
    A25: (l9 
    . i) 
    <= (l 
    . i); 
    
            (l9
    . i) 
    >= (l 
    . i) by 
    A15,
    A17,
    A18,
    Th13;
    
            then
    
            
    
    A26: (l9 
    . i) 
    = (l 
    . i) by 
    A25,
    XXREAL_0: 1;
    
            (l9
    . i) 
    >= (r 
    . i) by 
    A16,
    A17,
    A18,
    Th13;
    
            then (l9
    . i) 
    = (r 
    . i) by 
    A8,
    A26;
    
            hence thesis by
    A13,
    A26;
    
          end;
    
        end;
    
        consider x be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A27: for i holds 
    P[i, (x
    . i)] from 
    FUNCT_2:sch 3(
    A14);
    
        reconsider x as
    Element of ( 
    REAL d) by 
    Def3;
    
        
    
        
    
    A28: x 
    in ( 
    cell (l,r)) by 
    A27;
    
        for i st (r9
    . i) 
    < (l9 
    . i) holds (r9 
    . i) 
    < (x 
    . i) & (x 
    . i) 
    < (l9 
    . i) by 
    A27;
    
        hence contradiction by
    A6,
    A12,
    A28,
    Th22;
    
      end;
    
    end;
    
    theorem :: 
    
    CHAIN_1:47
    
    
    
    
    
    Th44: for X,X9 be 
    Subset of ( 
    Seg d) st ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)) & (for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) & (for i holds (i 
    in X9 & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X9 & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i))) holds X 
    c= X9 & (for i st i 
    in X or not i 
    in X9 holds (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i)) & for i st not i 
    in X & i 
    in X9 holds (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) 
    
    proof
    
      let X,X9 be
    Subset of ( 
    Seg d); 
    
      assume
    
      
    
    A1: ( 
    cell (l,r)) 
    c= ( 
    cell (l9,r9)); 
    
      assume
    
      
    
    A2: for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
      assume
    
      
    
    A3: for i holds i 
    in X9 & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X9 & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i); 
    
      
    
      
    
    A4: l 
    in ( 
    cell (l,r)) by 
    Th23;
    
      
    
      
    
    A5: r 
    in ( 
    cell (l,r)) by 
    Th23;
    
      
    
      
    
    A6: for i holds (l9 
    . i) 
    <= (r9 
    . i) by 
    A3;
    
      thus X
    c= X9 
    
      proof
    
        let i be
    object;
    
        assume that
    
        
    
    A7: i 
    in X and 
    
        
    
    A8: not i 
    in X9; 
    
        reconsider i as
    Element of ( 
    Seg d) by 
    A7;
    
        
    
        
    
    A9: (l 
    . i) 
    < (r 
    . i) by 
    A2,
    A7;
    
        
    
        
    
    A10: (l9 
    . i) 
    = (r9 
    . i) by 
    A3,
    A8;
    
        
    
        
    
    A11: (l9 
    . i) 
    <= (l 
    . i) by 
    A1,
    A4,
    A6,
    Th21;
    
        (r
    . i) 
    <= (r9 
    . i) by 
    A1,
    A5,
    A6,
    Th21;
    
        hence thesis by
    A9,
    A10,
    A11,
    XXREAL_0: 2;
    
      end;
    
      set k = (
    card X); 
    
      set k9 = (
    card X9); 
    
      
    
      
    
    A12: ( 
    card ( 
    Seg d)) 
    = d by 
    FINSEQ_1: 57;
    
      then
    
      
    
    A13: k 
    <= d by 
    NAT_1: 43;
    
      
    
      
    
    A14: k9 
    <= d by 
    A12,
    NAT_1: 43;
    
      
    
      
    
    A15: ( 
    cell (l,r)) 
    in ( 
    cells (k,G)) by 
    A2,
    A13,
    Th30;
    
      
    
      
    
    A16: ( 
    cell (l9,r9)) 
    in ( 
    cells (k9,G)) by 
    A3,
    A14,
    Th30;
    
      thus for i st i
    in X or not i 
    in X9 holds (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i) 
    
      proof
    
        let i;
    
        assume
    
        
    
    A17: i 
    in X or not i 
    in X9; 
    
        (l9
    . i) 
    <= (r9 
    . i) by 
    A3;
    
        then (l
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i) or (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) by 
    A1,
    A13,
    A14,
    A15,
    A16,
    Th42;
    
        hence thesis by
    A2,
    A3,
    A17;
    
      end;
    
      thus for i st not i
    in X & i 
    in X9 holds (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) 
    
      proof
    
        let i;
    
        assume that
    
        
    
    A18: not i 
    in X and 
    
        
    
    A19: i 
    in X9; 
    
        
    
        
    
    A20: (l 
    . i) 
    = (r 
    . i) by 
    A2,
    A18;
    
        (l9
    . i) 
    < (r9 
    . i) by 
    A3,
    A19;
    
        hence thesis by
    A1,
    A13,
    A14,
    A15,
    A16,
    A20,
    Th42;
    
      end;
    
    end;
    
    definition
    
      let d, G, k;
    
      mode
    
    Cell of k,G is 
    Element of ( 
    cells (k,G)); 
    
    end
    
    definition
    
      let d, G, k;
    
      mode
    
    Chain of k,G is 
    Subset of ( 
    cells (k,G)); 
    
    end
    
    definition
    
      let d, G, k;
    
      :: 
    
    CHAIN_1:def8
    
      func
    
    0_ (k,G) -> 
    Chain of k, G equals 
    {} ; 
    
      coherence by
    SUBSET_1: 1;
    
    end
    
    definition
    
      let d, G;
    
      :: 
    
    CHAIN_1:def9
    
      func
    
    Omega (G) -> 
    Chain of d, G equals ( 
    cells (d,G)); 
    
      coherence
    
      proof
    
        (
    cells (d,G)) 
    c= ( 
    cells (d,G)); 
    
        hence thesis;
    
      end;
    
    end
    
    notation
    
      let d, G, k;
    
      let C1,C2 be
    Chain of k, G; 
    
      synonym C1 
    
    + C2 for C1 
    \+\ C2; 
    
    end
    
    definition
    
      let d, G, k;
    
      let C1,C2 be
    Chain of k, G; 
    
      :: original:
    +
    
      redefine
    
      func C1
    
    + C2 -> 
    Chain of k, G ; 
    
      coherence
    
      proof
    
        (C1
    \+\ C2) 
    c= ( 
    cells (k,G)); 
    
        hence thesis;
    
      end;
    
    end
    
    definition
    
      let d, G;
    
      :: 
    
    CHAIN_1:def10
    
      func
    
    infinite-cell (G) -> 
    Cell of d, G means 
    
      :
    
    Def10: ex l, r st it 
    = ( 
    cell (l,r)) & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
      existence
    
      proof
    
        defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means $2 
    in (G 
    . $1) & for xi st xi 
    in (G 
    . $1) holds xi 
    <= $2; 
    
        
    
        
    
    A1: for i holds ex li be 
    Element of 
    REAL st 
    P[i, li] by
    Th9;
    
        consider l be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A2: for i holds 
    P[i, (l
    . i)] from 
    FUNCT_2:sch 3(
    A1);
    
        reconsider l as
    Element of ( 
    REAL d) by 
    Def3;
    
        defpred
    
    R[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means $2 
    in (G 
    . $1) & for xi st xi 
    in (G 
    . $1) holds xi 
    >= $2; 
    
        
    
        
    
    A3: for i holds ex ri be 
    Element of 
    REAL st 
    R[i, ri] by
    Th10;
    
        consider r be
    Function of ( 
    Seg d), 
    REAL such that 
    
        
    
    A4: for i holds 
    R[i, (r
    . i)] from 
    FUNCT_2:sch 3(
    A3);
    
        reconsider r as
    Element of ( 
    REAL d) by 
    Def3;
    
        
    
    A5: 
    
        now
    
          let i;
    
          (r
    . i) 
    in (G 
    . i) by 
    A4;
    
          then
    
          
    
    A6: (r 
    . i) 
    <= (l 
    . i) by 
    A2;
    
          now
    
            assume
    
            
    
    A7: (l 
    . i) 
    = (r 
    . i); 
    
            consider x1,x2 be
    object such that 
    
            
    
    A8: x1 
    in (G 
    . i) and 
    
            
    
    A9: x2 
    in (G 
    . i) and 
    
            
    
    A10: x1 
    <> x2 by 
    ZFMISC_1:def 10;
    
            reconsider x1, x2 as
    Element of 
    REAL by 
    A8,
    A9;
    
            
    
            
    
    A11: (r 
    . i) 
    <= x1 by 
    A4,
    A8;
    
            
    
            
    
    A12: x1 
    <= (l 
    . i) by 
    A2,
    A8;
    
            
    
            
    
    A13: (r 
    . i) 
    <= x2 by 
    A4,
    A9;
    
            
    
            
    
    A14: x2 
    <= (l 
    . i) by 
    A2,
    A9;
    
            x1
    = (l 
    . i) by 
    A7,
    A11,
    A12,
    XXREAL_0: 1;
    
            hence contradiction by
    A7,
    A10,
    A13,
    A14,
    XXREAL_0: 1;
    
          end;
    
          hence (r
    . i) 
    < (l 
    . i) by 
    A6,
    XXREAL_0: 1;
    
        end;
    
        
    
    A15: 
    
        now
    
          let i;
    
          
    
          
    
    A16: (l 
    . i) 
    in (G 
    . i) by 
    A2;
    
          
    
          
    
    A17: (r 
    . i) 
    in (G 
    . i) by 
    A4;
    
          
    
          
    
    A18: (r 
    . i) 
    < (l 
    . i) by 
    A5;
    
          for xi st xi
    in (G 
    . i) holds not ((l 
    . i) 
    < xi or xi 
    < (r 
    . i)) by 
    A2,
    A4;
    
          hence (r
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A16,
    A17,
    A18,
    Th13;
    
        end;
    
        then
    
        reconsider A = (
    cell (l,r)) as 
    Cell of d, G by 
    Th30;
    
        take A, l, r;
    
        thus thesis by
    A15;
    
      end;
    
      uniqueness
    
      proof
    
        let A,A9 be
    Cell of d, G; 
    
        given l, r such that
    
        
    
    A19: A 
    = ( 
    cell (l,r)) and 
    
        
    
    A20: for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
        given l9, r9 such that
    
        
    
    A21: A9 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A22: for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i); 
    
        reconsider l, r, l9, r9 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
        
    
    A23: 
    
        now
    
          let i;
    
          
    
          
    
    A24: (r 
    . i) 
    < (l 
    . i) by 
    A20;
    
          
    
          
    
    A25: 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A20;
    
          
    
          
    
    A26: (r9 
    . i) 
    < (l9 
    . i) by 
    A22;
    
          
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    A22;
    
          hence (l
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i) by 
    A24,
    A25,
    A26,
    Th19;
    
        end;
    
        then l
    = l9 by 
    FUNCT_2: 63;
    
        hence thesis by
    A19,
    A21,
    A23,
    FUNCT_2: 63;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:48
    
    
    
    
    
    Th45: ( 
    cell (l,r)) is 
    Cell of d, G implies (( 
    cell (l,r)) 
    = ( 
    infinite-cell G) iff for i holds (r 
    . i) 
    < (l 
    . i)) 
    
    proof
    
      assume
    
      
    
    A1: ( 
    cell (l,r)) is 
    Cell of d, G; 
    
      then
    
      reconsider A = (
    cell (l,r)) as 
    Cell of d, G; 
    
      hereby
    
        assume (
    cell (l,r)) 
    = ( 
    infinite-cell G); 
    
        then
    
        consider l9, r9 such that
    
        
    
    A2: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A3: for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    Def10;
    
        
    
        
    
    A4: l 
    = l9 by 
    A2,
    A3,
    Th28;
    
        r
    = r9 by 
    A2,
    A3,
    Th28;
    
        hence for i holds (r
    . i) 
    < (l 
    . i) by 
    A3,
    A4;
    
      end;
    
      set i0 = the
    Element of ( 
    Seg d); 
    
      assume for i holds (r
    . i) 
    < (l 
    . i); 
    
      then
    
      
    
    A5: (r 
    . i0) 
    < (l 
    . i0); 
    
      
    
      
    
    A6: A 
    = ( 
    cell (l,r)); 
    
      for i holds (r
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A1,
    A5,
    Th31;
    
      hence thesis by
    A6,
    Def10;
    
    end;
    
    theorem :: 
    
    CHAIN_1:49
    
    
    
    
    
    Th46: ( 
    cell (l,r)) 
    = ( 
    infinite-cell G) iff for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) 
    
    proof
    
      hereby
    
        assume (
    cell (l,r)) 
    = ( 
    infinite-cell G); 
    
        then
    
        consider l9, r9 such that
    
        
    
