:: Binary Arithmetics. Binary Sequences
:: by Robert Milewski
::
:: Received February 24, 1998
:: Copyright (c) 1998-2026 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies NUMBERS, XBOOLE_0, NAT_1, FINSEQ_2, MARGREL1, BINARITH, ARYTM_3,
POWER, SUBSET_1, ORDINAL4, FINSEQ_1, FUNCOP_1, XBOOLEAN, CARD_1, RELAT_1,
REAL_1, FINSEQ_5, EUCLID, XXREAL_0, FUNCT_1, PARTFUN1, ARYTM_1, BINARI_2,
ZFMISC_1, INT_1, ABIAN, BINARI_3, XCMPLX_0;
notations XBOOLE_0, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0, XREAL_0, XXREAL_0,
NAT_D, POWER, ABIAN, SERIES_1, MARGREL1, FUNCT_1, PARTFUN1, FUNCOP_1,
FINSEQ_1, FINSEQ_5, ZFMISC_1, FINSEQ_2, BINARITH, BINARI_2, EUCLID;
constructors NAT_D, FINSEQOP, SERIES_1, BINARITH, FINSEQ_5, BINARI_2, ABIAN,
EUCLID, TREES_3;
registrations RELSET_1, XREAL_0, NAT_1, XBOOLEAN, MARGREL1, NAT_2, ORDINAL1,
XBOOLE_0, FINSEQ_2, INT_1, FINSEQ_1;
requirements REAL, NUMERALS, SUBSET, ARITHM;
equalities FINSEQ_2, EUCLID, XBOOLEAN, ORDINAL1;
theorems TARSKI, NAT_1, NAT_2, MARGREL1, POWER, FUNCOP_1, FINSEQ_1, FINSEQ_2,
FINSEQ_4, FINSEQ_5, BINARITH, BINARI_2, XREAL_1, XCMPLX_1, XBOOLEAN,
NAT_D, PARTFUN1, XXREAL_0, CARD_1;
schemes NAT_1, NAT_2, FINSEQ_2;
begin :: Binary Arithmetics
theorem Th1:
for n be non zero Nat for F be Tuple of n,BOOLEAN holds Absval F
< 2 to_power n
proof
defpred P[non zero Nat] means for F be Tuple of $1,BOOLEAN holds Absval F <
2 to_power $1;
A1: for n be non zero Nat st P[n] holds P[n+1]
proof
let n be non zero Nat;
assume
A2: P[n];
n < n+1 by NAT_1:13;
then
A3: 2 to_power n < 2 to_power (n+1) by POWER:39;
let F be Tuple of n+1,BOOLEAN;
consider T be Element of n-tuples_on BOOLEAN,
d be Element of BOOLEAN such that
A4: F = T^<*d*> by FINSEQ_2:117;
A5: Absval F = Absval T + IFEQ(d,FALSE,0,2 to_power n) by A4,BINARITH:20;
A6: Absval T < 2 to_power n by A2;
per cases by XBOOLEAN:def 3;
suppose
d = FALSE;
then Absval F = Absval T + 0 by A5,FUNCOP_1:def 8;
then
Absval F + 2 to_power n < 2 to_power n + 2 to_power (n+1) by A2,A3,
XREAL_1:8;
hence thesis by XREAL_1:6;
end;
suppose
d = TRUE;
then Absval F = Absval T + 2 to_power n by A5,FUNCOP_1:def 8;
then Absval F < 2 to_power n + 2 to_power n by A6,XREAL_1:6;
then Absval F < 2 to_power n * 2;
then Absval F < 2 to_power n * 2 to_power 1 by POWER:25;
hence thesis by POWER:27;
end;
end;
A7: P[1]
proof
let F be Tuple of 1,BOOLEAN;
consider d be Element of BOOLEAN such that
A8: F = <*d*> by FINSEQ_2:97;
d = TRUE or d = FALSE by XBOOLEAN:def 3;
then Absval F = 1 or Absval F = 0 by A8,BINARITH:15,16;
then Absval F < 2;
hence thesis by POWER:25;
end;
thus for n being non zero Nat holds P[n] from NAT_1:sch 10(A7,A1);
end;
theorem Th2:
for n be non zero Nat for F1,F2 be Tuple of n,BOOLEAN st Absval
F1 = Absval F2 holds F1 = F2
proof
defpred P[non zero Nat] means for F1,F2 be Tuple of $1,BOOLEAN st Absval F1
= Absval F2 holds F1 = F2;
A1: for k be non zero Nat st P[k] holds P[k+1]
proof
let k be non zero Nat;
assume
A2: for F1,F2 be Tuple of k,BOOLEAN st Absval F1 = Absval F2 holds F1 = F2;
let F1,F2 be Tuple of k+1,BOOLEAN;
consider T1 be Element of k-tuples_on BOOLEAN,
d1 be Element of BOOLEAN such that
A3: F1 = T1^<*d1*> by FINSEQ_2:117;
assume
A4: Absval F1 = Absval F2;
consider T2 be Element of k-tuples_on BOOLEAN,
d2 be Element of BOOLEAN such that
A5: F2 = T2^<*d2*> by FINSEQ_2:117;
A6: Absval T1 + IFEQ(d1,FALSE,0,2 to_power k) = Absval F1 by A3,BINARITH:20
.= Absval T2 + IFEQ(d2,FALSE,0,2 to_power k) by A5,A4,BINARITH:20;
d1 = d2
proof
assume
A7: d1 <> d2;
per cases by XBOOLEAN:def 3;
suppose
A8: d1 = FALSE;
then
A9: IFEQ(d1,FALSE,0 qua Real,2 to_power k) = 0 by FUNCOP_1:def 8;
IFEQ(d2,FALSE,0 qua Real,2 to_power k) = 2 to_power k by A7,A8,
FUNCOP_1:def 8;
hence contradiction by A6