    A1: ( 
    cell (l,r)) 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A2: for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) by 
    Def10;
    
        
    
        
    
    A3: l 
    = l9 by 
    A1,
    A2,
    Th28;
    
        r
    = r9 by 
    A1,
    A2,
    Th28;
    
        hence for i holds (r
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A2,
    A3;
    
      end;
    
      assume
    
      
    
    A4: for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
      then (
    cell (l,r)) is 
    Cell of d, G by 
    Th30;
    
      hence thesis by
    A4,
    Def10;
    
    end;
    
    scheme :: 
    
    CHAIN_1:sch4
    
    ChainInd { d() -> non
    zero  
    Nat , G() -> 
    Grating of d() , k() -> 
    Nat , C() -> 
    Chain of k(), G() , P[ 
    set] } :
    
P[C()]
    
      provided
    
      
    
    A1: P[( 
    0_ (k(),G()))] 
    
       and 
    
      
    
    A2: for A be 
    Cell of k(), G() st A 
    in C() holds P[ 
    {A}]
    
       and 
    
      
    
    A3: for C1,C2 be 
    Chain of k(), G() st C1 
    c= C() & C2 
    c= C() & P[C1] & P[C2] holds P[(C1 
    + C2)]; 
    
      
    
      
    
    A4: C() is 
    finite;
    
      
    
      
    
    A5: P[ 
    {} ] by 
    A1;
    
      
    
      
    
    A6: for x,B be 
    set st x 
    in C() & B 
    c= C() & P[B] holds P[(B 
    \/  
    {x})]
    
      proof
    
        let A,C1 be
    set;
    
        assume that
    
        
    
    A7: A 
    in C() and 
    
        
    
    A8: C1 
    c= C() and 
    
        
    
    A9: P[C1]; 
    
        reconsider A9 = A as
    Cell of k(), G() by 
    A7;
    
        reconsider C19 = C1 as
    Chain of k(), G() by 
    A8,
    XBOOLE_1: 1;
    
        per cases ;
    
          suppose A
    in C1; 
    
          then
    {A}
    c= C1 by 
    ZFMISC_1: 31;
    
          hence thesis by
    A9,
    XBOOLE_1: 12;
    
        end;
    
          suppose
    
          
    
    A10: not A 
    in C1; 
    
          now
    
            let A9 be
    object;
    
            assume
    
            
    
    A11: A9 
    in (C1 
    /\  
    {A});
    
            then
    
            
    
    A12: A9 
    in C1 by 
    XBOOLE_0:def 4;
    
            A9
    in  
    {A} by
    A11,
    XBOOLE_0:def 4;
    
            hence contradiction by
    A10,
    A12,
    TARSKI:def 1;
    
          end;
    
          then (C1
    /\  
    {A})
    =  
    {} by 
    XBOOLE_0:def 1;
    
          
    
          then
    
          
    
    A13: (C19 
    +  
    {A9})
    = ((C1 
    \/  
    {A})
    \  
    {} ) by 
    XBOOLE_1: 101
    
          .= (C1
    \/  
    {A});
    
          
    
          
    
    A14: P[ 
    {A9}] by
    A2,
    A7;
    
          
    {A}
    c= C() by 
    A7,
    ZFMISC_1: 31;
    
          hence thesis by
    A3,
    A8,
    A9,
    A13,
    A14;
    
        end;
    
      end;
    
      thus P[C()] from
    FINSET_1:sch 2(
    A4,
    A5,
    A6);
    
    end;
    
    definition
    
      let d, G, k;
    
      let A be
    Cell of k, G; 
    
      :: 
    
    CHAIN_1:def11
    
      func
    
    star A -> 
    Chain of (k 
    + 1), G equals { B where B be 
    Cell of (k 
    + 1), G : A 
    c= B }; 
    
      coherence
    
      proof
    
        defpred
    
    P[
    set] means A
    c= $1; 
    
        { B where B be
    Cell of (k 
    + 1), G : 
    P[B] }
    c= ( 
    cells ((k 
    + 1),G)) from 
    FRAENKEL:sch 10;
    
        hence thesis;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:50
    
    
    
    
    
    Th47: for A be 
    Cell of k, G, B be 
    Cell of (k 
    + 1), G holds B 
    in ( 
    star A) iff A 
    c= B 
    
    proof
    
      let A be
    Cell of k, G, B be 
    Cell of (k 
    + 1), G; 
    
      defpred
    
    P[
    set] means A
    c= $1; 
    
      
    
      
    
    A1: ( 
    star A) 
    = { B9 where B9 be 
    Cell of (k 
    + 1), G : 
    P[B9] };
    
      thus B
    in ( 
    star A) iff 
    P[B] from
    LMOD_7:sch 7(
    A1);
    
    end;
    
    definition
    
      let d, G, k;
    
      let C be
    Chain of (k 
    + 1), G; 
    
      :: 
    
    CHAIN_1:def12
    
      func
    
    del C -> 
    Chain of k, G equals { A where A be 
    Cell of k, G : (k 
    + 1) 
    <= d & ( 
    card (( 
    star A) 
    /\ C)) is 
    odd };
    
      coherence
    
      proof
    
        defpred
    
    P[
    Cell of k, G] means (k 
    + 1) 
    <= d & ( 
    card (( 
    star $1) 
    /\ C)) is 
    odd;
    
        { A where A be
    Cell of k, G : 
    P[A] }
    c= ( 
    cells (k,G)) from 
    FRAENKEL:sch 10;
    
        hence thesis;
    
      end;
    
    end
    
    notation
    
      let d, G, k;
    
      let C be
    Chain of (k 
    + 1), G; 
    
      synonym 
    
    . C for 
    del C; 
    
    end
    
    definition
    
      let d, G, k;
    
      let C be
    Chain of (k 
    + 1), G, C9 be 
    Chain of k, G; 
    
      :: 
    
    CHAIN_1:def13
    
      pred C9
    
    bounds C means C9 
    = ( 
    del C); 
    
    end
    
    theorem :: 
    
    CHAIN_1:51
    
    
    
    
    
    Th48: for A be 
    Cell of k, G, C be 
    Chain of (k 
    + 1), G holds A 
    in ( 
    del C) iff (k 
    + 1) 
    <= d & ( 
    card (( 
    star A) 
    /\ C)) is 
    odd
    
    proof
    
      let A be
    Cell of k, G, C be 
    Chain of (k 
    + 1), G; 
    
      defpred
    
    P[
    Cell of k, G] means (k 
    + 1) 
    <= d & ( 
    card (( 
    star $1) 
    /\ C)) is 
    odd;
    
      
    
      
    
    A1: ( 
    del C) 
    = { A9 where A9 be 
    Cell of k, G : 
    P[A9] };
    
      thus A
    in ( 
    del C) iff 
    P[A] from
    LMOD_7:sch 7(
    A1);
    
    end;
    
    theorem :: 
    
    CHAIN_1:52
    
    
    
    
    
    Th49: (k 
    + 1) 
    > d implies for C be 
    Chain of (k 
    + 1), G holds ( 
    del C) 
    = ( 
    0_ (k,G)) 
    
    proof
    
      assume
    
      
    
    A1: (k 
    + 1) 
    > d; 
    
      let C be
    Chain of (k 
    + 1), G; 
    
      for A be
    object holds not A 
    in ( 
    del C) by 
    A1,
    Th48;
    
      hence thesis by
    XBOOLE_0:def 1;
    
    end;
    
    theorem :: 
    
    CHAIN_1:53
    
    
    
    
    
    Th50: (k 
    + 1) 
    <= d implies for A be 
    Cell of k, G, B be 
    Cell of (k 
    + 1), G holds A 
    in ( 
    del  
    {B}) iff A
    c= B 
    
    proof
    
      assume
    
      
    
    A1: (k 
    + 1) 
    <= d; 
    
      let A be
    Cell of k, G, B be 
    Cell of (k 
    + 1), G; 
    
      set X = ((
    star A) 
    /\  
    {B});
    
      (
    card X) is 
    odd iff B 
    in ( 
    star A) 
    
      proof
    
        per cases ;
    
          suppose
    
          
    
    A2: B 
    in ( 
    star A); 
    
          now
    
            let B9 be
    object;
    
            B9
    in  
    {B} iff B9
    = B by 
    TARSKI:def 1;
    
            hence B9
    in X iff B9 
    = B by 
    A2,
    XBOOLE_0:def 4;
    
          end;
    
          then X
    =  
    {B} by
    TARSKI:def 1;
    
          then (
    card X) 
    = ((2 
    *  
    0 ) 
    + 1) by 
    CARD_1: 30;
    
          hence thesis by
    A2;
    
        end;
    
          suppose
    
          
    
    A3: not B 
    in ( 
    star A); 
    
          now
    
            let B9 be
    object;
    
            B9
    = B or not B9 
    in  
    {B} by
    TARSKI:def 1;
    
            hence not B9
    in X by 
    A3,
    XBOOLE_0:def 4;
    
          end;
    
          then (
    card X) 
    = (2 
    *  
    0 ) by 
    CARD_1: 27,
    XBOOLE_0:def 1;
    
          hence thesis by
    A3;
    
        end;
    
      end;
    
      hence thesis by
    A1,
    Th47,
    Th48;
    
    end;
    
    theorem :: 
    
    CHAIN_1:54
    
    
    
    
    
    Th51: d 
    = (d9 
    + 1) implies for A be 
    Cell of d9, G holds ( 
    card ( 
    star A)) 
    = 2 
    
    proof
    
      assume
    
      
    
    A1: d 
    = (d9 
    + 1); 
    
      then
    
      
    
    A2: d9 
    < d by 
    NAT_1: 13;
    
      let A be
    Cell of d9, G; 
    
      consider l, r, i0 such that
    
      
    
    A3: A 
    = ( 
    cell (l,r)) and 
    
      
    
    A4: (l 
    . i0) 
    = (r 
    . i0) and 
    
      
    
    A5: (l 
    . i0) 
    in (G 
    . i0) and 
    
      
    
    A6: for i st i 
    <> i0 holds (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) by 
    A1,
    Th38;
    
      
    
    A7: 
    
      now
    
        let i;
    
        i
    = i0 or i 
    <> i0; 
    
        hence (l
    . i) 
    <= (r 
    . i) by 
    A4,
    A6;
    
      end;
    
      ex B1,B2 be
    set st B1 
    in ( 
    star A) & B2 
    in ( 
    star A) & B1 
    <> B2 & for B be 
    set st B 
    in ( 
    star A) holds B 
    = B1 or B 
    = B2 
    
      proof
    
        ex l1, r1 st
    [(l1
    . i0), (r1 
    . i0)] is 
    Gap of (G 
    . i0) & (r1 
    . i0) 
    = (l 
    . i0) & (((l1 
    . i0) 
    < (r1 
    . i0) & for i st i 
    <> i0 holds (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i)) or for i holds (r1 
    . i) 
    < (l1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i)) 
    
        proof
    
          consider l1i0 be
    Element of 
    REAL such that 
    
          
    
    A8: 
    [l1i0, (l
    . i0)] is 
    Gap of (G 
    . i0) by 
    A5,
    Th16;
    
          per cases by
    A8,
    Th13;
    
            suppose
    
            
    
    A9: l1i0 
    < (l 
    . i0); 
    
            defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    = i0 implies $2 
    = l1i0) & ($1 
    <> i0 implies $2 
    = (l 
    . $1)); 
    
            
    
            
    
    A10: for i holds ex li be 
    Element of 
    REAL st 
    P[i, li]
    
            proof
    
              let i;
    
              
    
              
    
    A11: (l 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
              i
    = i0 or i 
    <> i0; 
    
              hence thesis by
    A11;
    
            end;
    
            consider l1 be
    Function of ( 
    Seg d), 
    REAL such that 
    
            
    
    A12: for i holds 
    P[i, (l1
    . i)] from 
    FUNCT_2:sch 3(
    A10);
    
            reconsider l1 as
    Element of ( 
    REAL d) by 
    Def3;
    
            take l1, r;
    
            thus thesis by
    A4,
    A8,
    A9,
    A12;
    
          end;
    
            suppose
    
            
    
    A13: (l 
    . i0) 
    < l1i0; 
    
            consider l1, r1 such that (
    cell (l1,r1)) 
    = ( 
    infinite-cell G) and 
    
            
    
    A14: for i holds (r1 
    . i) 
    < (l1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i) by 
    Def10;
    
            take l1, r1;
    
            
    
            
    
    A15: (r1 
    . i0) 
    < (l1 
    . i0) by 
    A14;
    
            
    [(l1
    . i0), (r1 
    . i0)] is 
    Gap of (G 
    . i0) by 
    A14;
    
            hence thesis by
    A8,
    A13,
    A14,
    A15,
    Th19;
    
          end;
    
        end;
    
        then
    
        consider l1, r1 such that
    
        
    
    A16: 
    [(l1
    . i0), (r1 
    . i0)] is 
    Gap of (G 
    . i0) and 
    
        
    
    A17: (r1 
    . i0) 
    = (l 
    . i0) and 
    
        
    
    A18: ((l1 
    . i0) 
    < (r1 
    . i0) & for i st i 
    <> i0 holds (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i)) or for i holds (r1 
    . i) 
    < (l1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i); 
    
        
    
    A19: 
    
        now
    
          let i;
    
          
    
          
    
    A20: i 
    <> i0 & (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i) implies 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i) by 
    A6;
    
          i
    = i0 or i 
    <> i0; 
    
          hence
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i) by 
    A16,
    A18,
    A20;
    
        end;
    
        
    
        
    
    A21: (for i holds (l1 
    . i) 
    < (r1 
    . i)) or for i holds (r1 
    . i) 
    < (l1 
    . i) 
    
        proof
    
          per cases by
    A18;
    
            suppose
    
            
    
    A22: (l1 
    . i0) 
    < (r1 
    . i0) & for i st i 
    <> i0 holds (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i); 
    
            now
    
              let i;
    
              
    
              
    
    A23: i 
    <> i0 & (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i) implies (l1 
    . i) 
    < (r1 
    . i) by 
    A6;
    
              i
    = i0 or i 
    <> i0; 
    
              hence (l1
    . i) 
    < (r1 
    . i) by 
    A22,
    A23;
    
            end;
    
            hence thesis;
    
          end;
    
            suppose for i holds (r1
    . i) 
    < (l1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis;
    
          end;
    
        end;
    
        then
    
        reconsider B1 = (
    cell (l1,r1)) as 
    Cell of d, G by 
    A19,
    Th37;
    
        ex l2, r2 st
    [(l2
    . i0), (r2 
    . i0)] is 
    Gap of (G 
    . i0) & (l2 
    . i0) 
    = (l 
    . i0) & (((l2 
    . i0) 
    < (r2 
    . i0) & for i st i 
    <> i0 holds (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i)) or for i holds (r2 
    . i) 
    < (l2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i)) 
    
        proof
    
          consider r2i0 be
    Element of 
    REAL such that 
    
          
    
    A24: 
    [(l
    . i0), r2i0] is 
    Gap of (G 
    . i0) by 
    A5,
    Th15;
    
          per cases by
    A24,
    Th13;
    
            suppose
    
            
    
    A25: (l 
    . i0) 
    < r2i0; 
    
            defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    = i0 implies $2 
    = r2i0) & ($1 
    <> i0 implies $2 
    = (r 
    . $1)); 
    
            
    
            
    
    A26: for i holds ex ri be 
    Element of 
    REAL st 
    P[i, ri]
    
            proof
    
              let i;
    
              
    
              
    
    A27: (r 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
              i
    = i0 or i 
    <> i0; 
    
              hence thesis by
    A27;
    
            end;
    
            consider r2 be
    Function of ( 
    Seg d), 
    REAL such that 
    
            
    
    A28: for i holds 
    P[i, (r2
    . i)] from 
    FUNCT_2:sch 3(
    A26);
    
            reconsider r2 as
    Element of ( 
    REAL d) by 
    Def3;
    
            take l, r2;
    
            thus thesis by
    A24,
    A25,
    A28;
    
          end;
    
            suppose
    
            
    
    A29: r2i0 
    < (l 
    . i0); 
    
            consider l2, r2 such that (
    cell (l2,r2)) 
    = ( 
    infinite-cell G) and 
    
            
    
    A30: for i holds (r2 
    . i) 
    < (l2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i) by 
    Def10;
    
            take l2, r2;
    
            
    
            
    
    A31: (r2 
    . i0) 
    < (l2 
    . i0) by 
    A30;
    
            
    [(l2
    . i0), (r2 
    . i0)] is 
    Gap of (G 
    . i0) by 
    A30;
    
            hence thesis by
    A24,
    A29,
    A30,
    A31,
    Th19;
    
          end;
    
        end;
    
        then
    
        consider l2, r2 such that
    
        
    
    A32: 
    [(l2
    . i0), (r2 
    . i0)] is 
    Gap of (G 
    . i0) and 
    
        
    
    A33: (l2 
    . i0) 
    = (l 
    . i0) and 
    
        
    
    A34: ((l2 
    . i0) 
    < (r2 
    . i0) & for i st i 
    <> i0 holds (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i)) or for i holds (r2 
    . i) 
    < (l2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i); 
    
        
    
    A35: 
    
        now
    
          let i;
    
          
    
          
    
    A36: i 
    <> i0 & (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i) implies 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i) by 
    A6;
    
          i
    = i0 or i 
    <> i0; 
    
          hence
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i) by 
    A32,
    A34,
    A36;
    
        end;
    
        (for i holds (l2
    . i) 
    < (r2 
    . i)) or for i holds (r2 
    . i) 
    < (l2 
    . i) 
    
        proof
    
          per cases by
    A34;
    
            suppose
    
            
    
    A37: (l2 
    . i0) 
    < (r2 
    . i0) & for i st i 
    <> i0 holds (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i); 
    
            now
    
              let i;
    
              
    
              
    
    A38: i 
    <> i0 & (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i) implies (l2 
    . i) 
    < (r2 
    . i) by 
    A6;
    
              i
    = i0 or i 
    <> i0; 
    
              hence (l2
    . i) 
    < (r2 
    . i) by 
    A37,
    A38;
    
            end;
    
            hence thesis;
    
          end;
    
            suppose for i holds (r2
    . i) 
    < (l2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis;
    
          end;
    
        end;
    
        then
    
        reconsider B2 = (
    cell (l2,r2)) as 
    Cell of d, G by 
    A35,
    Th37;
    
        take B1, B2;
    
        A
    c= B1 
    
        proof
    
          per cases by
    A18;
    
            suppose
    
            
    
    A39: (l1 
    . i0) 
    < (r1 
    . i0) & for i st i 
    <> i0 holds (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i); 
    
            
    
    A40: 
    
            now
    
              let i;
    
              i
    = i0 or i 
    <> i0 & (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i) by 
    A39;
    
              hence (l1
    . i) 
    <= (r1 
    . i) by 
    A6,
    A39;
    
            end;
    
            now
    
              let i;
    
              i
    = i0 or i 
    <> i0 & (l1 
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i) by 
    A39;
    
              hence (l1
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r1 
    . i) by 
    A4,
    A17,
    A40;
    
            end;
    
            hence thesis by
    A3,
    A40,
    Th25;
    
          end;
    
            suppose for i holds (r1
    . i) 
    < (l1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis by
    A3,
    A4,
    A7,
    A17,
    Th27;
    
          end;
    
        end;
    
        hence B1
    in ( 
    star A) by 
    A1;
    
        A
    c= B2 
    
        proof
    
          per cases by
    A34;
    
            suppose
    
            
    
    A41: (l2 
    . i0) 
    < (r2 
    . i0) & for i st i 
    <> i0 holds (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i); 
    
            
    
    A42: 
    
            now
    
              let i;
    
              i
    = i0 or i 
    <> i0 & (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i) by 
    A41;
    
              hence (l2
    . i) 
    <= (r2 
    . i) by 
    A6,
    A41;
    
            end;
    
            now
    
              let i;
    
              i
    = i0 or i 
    <> i0 & (l2 
    . i) 
    = (l 
    . i) & (r2 
    . i) 
    = (r 
    . i) by 
    A41;
    
              hence (l2
    . i) 
    <= (l 
    . i) & (l 
    . i) 
    <= (r 
    . i) & (r 
    . i) 
    <= (r2 
    . i) by 
    A4,
    A33,
    A42;
    
            end;
    
            hence thesis by
    A3,
    A42,
    Th25;
    
          end;
    
            suppose for i holds (r2
    . i) 
    < (l2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i); 
    
            hence thesis by
    A3,
    A7,
    A33,
    Th27;
    
          end;
    
        end;
    
        hence B2
    in ( 
    star A) by 
    A1;
    
        
    
        
    
    A43: l1 
    <> l2 by 
    A17,
    A18,
    A33;
    
        (for i holds (l1
    . i) 
    <= (r1 
    . i)) or for i holds (r1 
    . i) 
    < (l1 
    . i) by 
    A21;
    
        hence B1
    <> B2 by 
    A43,
    Th28;
    
        let B be
    set;
    
        assume
    
        
    
    A44: B 
    in ( 
    star A); 
    
        then
    
        reconsider B as
    Cell of d, G by 
    A1;
    
        consider l9, r9 such that
    
        
    
    A45: B 
    = ( 
    cell (l9,r9)) and 
    
        
    
    A46: for i holds 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) and 
    
        
    
    A47: (for i holds (l9 
    . i) 
    < (r9 
    . i)) or for i holds (r9 
    . i) 
    < (l9 
    . i) by 
    Th36;
    
        
    
        
    
    A48: 
    [(l9
    . i0), (r9 
    . i0)] is 
    Gap of (G 
    . i0) by 
    A46;
    
        
    
        
    
    A49: A 
    c= B by 
    A44,
    Th47;
    
        per cases by
    A47;
    
          suppose
    
          
    
    A50: for i holds (l9 
    . i) 
    < (r9 
    . i); 
    
          
    
    A51: 
    
          now
    
            let i;
    
            assume
    
            
    
    A52: i 
    <> i0; 
    
            (l9
    . i) 
    < (r9 
    . i) by 
    A50;
    
            then (l
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (r9 
    . i) or (l 
    . i) 
    = (l9 
    . i) & (r 
    . i) 
    = (l9 
    . i) or (l 
    . i) 
    = (r9 
    . i) & (r 
    . i) 
    = (r9 
    . i) by 
    A2,
    A3,
    A45,
    A49,
    Th42;
    
            hence (l9
    . i) 
    = (l 
    . i) & (r9 
    . i) 
    = (r 
    . i) by 
    A6,
    A52;
    
          end;
    
          thus thesis
    
          proof
    
            
    
            
    
    A53: (l9 
    . i0) 
    < (r9 
    . i0) by 
    A50;
    
            per cases by
    A2,
    A3,
    A4,
    A45,
    A49,
    A53,
    Th42;
    
              suppose
    
              
    
    A54: (l 
    . i0) 
    = (r9 
    . i0) & (r 
    . i0) 
    = (r9 
    . i0); 
    
              then
    
              
    
    A55: (l9 
    . i0) 
    = (l1 
    . i0) by 
    A16,
    A17,
    A48,
    Th18;
    
              reconsider l9, r9, l1, r1 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              
    
    A56: 
    
              now
    
                let i;
    
                
    
                
    
    A57: (l1 
    . i0) 
    < (l 
    . i0) by 
    A50,
    A54,
    A55;
    
                then i
    = i0 or i 
    <> i0 & (l9 
    . i) 
    = (l 
    . i) & (l1 
    . i) 
    = (l 
    . i) by 
    A17,
    A18,
    A51;
    
                hence (l9
    . i) 
    = (l1 
    . i) by 
    A16,
    A17,
    A48,
    A54,
    Th18;
    
                i
    = i0 or i 
    <> i0 & (r9 
    . i) 
    = (r 
    . i) & (r1 
    . i) 
    = (r 
    . i) by 
    A17,
    A18,
    A51,
    A57;
    
                hence (r9
    . i) 
    = (r1 
    . i) by 
    A17,
    A54;
    
              end;
    
              then l9
    = l1 by 
    FUNCT_2: 63;
    
              hence thesis by
    A45,
    A56,
    FUNCT_2: 63;
    
            end;
    
              suppose
    
              
    
    A58: (l 
    . i0) 
    = (l9 
    . i0) & (r 
    . i0) 
    = (l9 
    . i0); 
    
              then
    
              
    
    A59: (r9 
    . i0) 
    = (r2 
    . i0) by 
    A32,
    A33,
    A48,
    Th17;
    
              reconsider l9, r9, l2, r2 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              
    
    A60: 
    
              now
    
                let i;
    
                
    
                
    
    A61: (l 
    . i0) 
    < (r2 
    . i0) by 
    A50,
    A58,
    A59;
    
                then i
    = i0 or i 
    <> i0 & (r9 
    . i) 
    = (r 
    . i) & (r2 
    . i) 
    = (r 
    . i) by 
    A33,
    A34,
    A51;
    
                hence (r9
    . i) 
    = (r2 
    . i) by 
    A32,
    A33,
    A48,
    A58,
    Th17;
    
                i
    = i0 or i 
    <> i0 & (l9 
    . i) 
    = (l 
    . i) & (l2 
    . i) 
    = (l 
    . i) by 
    A33,
    A34,
    A51,
    A61;
    
                hence (l9
    . i) 
    = (l2 
    . i) by 
    A33,
    A58;
    
              end;
    
              then l9
    = l2 by 
    FUNCT_2: 63;
    
              hence thesis by
    A45,
    A60,
    FUNCT_2: 63;
    
            end;
    
          end;
    
        end;
    
          suppose
    
          
    
    A62: for i holds (r9 
    . i) 
    < (l9 
    . i); 
    
          consider i1 such that
    
          
    
    A63: (l 
    . i1) 
    = (l9 
    . i1) & (r 
    . i1) 
    = (l9 
    . i1) or (l 
    . i1) 
    = (r9 
    . i1) & (r 
    . i1) 
    = (r9 
    . i1) by 
    A2,
    A3,
    A45,
    A49,
    Th43;
    
          
    
          
    
    A64: i0 
    = i1 by 
    A6,
    A63;
    
          thus thesis
    
          proof
    
            per cases by
    A63,
    A64;
    
              suppose
    
              
    
    A65: (l 
    . i0) 
    = (r9 
    . i0) & (r 
    . i0) 
    = (r9 
    . i0); 
    
              then (l9
    . i0) 
    = (l1 
    . i0) by 
    A16,
    A17,
    A48,
    Th18;
    
              then B1
    = ( 
    infinite-cell G) by 
    A17,
    A18,
    A62,
    A65,
    Th45;
    
              hence thesis by
    A45,
    A62,
    Th45;
    
            end;
    
              suppose
    
              
    
    A66: (l 
    . i0) 
    = (l9 
    . i0) & (r 
    . i0) 
    = (l9 
    . i0); 
    
              then (r9
    . i0) 
    = (r2 
    . i0) by 
    A32,
    A33,
    A48,
    Th17;
    
              then B2
    = ( 
    infinite-cell G) by 
    A33,
    A34,
    A62,
    A66,
    Th45;
    
              hence thesis by
    A45,
    A62,
    Th45;
    
            end;
    
          end;
    
        end;
    
      end;
    
      hence thesis by
    Th5;
    
    end;
    
    theorem :: 
    
    CHAIN_1:55
    
    
    
    
    
    Th52: for G be 
    Grating of d, B be 
    Cell of ( 
    0  
    + 1), G holds ( 
    card ( 
    del  
    {B}))
    = 2 
    
    proof
    
      
    
      
    
    A1: ( 
    0  
    + 1) 
    <= d by 
    Def2;
    
      let G be
    Grating of d, B be 
    Cell of ( 
    0  
    + 1), G; 
    
      consider l, r, i0 such that
    
      
    
    A2: B 
    = ( 
    cell (l,r)) and 
    
      
    
    A3: (l 
    . i0) 
    < (r 
    . i0) or d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0) and 
    
      
    
    A4: 
    [(l
    . i0), (r 
    . i0)] is 
    Gap of (G 
    . i0) and 
    
      
    
    A5: for i st i 
    <> i0 holds (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    Th40;
    
      ex A1,A2 be
    set st A1 
    in ( 
    del  
    {B}) & A2
    in ( 
    del  
    {B}) & A1
    <> A2 & for A be 
    set st A 
    in ( 
    del  
    {B}) holds A
    = A1 or A 
    = A2 
    
      proof
    
        for i holds (l
    . i) 
    in (G 
    . i) & (r 
    . i) 
    in (G 
    . i) by 
    A1,
    A2,
    Th32;
    
        then
    
        reconsider A1 = (
    cell (l,l)), A2 = ( 
    cell (r,r)) as 
    Cell of 
    0 , G by 
    Th35;
    
        take A1, A2;
    
        
    
        
    
    A6: A1 
    =  
    {l} by
    Th24;
    
        
    
        
    
    A7: A2 
    =  
    {r} by
    Th24;
    
        
    
        
    
    A8: l 
    in B by 
    A2,
    Th23;
    
        
    
        
    
    A9: r 
    in B by 
    A2,
    Th23;
    
        
    
        
    
    A10: 
    {l}
    c= B by 
    A8,
    ZFMISC_1: 31;
    
        
    {r}
    c= B by 
    A9,
    ZFMISC_1: 31;
    
        hence A1
    in ( 
    del  
    {B}) & A2
    in ( 
    del  
    {B}) by
    A1,
    A6,
    A7,
    A10,
    Th50;
    
        thus A1
    <> A2 by 
    A3,
    A6,
    A7,
    ZFMISC_1: 3;
    
        let A be
    set;
    
        assume
    
        
    
    A11: A 
    in ( 
    del  
    {B});
    
        then
    
        reconsider A as
    Cell of 
    0 , G; 
    
        
    
        
    
    A12: A 
    c= B by 
    A1,
    A11,
    Th50;
    
        consider x such that
    
        
    
    A13: A 
    = ( 
    cell (x,x)) and 
    
        
    
    A14: for i holds (x 
    . i) 
    in (G 
    . i) by 
    Th34;
    
        
    
        
    
    A15: x 
    in A by 
    A13,
    Th23;
    
        per cases by
    A3;
    
          suppose
    
          
    
    A16: (l 
    . i0) 
    < (r 
    . i0); 
    
          
    
    A17: 
    
          now
    
            let i;
    
            i
    = i0 or i 
    <> i0; 
    
            hence (l
    . i) 
    <= (r 
    . i) by 
    A5,
    A16;
    
          end;
    
          
    
          
    
    A18: (x 
    . i0) 
    in (G 
    . i0) by 
    A14;
    
          
    
          
    
    A19: (l 
    . i0) 
    <= (x 
    . i0) by 
    A2,
    A12,
    A15,
    A17,
    Th21;
    
          
    
          
    
    A20: (x 
    . i0) 
    <= (r 
    . i0) by 
    A2,
    A12,
    A15,
    A17,
    Th21;
    
          
    
          
    
    A21: not ((l 
    . i0) 
    < (x 
    . i0) & (x 
    . i0) 
    < (r 
    . i0)) by 
    A4,
    A18,
    Th13;
    
          
    
    A22: 
    
          now
    
            let i;
    
            assume i
    <> i0; 
    
            then
    
            
    
    A23: (l 
    . i) 
    = (r 
    . i) by 
    A5;
    
            
    
            
    
    A24: (l 
    . i) 
    <= (x 
    . i) by 
    A2,
    A12,
    A15,
    A17,
    Th21;
    
            (x
    . i) 
    <= (r 
    . i) by 
    A2,
    A12,
    A15,
    A17,
    Th21;
    
            hence (x
    . i) 
    = (l 
    . i) & (x 
    . i) 
    = (r 
    . i) by 
    A23,
    A24,
    XXREAL_0: 1;
    
          end;
    
          thus thesis
    
          proof
    
            per cases by
    A19,
    A20,
    A21,
    XXREAL_0: 1;
    
              suppose
    
              
    
    A25: (x 
    . i0) 
    = (l 
    . i0); 
    
              reconsider x, l as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              now
    
                let i;
    
                i
    = i0 or i 
    <> i0; 
    
                hence (x
    . i) 
    = (l 
    . i) by 
    A22,
    A25;
    
              end;
    
              then x
    = l by 
    FUNCT_2: 63;
    
              hence thesis by
    A13;
    
            end;
    
              suppose
    
              
    
    A26: (x 
    . i0) 
    = (r 
    . i0); 
    
              reconsider x, r as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              now
    
                let i;
    
                i
    = i0 or i 
    <> i0; 
    
                hence (x
    . i) 
    = (r 
    . i) by 
    A22,
    A26;
    
              end;
    
              then x
    = r by 
    FUNCT_2: 63;
    
              hence thesis by
    A13;
    
            end;
    
          end;
    
        end;
    
          suppose
    
          
    
    A27: d 
    = 1 & (r 
    . i0) 
    < (l 
    . i0); 
    
          
    
          
    
    A28: for i holds i 
    = i0 
    
          proof
    
            let i;
    
            
    
            
    
    A29: 1 
    <= i by 
    FINSEQ_1: 1;
    
            
    
            
    
    A30: i 
    <= d by 
    FINSEQ_1: 1;
    
            
    
            
    
    A31: 1 
    <= i0 by 
    FINSEQ_1: 1;
    
            
    
            
    
    A32: i0 
    <= d by 
    FINSEQ_1: 1;
    
            i
    = 1 by 
    A27,
    A29,
    A30,
    XXREAL_0: 1;
    
            hence thesis by
    A27,
    A31,
    A32,
    XXREAL_0: 1;
    
          end;
    
          consider i1 such that (r
    . i1) 
    < (l 
    . i1) and 
    
          
    
    A33: (x 
    . i1) 
    <= (r 
    . i1) or (l 
    . i1) 
    <= (x 
    . i1) by 
    A2,
    A12,
    A15,
    A27,
    Th22;
    
          
    
          
    
    A34: i1 
    = i0 by 
    A28;
    
          
    
          
    
    A35: (x 
    . i0) 
    in (G 
    . i0) by 
    A14;
    
          then
    
          
    
    A36: not (x 
    . i0) 
    < (r 
    . i0) by 
    A4,
    A27,
    Th13;
    
          
    
          
    
    A37: not (l 
    . i0) 
    < (x 
    . i0) by 
    A4,
    A27,
    A35,
    Th13;
    
          thus thesis
    
          proof
    
            per cases by
    A33,
    A34,
    A36,
    A37,
    XXREAL_0: 1;
    
              suppose
    
              
    
    A38: (x 
    . i0) 
    = (r 
    . i0); 
    
              reconsider x, r as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              now
    
                let i;
    
                i
    = i0 by 
    A28;
    
                hence (x
    . i) 
    = (r 
    . i) by 
    A38;
    
              end;
    
              then x
    = r by 
    FUNCT_2: 63;
    
              hence thesis by
    A13;
    
            end;
    
              suppose
    
              
    
    A39: (x 
    . i0) 
    = (l 
    . i0); 
    
              reconsider x, l as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
              now
    
                let i;
    
                i
    = i0 by 
    A28;
    
                hence (x
    . i) 
    = (l 
    . i) by 
    A39;
    
              end;
    
              then x
    = l by 
    FUNCT_2: 63;
    
              hence thesis by
    A13;
    
            end;
    
          end;
    
        end;
    
      end;
    
      hence thesis by
    Th5;
    
    end;
    
    theorem :: 
    
    CHAIN_1:56
    
    (
    Omega G) 
    = (( 
    0_ (d,G)) 
    ` ) & ( 
    0_ (d,G)) 
    = (( 
    Omega G) 
    ` ) 
    
    proof
    
      (
    Omega G) 
    = (( 
    0_ (d,G)) 
    ` ); 
    
      hence thesis;
    
    end;
    
    theorem :: 
    
    CHAIN_1:57
    
    for C be
    Chain of k, G holds (C 
    + ( 
    0_ (k,G))) 
    = C; 
    
    theorem :: 
    
    CHAIN_1:58
    
    
    
    
    
    Th55: for C be 
    Chain of d, G holds (C 
    ` ) 
    = (C 
    + ( 
    Omega G)) 
    
    proof
    
      let C be
    Chain of d, G; 
    
      (C
    /\ ( 
    cells (d,G))) 
    = C by 
    XBOOLE_1: 28;
    
      hence thesis by
    XBOOLE_1: 100;
    
    end;
    
    theorem :: 
    
    CHAIN_1:59
    
    
    
    
    
    Th56: ( 
    del ( 
    0_ ((k 
    + 1),G))) 
    = ( 
    0_ (k,G)) 
    
    proof
    
      now
    
        let A be
    Cell of k, G; 
    
        (
    card (( 
    star A) 
    /\ ( 
    0_ ((k 
    + 1),G)))) 
    = (2 
    *  
    0 ); 
    
        hence A
    in ( 
    del ( 
    0_ ((k 
    + 1),G))) iff A 
    in ( 
    0_ (k,G)) by 
    Th48;
    
      end;
    
      hence thesis by
    SUBSET_1: 3;
    
    end;
    
    theorem :: 
    
    CHAIN_1:60
    
    
    
    
    
    Th57: for G be 
    Grating of (d9 
    + 1) holds ( 
    del ( 
    Omega G)) 
    = ( 
    0_ (d9,G)) 
    
    proof
    
      let G be
    Grating of (d9 
    + 1); 
    
      now
    
        let A be
    Cell of d9, G; 
    
        ((
    star A) 
    /\ ( 
    Omega G)) 
    = ( 
    star A) by 
    XBOOLE_1: 28;
    
        then (
    card (( 
    star A) 
    /\ ( 
    Omega G))) 
    = (2 
    * 1) by 
    Th51;
    
        hence A
    in ( 
    del ( 
    Omega G)) iff A 
    in ( 
    0_ (d9,G)) by 
    Th48;
    
      end;
    
      hence thesis by
    SUBSET_1: 3;
    
    end;
    
    theorem :: 
    
    CHAIN_1:61
    
    
    
    
    
    Th58: for C1,C2 be 
    Chain of (k 
    + 1), G holds ( 
    del (C1 
    + C2)) 
    = (( 
    del C1) 
    + ( 
    del C2)) 
    
    proof
    
      let C1,C2 be
    Chain of (k 
    + 1), G; 
    
      now
    
        let A be
    Cell of k, G; 
    
        
    
        
    
    A1: (( 
    star A) 
    /\ (C1 
    \+\ C2)) 
    = ((( 
    star A) 
    /\ C1) 
    \+\ (( 
    star A) 
    /\ C2)) by 
    XBOOLE_1: 112;
    
        
    
        
    
    A2: A 
    in ( 
    del (C1 
    + C2)) iff (k 
    + 1) 
    <= d & ( 
    card (( 
    star A) 
    /\ (C1 
    \+\ C2))) is 
    odd by 
    Th48;
    
        
    
        
    
    A3: A 
    in ( 
    del C1) iff (k 
    + 1) 
    <= d & ( 
    card (( 
    star A) 
    /\ C1)) is 
    odd by 
    Th48;
    
        A
    in ( 
    del C2) iff (k 
    + 1) 
    <= d & ( 
    card (( 
    star A) 
    /\ C2)) is 
    odd by 
    Th48;
    
        hence A
    in ( 
    del (C1 
    + C2)) iff A 
    in (( 
    del C1) 
    + ( 
    del C2)) by 
    A1,
    A2,
    A3,
    Th7,
    XBOOLE_0: 1;
    
      end;
    
      hence thesis by
    SUBSET_1: 3;
    
    end;
    
    theorem :: 
    
    CHAIN_1:62
    
    
    
    
    
    Th59: for G be 
    Grating of (d9 
    + 1), C be 
    Chain of (d9 
    + 1), G holds ( 
    del (C 
    ` )) 
    = ( 
    del C) 
    
    proof
    
      let G be
    Grating of (d9 
    + 1), C be 
    Chain of (d9 
    + 1), G; 
    
      
    
      thus (
    del (C 
    ` )) 
    = ( 
    del (C 
    + ( 
    Omega G))) by 
    Th55
    
      .= ((
    del C) 
    + ( 
    del ( 
    Omega G))) by 
    Th58
    
      .= ((
    del C) 
    + ( 
    0_ (d9,G))) by 
    Th57
    
      .= (
    del C); 
    
    end;
    
    theorem :: 
    
    CHAIN_1:63
    
    
    
    
    
    Th60: for C be 
    Chain of ((k 
    + 1) 
    + 1), G holds ( 
    del ( 
    del C)) 
    = ( 
    0_ (k,G)) 
    
    proof
    
      let C be
    Chain of ((k 
    + 1) 
    + 1), G; 
    
      per cases ;
    
        suppose
    
        
    
    A1: ((k 
    + 1) 
    + 1) 
    <= d; 
    
        then
    
        
    
    A2: (k 
    + 1) 
    < d by 
    NAT_1: 13;
    
        then
    
        
    
    A3: k 
    < d by 
    NAT_1: 13;
    
        
    
        
    
    A4: for C be 
    Cell of ((k 
    + 1) 
    + 1), G, l, r st C 
    = ( 
    cell (l,r)) & for i holds (l 
    . i) 
    <= (r 
    . i) holds ( 
    del ( 
    del  
    {C}))
    = ( 
    0_ (k,G)) 
    
        proof
    
          let C be
    Cell of ((k 
    + 1) 
    + 1), G, l, r; 
    
          assume that
    
          
    
    A5: C 
    = ( 
    cell (l,r)) and 
    
          
    
    A6: for i holds (l 
    . i) 
    <= (r 
    . i); 
    
          now
    
            let A be
    object;
    
            assume
    
            
    
    A7: A 
    in ( 
    del ( 
    del  
    {C}));
    
            then
    
            reconsider A as
    Cell of k, G; 
    
            set BB = ((
    star A) 
    /\ ( 
    del  
    {C}));
    
            
    
    A8: 
    
            now
    
              let B be
    Cell of (k 
    + 1), G; 
    
              B
    in BB iff B 
    in ( 
    star A) & B 
    in ( 
    del  
    {C}) by
    XBOOLE_0:def 4;
    
              hence B
    in BB iff A 
    c= B & B 
    c= C by 
    A1,
    Th47,
    Th50;
    
            end;
    
            
    
            
    
    A9: ( 
    card BB) is 
    odd by 
    A7,
    Th48;
    
            consider B be
    object such that 
    
            
    
    A10: B 
    in BB by 
    A9,
    CARD_1: 27,
    XBOOLE_0:def 1;
    
            reconsider B as
    Cell of (k 
    + 1), G by 
    A10;
    
            
    
            
    
    A11: A 
    c= B by 
    A8,
    A10;
    
            B
    c= C by 
    A8,
    A10;
    
            then
    
            
    
    A12: A 
    c= C by 
    A11;
    
            set i0 = the
    Element of ( 
    Seg d); 
    
            (l
    . i0) 
    <= (r 
    . i0) by 
    A6;
    
            then
    
            consider Z be
    Subset of ( 
    Seg d) such that 
    
            
    
    A13: ( 
    card Z) 
    = ((k 
    + 1) 
    + 1) and 
    
            
    
    A14: for i holds i 
    in Z & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in Z & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i) by 
    A1,
    A5,
    Th30;
    
            consider l9, r9 such that
    
            
    
    A15: A 
    = ( 
    cell (l9,r9)) and 
    
            
    
    A16: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = k & for i holds (i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i))) or (k 
    = d & for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) by 
    A3,
    Th29;
    
            (l9
    . i0) 
    <= (r9 
    . i0) by 
    A5,
    A6,
    A12,
    A15,
    Th25;
    
            then
    
            consider X be
    Subset of ( 
    Seg d) such that 
    
            
    
    A17: ( 
    card X) 
    = k and 
    
            
    
    A18: for i holds i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i) by 
    A16;
    
            ex B1,B2 be
    set st B1 
    in BB & B2 
    in BB & B1 
    <> B2 & for B be 
    set st B 
    in BB holds B 
    = B1 or B 
    = B2 
    
            proof
    
              
    
              
    
    A19: X 
    c= Z by 
    A5,
    A12,
    A14,
    A15,
    A18,
    Th44;
    
              
    
              then (
    card (Z 
    \ X)) 
    = ((k 
    + (1 
    + 1)) 
    - k) by 
    A13,
    A17,
    CARD_2: 44
    
              .= 2;
    
              then
    
              consider i1,i2 be
    set such that 
    
              
    
    A20: i1 
    in (Z 
    \ X) and 
    
              
    
    A21: i2 
    in (Z 
    \ X) and 
    
              
    
    A22: i1 
    <> i2 and 
    
              
    
    A23: for i be 
    set st i 
    in (Z 
    \ X) holds i 
    = i1 or i 
    = i2 by 
    Th5;
    
              
    
              
    
    A24: i1 
    in Z by 
    A20,
    XBOOLE_0:def 5;
    
              
    
              
    
    A25: i2 
    in Z by 
    A21,
    XBOOLE_0:def 5;
    
              
    
              
    
    A26: not i1 
    in X by 
    A20,
    XBOOLE_0:def 5;
    
              
    
              
    
    A27: not i2 
    in X by 
    A21,
    XBOOLE_0:def 5;
    
              reconsider i1, i2 as
    Element of ( 
    Seg d) by 
    A20,
    A21;
    
              set Y1 = (X
    \/  
    {i1});
    
              
    
              
    
    A28: X 
    c= Y1 by 
    XBOOLE_1: 7;
    
              
    {i1}
    c= Z by 
    A24,
    ZFMISC_1: 31;
    
              then
    
              
    
    A29: Y1 
    c= Z by 
    A19,
    XBOOLE_1: 8;
    
              defpred
    
    S[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    in Y1 implies $2 
    = (l 
    . $1)) & ( not $1 
    in Y1 implies $2 
    = (l9 
    . $1)); 
    
              
    
              
    
    A30: for i holds ex xi be 
    Element of 
    REAL st 
    S[i, xi]
    
              proof
    
                let i;
    
                
    
                
    
    A31: (l 
    . i) 
    in  
    REAL & (l9 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
                i
    in Y1 or not i 
    in Y1; 
    
                hence thesis by
    A31;
    
              end;
    
              consider l1 be
    Function of ( 
    Seg d), 
    REAL such that 
    
              
    
    A32: for i holds 
    S[i, (l1
    . i)] from 
    FUNCT_2:sch 3(
    A30);
    
              defpred
    
    R[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    in Y1 implies $2 
    = (r 
    . $1)) & ( not $1 
    in Y1 implies $2 
    = (r9 
    . $1)); 
    
              
    
              
    
    A33: for i holds ex xi be 
    Element of 
    REAL st 
    R[i, xi]
    
              proof
    
                let i;
    
                
    
                
    
    A34: (r 
    . i) 
    in  
    REAL & (r9 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
                i
    in Y1 or not i 
    in Y1; 
    
                hence thesis by
    A34;
    
              end;
    
              consider r1 be
    Function of ( 
    Seg d), 
    REAL such that 
    
              
    
    A35: for i holds 
    R[i, (r1
    . i)] from 
    FUNCT_2:sch 3(
    A33);
    
              reconsider l1, r1 as
    Element of ( 
    REAL d) by 
    Def3;
    
              
    
              
    
    A36: for i holds (l1 
    . i) 
    <= (r1 
    . i) 
    
              proof
    
                let i;
    
                (l1
    . i) 
    = (l 
    . i) & (r1 
    . i) 
    = (r 
    . i) or (l1 
    . i) 
    = (l9 
    . i) & (r1 
    . i) 
    = (r9 
    . i) by 
    A32,
    A35;
    
                hence thesis by
    A14,
    A18;
    
              end;
    
              
    
              
    
    A37: ( 
    card Y1) 
    = (( 
    card X) 
    + ( 
    card  
    {i1})) by
    A26,
    CARD_2: 40,
    ZFMISC_1: 50
    
              .= (k
    + 1) by 
    A17,
    CARD_1: 30;
    
              for i holds i
    in Y1 & (l1 
    . i) 
    < (r1 
    . i) & 
    [(l1
    . i), (r1 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in Y1 & (l1 
    . i) 
    = (r1 
    . i) & (l1 
    . i) 
    in (G 
    . i) 
    
              proof
    
                let i;
    
                per cases ;
    
                  suppose
    
                  
    
    A38: i 
    in Y1; 
    
                  then
    
                  
    
    A39: (l1 
    . i) 
    = (l 
    . i) by 
    A32;
    
                  (r1
    . i) 
    = (r 
    . i) by 
    A35,
    A38;
    
                  hence thesis by
    A14,
    A29,
    A38,
    A39;
    
                end;
    
                  suppose
    
                  
    
    A40: not i 
    in Y1; 
    
                  then
    
                  
    
    A41: (l1 
    . i) 
    = (l9 
    . i) by 
    A32;
    
                  
    
                  
    
    A42: (r1 
    . i) 
    = (r9 
    . i) by 
    A35,
    A40;
    
                   not i
    in X by 
    A28,
    A40;
    
                  hence thesis by
    A18,
    A40,
    A41,
    A42;
    
                end;
    
              end;
    
              then
    
              reconsider B1 = (
    cell (l1,r1)) as 
    Cell of (k 
    + 1), G by 
    A2,
    A37,
    Th30;
    
              set Y2 = (X
    \/  
    {i2});
    
              
    
              
    
    A43: X 
    c= Y2 by 
    XBOOLE_1: 7;
    
              
    {i2}
    c= Z by 
    A25,
    ZFMISC_1: 31;
    
              then
    
              
    
    A44: Y2 
    c= Z by 
    A19,
    XBOOLE_1: 8;
    
              defpred
    
    P[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    in Y2 implies $2 
    = (l 
    . $1)) & ( not $1 
    in Y2 implies $2 
    = (l9 
    . $1)); 
    
              
    
              
    
    A45: for i holds ex xi be 
    Element of 
    REAL st 
    P[i, xi]
    
              proof
    
                let i;
    
                
    
                
    
    A46: (l 
    . i) 
    in  
    REAL & (l9 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
                i
    in Y2 or not i 
    in Y2; 
    
                hence thesis by
    A46;
    
              end;
    
              consider l2 be
    Function of ( 
    Seg d), 
    REAL such that 
    
              
    
    A47: for i holds 
    P[i, (l2
    . i)] from 
    FUNCT_2:sch 3(
    A45);
    
              defpred
    
    R[
    Element of ( 
    Seg d), 
    Element of 
    REAL ] means ($1 
    in Y2 implies $2 
    = (r 
    . $1)) & ( not $1 
    in Y2 implies $2 
    = (r9 
    . $1)); 
    
              
    
              
    
    A48: for i holds ex xi be 
    Element of 
    REAL st 
    R[i, xi]
    
              proof
    
                let i;
    
                
    
                
    
    A49: (r 
    . i) 
    in  
    REAL & (r9 
    . i) 
    in  
    REAL by 
    XREAL_0:def 1;
    
                i
    in Y2 or not i 
    in Y2; 
    
                hence thesis by
    A49;
    
              end;
    
              consider r2 be
    Function of ( 
    Seg d), 
    REAL such that 
    
              
    
    A50: for i holds 
    R[i, (r2
    . i)] from 
    FUNCT_2:sch 3(
    A48);
    
              reconsider l2, r2 as
    Element of ( 
    REAL d) by 
    Def3;
    
              
    
              
    
    A51: ( 
    card Y2) 
    = (( 
    card X) 
    + ( 
    card  
    {i2})) by
    A27,
    CARD_2: 40,
    ZFMISC_1: 50
    
              .= (k
    + 1) by 
    A17,
    CARD_1: 30;
    
              for i holds i
    in Y2 & (l2 
    . i) 
    < (r2 
    . i) & 
    [(l2
    . i), (r2 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in Y2 & (l2 
    . i) 
    = (r2 
    . i) & (l2 
    . i) 
    in (G 
    . i) 
    
              proof
    
                let i;
    
                per cases ;
    
                  suppose
    
                  
    
    A52: i 
    in Y2; 
    
                  then
    
                  
    
    A53: (l2 
    . i) 
    = (l 
    . i) by 
    A47;
    
                  (r2
    . i) 
    = (r 
    . i) by 
    A50,
    A52;
    
                  hence thesis by
    A14,
    A44,
    A52,
    A53;
    
                end;
    
                  suppose
    
                  
    
    A54: not i 
    in Y2; 
    
                  then
    
                  
    
    A55: (l2 
    . i) 
    = (l9 
    . i) by 
    A47;
    
                  
    
                  
    
    A56: (r2 
    . i) 
    = (r9 
    . i) by 
    A50,
    A54;
    
                   not i
    in X by 
    A43,
    A54;
    
                  hence thesis by
    A18,
    A54,
    A55,
    A56;
    
                end;
    
              end;
    
              then
    
              reconsider B2 = (
    cell (l2,r2)) as 
    Cell of (k 
    + 1), G by 
    A2,
    A51,
    Th30;
    
              take B1, B2;
    
              
    
              
    
    A57: for i holds (l1 
    . i) 
    <= (l9 
    . i) & (l9 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    <= (r1 
    . i) & (l 
    . i) 
    <= (l1 
    . i) & (l1 
    . i) 
    <= (r1 
    . i) & (r1 
    . i) 
    <= (r 
    . i) 
    
              proof
    
                let i;
    
                per cases ;
    
                  suppose
    
                  
    
    A58: i 
    in Y1; 
    
                  then
    
                  
    
    A59: (l1 
    . i) 
    = (l 
    . i) by 
    A32;
    
                  (r1
    . i) 
    = (r 
    . i) by 
    A35,
    A58;
    
                  hence thesis by
    A5,
    A6,
    A12,
    A15,
    A59,
    Th25;
    
                end;
    
                  suppose
    
                  
    
    A60: not i 
    in Y1; 
    
                  then
    
                  
    
    A61: (l1 
    . i) 
    = (l9 
    . i) by 
    A32;
    
                  (r1
    . i) 
    = (r9 
    . i) by 
    A35,
    A60;
    
                  hence thesis by
    A5,
    A6,
    A12,
    A15,
    A61,
    Th25;
    
                end;
    
              end;
    
              then
    
              
    
    A62: A 
    c= B1 by 
    A15,
    Th25;
    
              B1
    c= C by 
    A5,
    A6,
    A57,
    Th25;
    
              hence B1
    in BB by 
    A8,
    A62;
    
              
    
              
    
    A63: for i holds (l2 
    . i) 
    <= (l9 
    . i) & (l9 
    . i) 
    <= (r9 
    . i) & (r9 
    . i) 
    <= (r2 
    . i) & (l 
    . i) 
    <= (l2 
    . i) & (l2 
    . i) 
    <= (r2 
    . i) & (r2 
    . i) 
    <= (r 
    . i) 
    
              proof
    
                let i;
    
                per cases ;
    
                  suppose
    
                  
    
    A64: i 
    in Y2; 
    
                  then
    
                  
    
    A65: (l2 
    . i) 
    = (l 
    . i) by 
    A47;
    
                  (r2
    . i) 
    = (r 
    . i) by 
    A50,
    A64;
    
                  hence thesis by
    A5,
    A6,
    A12,
    A15,
    A65,
    Th25;
    
                end;
    
                  suppose
    
                  
    
    A66: not i 
    in Y2; 
    
                  then
    
                  
    
    A67: (l2 
    . i) 
    = (l9 
    . i) by 
    A47;
    
                  (r2
    . i) 
    = (r9 
    . i) by 
    A50,
    A66;
    
                  hence thesis by
    A5,
    A6,
    A12,
    A15,
    A67,
    Th25;
    
                end;
    
              end;
    
              then
    
              
    
    A68: A 
    c= B2 by 
    A15,
    Th25;
    
              B2
    c= C by 
    A5,
    A6,
    A63,
    Th25;
    
              hence B2
    in BB by 
    A8,
    A68;
    
              i1
    in  
    {i1} by
    TARSKI:def 1;
    
              then
    
              
    
    A69: i1 
    in Y1 by 
    XBOOLE_0:def 3;
    
              
    
              
    
    A70: not i1 
    in X by 
    A20,
    XBOOLE_0:def 5;
    
               not i1
    in  
    {i2} by
    A22,
    TARSKI:def 1;
    
              then
    
              
    
    A71: not i1 
    in Y2 by 
    A70,
    XBOOLE_0:def 3;
    
              
    
              
    
    A72: (l1 
    . i1) 
    = (l 
    . i1) by 
    A32,
    A69;
    
              
    
              
    
    A73: (r1 
    . i1) 
    = (r 
    . i1) by 
    A35,
    A69;
    
              
    
              
    
    A74: (l2 
    . i1) 
    = (l9 
    . i1) by 
    A47,
    A71;
    
              
    
              
    
    A75: (r2 
    . i1) 
    = (r9 
    . i1) by 
    A50,
    A71;
    
              (l
    . i1) 
    < (r 
    . i1) by 
    A14,
    A24;
    
              then l1
    <> l2 or r1 
    <> r2 by 
    A18,
    A26,
    A72,
    A73,
    A74,
    A75;
    
              hence B1
    <> B2 by 
    A36,
    Th28;
    
              let B be
    set;
    
              assume
    
              
    
    A76: B 
    in BB; 
    
              then
    
              reconsider B as
    Cell of (k 
    + 1), G; 
    
              
    
              
    
    A77: A 
    c= B by 
    A8,
    A76;
    
              
    
              
    
    A78: B 
    c= C by 
    A8,
    A76;
    
              consider l99, r99 such that
    
              
    
    A79: B 
    = ( 
    cell (l99,r99)) and 
    
              
    
    A80: (ex Y be 
    Subset of ( 
    Seg d) st ( 
    card Y) 
    = (k 
    + 1) & for i holds (i 
    in Y & (l99 
    . i) 
    < (r99 
    . i) & 
    [(l99
    . i), (r99 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in Y & (l99 
    . i) 
    = (r99 
    . i) & (l99 
    . i) 
    in (G 
    . i))) or ((k 
    + 1) 
    = d & for i holds (r99 
    . i) 
    < (l99 
    . i) & 
    [(l99
    . i), (r99 
    . i)] is 
    Gap of (G 
    . i)) by 
    A2,
    Th29;
    
              (l99
    . i0) 
    <= (r99 
    . i0) by 
    A5,
    A6,
    A78,
    A79,
    Th25;
    
              then
    
              consider Y be
    Subset of ( 
    Seg d) such that 
    
              
    
    A81: ( 
    card Y) 
    = (k 
    + 1) and 
    
              
    
    A82: for i holds i 
    in Y & (l99 
    . i) 
    < (r99 
    . i) & 
    [(l99
    . i), (r99 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in Y & (l99 
    . i) 
    = (r99 
    . i) & (l99 
    . i) 
    in (G 
    . i) by 
    A80;
    
              
    
              
    
    A83: X 
    c= Y by 
    A15,
    A18,
    A77,
    A79,
    A82,
    Th44;
    
              
    
              
    
    A84: Y 
    c= Z by 
    A5,
    A14,
    A78,
    A79,
    A82,
    Th44;
    
              (
    card (Y 
    \ X)) 
    = ((k 
    + 1) 
    - k) by 
    A17,
    A81,
    A83,
    CARD_2: 44
    
              .= 1;
    
              then
    
              consider i9 be
    object such that 
    
              
    
    A85: (Y 
    \ X) 
    =  
    {i9} by
    CARD_2: 42;
    
              
    
              
    
    A86: i9 
    in (Y 
    \ X) by 
    A85,
    TARSKI:def 1;
    
              then
    
              reconsider i9 as
    Element of ( 
    Seg d); 
    
              
    
              
    
    A87: i9 
    in Y by 
    A86,
    XBOOLE_0:def 5;
    
               not i9
    in X by 
    A86,
    XBOOLE_0:def 5;
    
              then
    
              
    
    A88: i9 
    in (Z 
    \ X) by 
    A84,
    A87,
    XBOOLE_0:def 5;
    
              
    
              
    
    A89: Y 
    = (X 
    \/ Y) by 
    A83,
    XBOOLE_1: 12
    
              .= (X
    \/  
    {i9}) by
    A85,
    XBOOLE_1: 39;
    
              per cases by
    A23,
    A88,
    A89;
    
                suppose
    
                
    
    A90: Y 
    = Y1; 
    
                reconsider l99, r99, l1, r1 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
                
    
    A91: 
    
                now
    
                  let i;
    
                  i
    in Y or not i 
    in Y; 
    
                  then (l99
    . i) 
    = (l 
    . i) & (l1 
    . i) 
    = (l 
    . i) & (r99 
    . i) 
    = (r 
    . i) & (r1 
    . i) 
    = (r 
    . i) or (l99 
    . i) 
    = (l9 
    . i) & (l1 
    . i) 
    = (l9 
    . i) & (r99 
    . i) 
    = (r9 
    . i) & (r1 
    . i) 
    = (r9 
    . i) by 
    A5,
    A14,
    A15,
    A18,
    A32,
    A35,
    A77,
    A78,
    A79,
    A82,
    A90,
    Th44;
    
                  hence (l99
    . i) 
    = (l1 
    . i) & (r99 
    . i) 
    = (r1 
    . i); 
    
                end;
    
                then l99
    = l1 by 
    FUNCT_2: 63;
    
                hence thesis by
    A79,
    A91,
    FUNCT_2: 63;
    
              end;
    
                suppose
    
                
    
    A92: Y 
    = Y2; 
    
                reconsider l99, r99, l2, r2 as
    Function of ( 
    Seg d), 
    REAL by 
    Def3;
    
                
    
    A93: 
    
                now
    
                  let i;
    
                  i
    in Y or not i 
    in Y; 
    
                  then (l99
    . i) 
    = (l 
    . i) & (l2 
    . i) 
    = (l 
    . i) & (r99 
    . i) 
    = (r 
    . i) & (r2 
    . i) 
    = (r 
    . i) or (l99 
    . i) 
    = (l9 
    . i) & (l2 
    . i) 
    = (l9 
    . i) & (r99 
    . i) 
    = (r9 
    . i) & (r2 
    . i) 
    = (r9 
    . i) by 
    A5,
    A14,
    A15,
    A18,
    A47,
    A50,
    A77,
    A78,
    A79,
    A82,
    A92,
    Th44;
    
                  hence (l99
    . i) 
    = (l2 
    . i) & (r99 
    . i) 
    = (r2 
    . i); 
    
                end;
    
                then l99
    = l2 by 
    FUNCT_2: 63;
    
                hence thesis by
    A79,
    A93,
    FUNCT_2: 63;
    
              end;
    
            end;
    
            then (
    card BB) 
    = (2 
    * 1) by 
    Th5;
    
            hence contradiction by
    A7,
    Th48;
    
          end;
    
          hence thesis by
    XBOOLE_0:def 1;
    
        end;
    
        
    
        
    
    A94: for C1,C2 be 
    Chain of ((k 
    + 1) 
    + 1), G st ( 
    del ( 
    del C1)) 
    = ( 
    0_ (k,G)) & ( 
    del ( 
    del C2)) 
    = ( 
    0_ (k,G)) holds ( 
    del ( 
    del (C1 
    + C2))) 
    = ( 
    0_ (k,G)) 
    
        proof
    
          let C1,C2 be
    Chain of ((k 
    + 1) 
    + 1), G; 
    
          assume that
    
          
    
    A95: ( 
    del ( 
    del C1)) 
    = ( 
    0_ (k,G)) and 
    
          
    
    A96: ( 
    del ( 
    del C2)) 
    = ( 
    0_ (k,G)); 
    
          
    
          thus (
    del ( 
    del (C1 
    + C2))) 
    = ( 
    del (( 
    del C1) 
    + ( 
    del C2))) by 
    Th58
    
          .= ((
    0_ (k,G)) 
    + ( 
    0_ (k,G))) by 
    A95,
    A96,
    Th58
    
          .= (
    0_ (k,G)); 
    
        end;
    
        defpred
    
    P[
    Chain of ((k 
    + 1) 
    + 1), G] means ( 
    del ( 
    del $1)) 
    = ( 
    0_ (k,G)); 
    
        (
    del ( 
    del ( 
    0_ (((k 
    + 1) 
    + 1),G)))) 
    = ( 
    del ( 
    0_ ((k 
    + 1),G))) by 
    Th56
    
        .= (
    0_ (k,G)) by 
    Th56;
    
        then
    
        
    
    A97: 
    P[(
    0_ (((k 
    + 1) 
    + 1),G))]; 
    
        for A be
    Cell of ((k 
    + 1) 
    + 1), G holds ( 
    del ( 
    del  
    {A}))
    = ( 
    0_ (k,G)) 
    
        proof
    
          let A be
    Cell of ((k 
    + 1) 
    + 1), G; 
    
          consider l, r such that
    
          
    
    A98: A 
    = ( 
    cell (l,r)) and 
    
          
    
    A99: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = ((k 
    + 1) 
    + 1) & for i holds (i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i))) or (((k 
    + 1) 
    + 1) 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i)) by 
    A1,
    Th29;
    
          per cases by
    A99;
    
            suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = ((k 
    + 1) 
    + 1) & for i holds i 
    in X & (l 
    . i) 
    < (r 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l 
    . i) 
    = (r 
    . i) & (l 
    . i) 
    in (G 
    . i); 
    
            then for i holds (l
    . i) 
    <= (r 
    . i); 
    
            hence thesis by
    A4,
    A98;
    
          end;
    
            suppose
    
            
    
    A100: ((k 
    + 1) 
    + 1) 
    = d & for i holds (r 
    . i) 
    < (l 
    . i) & 
    [(l
    . i), (r 
    . i)] is 
    Gap of (G 
    . i); 
    
            then
    
            
    
    A101: A 
    = ( 
    infinite-cell G) by 
    A98,
    Th46;
    
            set C = (
    {A}
    ` ); 
    
            
    
            
    
    A102: for A be 
    Cell of ((k 
    + 1) 
    + 1), G st A 
    in C holds 
    P[
    {A}]
    
            proof
    
              let A9 be
    Cell of ((k 
    + 1) 
    + 1), G; 
    
              assume A9
    in C; 
    
              then not A9
    in  
    {A} by
    XBOOLE_0:def 5;
    
              then
    
              
    
    A103: A9 
    <> ( 
    infinite-cell G) by 
    A101,
    TARSKI:def 1;
    
              consider l9, r9 such that
    
              
    
    A104: A9 
    = ( 
    cell (l9,r9)) and 
    
              
    
    A105: (ex X be 
    Subset of ( 
    Seg d) st ( 
    card X) 
    = ((k 
    + 1) 
    + 1) & for i holds (i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) or ( not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i))) or (((k 
    + 1) 
    + 1) 
    = d & for i holds (r9 
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i)) by 
    A1,
    Th29;
    
              per cases by
    A105;
    
                suppose ex X be
    Subset of ( 
    Seg d) st ( 
    card X) 
    = ((k 
    + 1) 
    + 1) & for i holds i 
    in X & (l9 
    . i) 
    < (r9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i) or not i 
    in X & (l9 
    . i) 
    = (r9 
    . i) & (l9 
    . i) 
    in (G 
    . i); 
    
                then for i holds (l9
    . i) 
    <= (r9 
    . i); 
    
                hence thesis by
    A4,
    A104;
    
              end;
    
                suppose for i holds (r9
    . i) 
    < (l9 
    . i) & 
    [(l9
    . i), (r9 
    . i)] is 
    Gap of (G 
    . i); 
    
                hence thesis by
    A103,
    A104,
    Th46;
    
              end;
    
            end;
    
            
    
            
    
    A106: for C1,C2 be 
    Chain of ((k 
    + 1) 
    + 1), G st C1 
    c= C & C2 
    c= C & 
    P[C1] &
    P[C2] holds
    P[(C1
    + C2)] by 
    A94;
    
            
    P[C] from
    ChainInd(
    A97,
    A102,
    A106);
    
            hence thesis by
    A100,
    Th59;
    
          end;
    
        end;
    
        then
    
        
    
    A107: for A be 
    Cell of ((k 
    + 1) 
    + 1), G st A 
    in C holds 
    P[
    {A}];
    
        
    
        
    
    A108: for C1,C2 be 
    Chain of ((k 
    + 1) 
    + 1), G st C1 
    c= C & C2 
    c= C & 
    P[C1] &
    P[C2] holds
    P[(C1
    + C2)] by 
    A94;
    
        thus
    P[C] from
    ChainInd(
    A97,
    A107,
    A108);
    
      end;
    
        suppose ((k
    + 1) 
    + 1) 
    > d; 
    
        then (
    del C) 
    = ( 
    0_ ((k 
    + 1),G)) by 
    Th49;
    
        hence thesis by
    Th56;
    
      end;
    
    end;
    
    definition
    
      let d, G, k;
    
      :: 
    
    CHAIN_1:def14
    
      mode
    
    Cycle of k,G -> 
    Chain of k, G means 
    
      :
    
    Def14: k 
    =  
    0 & ( 
    card it ) is 
    even or ex k9 st k 
    = (k9 
    + 1) & ex C be 
    Chain of (k9 
    + 1), G st C 
    = it & ( 
    del C) 
    = ( 
    0_ (k9,G)); 
    
      existence
    
      proof
    
        per cases by
    NAT_1: 6;
    
          suppose
    
          
    
    A1: k 
    =  
    0 ; 
    
          take (
    0_ (k,G)); 
    
          thus thesis by
    A1;
    
        end;
    
          suppose ex k9 be
    Nat st k 
    = (k9 
    + 1); 
    
          then
    
          consider k9 be
    Nat such that 
    
          
    
    A2: k 
    = (k9 
    + 1); 
    
          reconsider k9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
          take C9 = (
    0_ (k,G)); 
    
          now
    
            take k9;
    
            thus k
    = (k9 
    + 1) by 
    A2;
    
            reconsider C = C9 as
    Chain of (k9 
    + 1), G by 
    A2;
    
            take C;
    
            thus C
    = C9 & ( 
    del C) 
    = ( 
    0_ (k9,G)) by 
    A2,
    Th56;
    
          end;
    
          hence thesis;
    
        end;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:64
    
    
    
    
    
    Th61: for C be 
    Chain of (k 
    + 1), G holds C is 
    Cycle of (k 
    + 1), G iff ( 
    del C) 
    = ( 
    0_ (k,G)) 
    
    proof
    
      let C be
    Chain of (k 
    + 1), G; 
    
      hereby
    
        assume C is
    Cycle of (k 
    + 1), G; 
    
        then ex k9 st ((k
    + 1) 
    = (k9 
    + 1)) & (ex C9 be 
    Chain of (k9 
    + 1), G st C9 
    = C & ( 
    del C9) 
    = ( 
    0_ (k9,G))) by 
    Def14;
    
        hence (
    del C) 
    = ( 
    0_ (k,G)); 
    
      end;
    
      thus thesis by
    Def14;
    
    end;
    
    theorem :: 
    
    CHAIN_1:65
    
    k
    > d implies for C be 
    Chain of k, G holds C is 
    Cycle of k, G 
    
    proof
    
      assume
    
      
    
    A1: k 
    > d; 
    
      let C be
    Chain of k, G; 
    
      consider k9 be
    Nat such that 
    
      
    
    A2: k 
    = (k9 
    + 1) by 
    A1,
    NAT_1: 6;
    
      reconsider k9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
      reconsider C9 = C as
    Chain of (k9 
    + 1), G by 
    A2;
    
      (
    del C9) 
    = ( 
    0_ (k9,G)) by 
    A1,
    A2,
    Th49;
    
      hence thesis by
    A2,
    Def14;
    
    end;
    
    theorem :: 
    
    CHAIN_1:66
    
    
    
    
    
    Th63: for C be 
    Chain of 
    0 , G holds C is 
    Cycle of 
    0 , G iff ( 
    card C) is 
    even
    
    proof
    
      let C be
    Chain of 
    0 , G; 
    
      hereby
    
        assume C is
    Cycle of 
    0 , G; 
    
        then
    0  
    =  
    0 & ( 
    card C) is 
    even or ex k9 st 
    0  
    = (k9 
    + 1) & ex C9 be 
    Chain of (k9 
    + 1), G st C9 
    = C & ( 
    del C9) 
    = ( 
    0_ (k9,G)) by 
    Def14;
    
        hence (
    card C) is 
    even;
    
      end;
    
      thus thesis by
    Def14;
    
    end;
    
    definition
    
      let d, G, k;
    
      let C be
    Cycle of (k 
    + 1), G; 
    
      :: original:
    del
    
      redefine
    
      :: 
    
    CHAIN_1:def15
    
      func
    
    del C equals ( 
    0_ (k,G)); 
    
      compatibility by
    Th61;
    
    end
    
    definition
    
      let d, G, k;
    
      :: original:
    0_
    
      redefine
    
      func
    
    0_ (k,G) -> 
    Cycle of k, G ; 
    
      coherence
    
      proof
    
        per cases by
    NAT_1: 6;
    
          suppose
    
          
    
    A1: k 
    =  
    0 ; 
    
          (
    card  
    {} ) 
    = (2 
    *  
    0 ); 
    
          hence thesis by
    A1,
    Def14;
    
        end;
    
          suppose ex k9 be
    Nat st k 
    = (k9 
    + 1); 
    
          then
    
          consider k9 be
    Nat such that 
    
          
    
    A2: k 
    = (k9 
    + 1); 
    
          reconsider k9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
          (
    del ( 
    0_ ((k9 
    + 1),G))) 
    = ( 
    0_ (k9,G)) by 
    Th56;
    
          hence thesis by
    A2,
    Def14;
    
        end;
    
      end;
    
    end
    
    definition
    
      let d, G;
    
      :: original:
    Omega
    
      redefine
    
      func
    
    Omega (G) -> 
    Cycle of d, G ; 
    
      coherence
    
      proof
    
        consider d9 be
    Nat such that 
    
        
    
    A1: d 
    = (d9 
    + 1) by 
    NAT_1: 6;
    
        reconsider d9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
        reconsider G as
    Grating of (d9 
    + 1) by 
    A1;
    
        (
    del ( 
    Omega G)) 
    = ( 
    0_ (d9,G)) by 
    Th57;
    
        hence thesis by
    A1,
    Def14;
    
      end;
    
    end
    
    definition
    
      let d, G, k;
    
      let C1,C2 be
    Cycle of k, G; 
    
      :: original:
    +
    
      redefine
    
      func C1
    
    + C2 -> 
    Cycle of k, G ; 
    
      coherence
    
      proof
    
        per cases by
    NAT_1: 6;
    
          suppose
    
          
    
    A1: k 
    =  
    0 ; 
    
          then
    
          
    
    A2: ( 
    card C1) is 
    even by 
    Th63;
    
          (
    card C2) is 
    even by 
    A1,
    Th63;
    
          then (
    card (C1 
    + C2)) is 
    even by 
    A2,
    Th7;
    
          hence (C1
    + C2) is 
    Cycle of k, G by 
    A1,
    Th63;
    
        end;
    
          suppose ex k9 be
    Nat st k 
    = (k9 
    + 1); 
    
          then
    
          consider k9 be
    Nat such that 
    
          
    
    A3: k 
    = (k9 
    + 1); 
    
          reconsider k9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
          reconsider C1, C2 as
    Cycle of (k9 
    + 1), G by 
    A3;
    
          
    
          
    
    A4: ( 
    del C1) 
    = ( 
    0_ (k9,G)); 
    
          (
    del C2) 
    = ( 
    0_ (k9,G)); 
    
          
    
          then (
    del (C1 
    + C2)) 
    = (( 
    0_ (k9,G)) 
    + ( 
    0_ (k9,G))) by 
    A4,
    Th58
    
          .= (
    0_ (k9,G)); 
    
          hence thesis by
    A3,
    Th61;
    
        end;
    
      end;
    
    end
    
    theorem :: 
    
    CHAIN_1:67
    
    for C be
    Cycle of d, G holds (C 
    ` ) is 
    Cycle of d, G 
    
    proof
    
      let C be
    Cycle of d, G; 
    
      consider d9 be
    Nat such that 
    
      
    
    A1: d 
    = (d9 
    + 1) by 
    NAT_1: 6;
    
      reconsider d9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
      reconsider G as
    Grating of (d9 
    + 1) by 
    A1;
    
      reconsider C as
    Cycle of (d9 
    + 1), G by 
    A1;
    
      (
    del C) 
    = ( 
    0_ (d9,G)); 
    
      then (
    del (C 
    ` )) 
    = ( 
    0_ (d9,G)) by 
    Th59;
    
      hence thesis by
    A1,
    Th61;
    
    end;
    
    definition
    
      let d, G, k;
    
      let C be
    Chain of (k 
    + 1), G; 
    
      :: original:
    del
    
      redefine
    
      func
    
    del C -> 
    Cycle of k, G ; 
    
      coherence
    
      proof
    
        per cases by
    NAT_1: 6;
    
          suppose
    
          
    
    A1: k 
    =  
    0 ; 
    
          defpred
    
    P[
    Chain of (k 
    + 1), G] means ( 
    del $1) is 
    Cycle of k, G; 
    
          (
    del ( 
    0_ ((k 
    + 1),G))) 
    = ( 
    0_ (k,G)); 
    
          then
    
          
    
    A2: 
    P[(
    0_ ((k 
    + 1),G))]; 
    
          now
    
            let B be
    Cell of ( 
    0  
    + 1), G; 
    
            assume B
    in C; 
    
            (
    card ( 
    del  
    {B}))
    = (2 
    * 1) by 
    Th52;
    
            hence (
    del  
    {B}) is
    Cycle of 
    0 , G by 
    Def14;
    
          end;
    
          then
    
          
    
    A3: for A be 
    Cell of (k 
    + 1), G st A 
    in C holds 
    P[
    {A}] by
    A1;
    
          
    
          
    
    A4: for C1,C2 be 
    Chain of (k 
    + 1), G st C1 
    c= C & C2 
    c= C & 
    P[C1] &
    P[C2] holds
    P[(C1
    + C2)] 
    
          proof
    
            let C1,C2 be
    Chain of (k 
    + 1), G; 
    
            assume that C1
    c= C and C2 
    c= C and 
    
            
    
    A5: 
    P[C1] and
    
            
    
    A6: 
    P[C2];
    
            reconsider C19 = (
    del C1), C29 = ( 
    del C2) as 
    Cycle of k, G by 
    A5,
    A6;
    
            (
    del (C1 
    + C2)) 
    = (C19 
    + C29) by 
    Th58;
    
            hence thesis;
    
          end;
    
          thus
    P[C] from
    ChainInd(
    A2,
    A3,
    A4);
    
        end;
    
          suppose ex k9 be
    Nat st k 
    = (k9 
    + 1); 
    
          then
    
          consider k9 be
    Nat such that 
    
          
    
    A7: k 
    = (k9 
    + 1); 
    
          reconsider k9 as
    Element of 
    NAT by 
    ORDINAL1:def 12;
    
          reconsider C as
    Chain of ((k9 
    + 1) 
    + 1), G by 
    A7;
    
          (
    del ( 
    del C)) 
    = ( 
    0_ (k9,G)) by 
    Th60;
    
          hence thesis by
    A7,
    Th61;
    
        end;
    
      end;
    
    end
    
    begin
    
    definition
    
      let d, G, k;
    
      :: 
    
    CHAIN_1:def16
    
      func
    
    Chains (k,G) -> 
    strict  
    AbGroup means 
    
      :
    
    Def16: the 
    carrier of it 
    = ( 
    bool ( 
    cells (k,G))) & ( 
    0. it ) 
    = ( 
    0_ (k,G)) & for A,B be 
    Element of it , A9,B9 be 
    Chain of k, G st A 
    = A9 & B 
    = B9 holds (A 
    + B) 
    = (A9 
    + B9); 
    
      existence
    
      proof
    
        deffunc
    
    f(
    Chain of k, G, 
    Chain of k, G) = ($1 
    + $2); 
    
        consider op be
    BinOp of ( 
    bool ( 
    cells (k,G))) such that 
    
        
    
    A1: for A,B be 
    Chain of k, G holds (op 
    . (A,B)) 
    =  
    f(A,B) from
    BINOP_1:sch 4;
    
        set ch =
    addLoopStr (# ( 
    bool ( 
    cells (k,G))), op, ( 
    0_ (k,G)) #); 
    
        
    
        
    
    A2: ch is 
    add-associative
    
        proof
    
          let A,B,C be
    Element of ch; 
    
          reconsider A9 = A, B9 = B, C9 = C as
    Chain of k, G; 
    
          
    
          thus ((A
    + B) 
    + C) 
    = (op 
    . ((A9 
    + B9),C)) by 
    A1
    
          .= ((A9
    + B9) 
    + C9) by 
    A1
    
          .= (A9
    + (B9 
    + C9)) by 
    XBOOLE_1: 91
    
          .= (op
    . (A,(B9 
    + C9))) by 
    A1
    
          .= (A
    + (B 
    + C)) by 
    A1;
    
        end;
    
        
    
        
    
    A3: ch is 
    right_zeroed
    
        proof
    
          let A be
    Element of ch; 
    
          reconsider A9 = A as
    Chain of k, G; 
    
          
    
          thus (A
    + ( 
    0. ch)) 
    = (A9 
    + ( 
    0_ (k,G))) by 
    A1
    
          .= A;
    
        end;
    
        
    
        
    
    A4: ch is 
    right_complementable
    
        proof
    
          let A be
    Element of ch; 
    
          reconsider A9 = A as
    Chain of k, G; 
    
          take A;
    
          
    
          thus (A
    + A) 
    = (A9 
    + A9) by 
    A1
    
          .= (
    0. ch) by 
    XBOOLE_1: 92;
    
        end;
    
        ch is
    Abelian
    
        proof
    
          let A,B be
    Element of ch; 
    
          reconsider A9 = A, B9 = B as
    Chain of k, G; 
    
          
    
          thus (A
    + B) 
    = (A9 
    + B9) by 
    A1
    
          .= (B
    + A) by 
    A1;
    
        end;
    
        then
    
        reconsider ch as
    strict  
    AbGroup by 
    A2,
    A3,
    A4;
    
        take ch;
    
        thus the
    carrier of ch 
    = ( 
    bool ( 
    cells (k,G))); 
    
        thus (
    0. ch) 
    = ( 
    0_ (k,G)); 
    
        let A,B be
    Element of ch, A9,B9 be 
    Chain of k, G; 
    
        assume that
    
        
    
    A5: A 
    = A9 and 
    
        
    
    A6: B 
    = B9; 
    
        thus thesis by
    A1,
    A5,
    A6;
    
      end;
    
      uniqueness
    
      proof
    
        let C1,C2 be
    strict  
    AbGroup;
    
        assume that
    
        
    
    A7: the 
    carrier of C1 
    = ( 
    bool ( 
    cells (k,G))) and 
    
        
    
    A8: ( 
    0. C1) 
    = ( 
    0_ (k,G)) and 
    
        
    
    A9: for A,B be 
    Element of C1, A9,B9 be 
    Chain of k, G st A 
    = A9 & B 
    = B9 holds (A 
    + B) 
    = (A9 
    + B9) and 
    
        
    
    A10: the 
    carrier of C2 
    = ( 
    bool ( 
    cells (k,G))) and 
    
        
    
    A11: ( 
    0. C2) 
    = ( 
    0_ (k,G)) and 
    
        
    
    A12: for A,B be 
    Element of C2, A9,B9 be 
    Chain of k, G st A 
    = A9 & B 
    = B9 holds (A 
    + B) 
    = (A9 
    + B9); 
    
        set X =
    [:(
    bool ( 
    cells (k,G))), ( 
    bool ( 
    cells (k,G))):]; 
    
        reconsider op1 = the
    addF of C1, op2 = the 
    addF of C2 as 
    Function of X, ( 
    bool ( 
    cells (k,G))) by 
    A7,
    A10;
    
        now
    
          let AB be
    Element of X; 
    
          consider A9,B9 be
    Chain of k, G such that 
    
          
    
    A13: AB 
    =  
    [A9, B9] by
    DOMAIN_1: 1;
    
          reconsider A1 = A9, B1 = B9 as
    Element of C1 by 
    A7;
    
          reconsider A2 = A9, B2 = B9 as
    Element of C2 by 
    A10;
    
          
    
          thus (op1
    . AB) 
    = (A1 
    + B1) by 
    A13
    
          .= (A9
    + B9) by 
    A9
    
          .= (A2
    + B2) by 
    A12
    
          .= (op2
    . AB) by 
    A13;
    
        end;
    
        hence thesis by
    A7,
    A8,
    A10,
    A11,
    FUNCT_2: 63;
    
      end;
    
    end
    
    definition
    
      let d, G, k;
    
      mode
    
    GrChain of k,G is 
    Element of ( 
    Chains (k,G)); 
    
    end
    
    theorem :: 
    
    CHAIN_1:68
    
    for x be
    set holds x is 
    Chain of k, G iff x is 
    GrChain of k, G by 
    Def16;
    
    definition
    
      let d, G, k;
    
      :: 
    
    CHAIN_1:def17
    
      func
    
    del (k,G) -> 
    Homomorphism of ( 
    Chains ((k 
    + 1),G)), ( 
    Chains (k,G)) means for A be 
    Element of ( 
    Chains ((k 
    + 1),G)), A9 be 
    Chain of (k 
    + 1), G st A 
    = A9 holds (it 
    . A) 
    = ( 
    del A9); 
    
      existence
    
      proof
    
        deffunc
    
    f(
    Subset of ( 
    cells ((k 
    + 1),G))) = ( 
    del $1); 
    
        consider f be
    Function of ( 
    bool ( 
    cells ((k 
    + 1),G))), ( 
    bool ( 
    cells (k,G))) such that 
    
        
    
    A1: for A be 
    Subset of ( 
    cells ((k 
    + 1),G)) holds (f 
    . A) 
    =  
    f(A) from
    FUNCT_2:sch 4;
    
        
    
        
    
    A2: the 
    carrier of ( 
    Chains ((k 
    + 1),G)) 
    = ( 
    bool ( 
    cells ((k 
    + 1),G))) by 
    Def16;
    
        the
    carrier of ( 
    Chains (k,G)) 
    = ( 
    bool ( 
    cells (k,G))) by 
    Def16;
    
        then
    
        reconsider f9 = f as
    Function of ( 
    Chains ((k 
    + 1),G)), ( 
    Chains (k,G)) by 
    A2;
    
        now
    
          let A,B be
    Element of ( 
    Chains ((k 
    + 1),G)); 
    
          reconsider A9 = A, B9 = B as
    Chain of (k 
    + 1), G by 
    Def16;
    
          
    
          thus (f
    . (A 
    + B)) 
    = (f 
    . (A9 
    + B9)) by 
    Def16
    
          .= (
    del (A9 
    + B9)) by 
    A1
    
          .= ((
    del A9) 
    + ( 
    del B9)) by 
    Th58
    
          .= ((
    del A9) 
    + (f 
    . B9)) by 
    A1
    
          .= ((f
    . A9) 
    + (f 
    . B9)) by 
    A1
    
          .= ((f9
    . A) 
    + (f9 
    . B)) by 
    Def16;
    
        end;
    
        then f9 is
    additive by 
    VECTSP_1:def 20;
    
        then
    
        reconsider f9 as
    Homomorphism of ( 
    Chains ((k 
    + 1),G)), ( 
    Chains (k,G)); 
    
        take f9;
    
        thus thesis by
    A1;
    
      end;
    
      uniqueness
    
      proof
    
        let f,g be
    Homomorphism of ( 
    Chains ((k 
    + 1),G)), ( 
    Chains (k,G)); 
    
        assume
    
        
    
    A3: for A be 
    Element of ( 
    Chains ((k 
    + 1),G)), A9 be 
    Chain of (k 
    + 1), G st A 
    = A9 holds (f 
    . A) 
    = ( 
    del A9); 
    
        assume
    
        
    
    A4: for A be 
    Element of ( 
    Chains ((k 
    + 1),G)), A9 be 
    Chain of (k 
    + 1), G st A 
    = A9 holds (g 
    . A) 
    = ( 
    del A9); 
    
        now
    
          let A be
    Element of ( 
    Chains ((k 
    + 1),G)); 
    
          reconsider A9 = A as
    Element of ( 
    Chains ((k 
    + 1),G)); 
    
          reconsider A99 = A as
    Chain of (k 
    + 1), G by 
    Def16;
    
          (f
    . A9) 
    = ( 
    del A99) by 
    A3
    
          .= (g
    . A9) by 
    A4;
    
          hence (f
    . A) 
    = (g 
    . A); 
    
        end;
    
        hence f
    = g by 
    FUNCT_2: 63;
    
      end;
    
    end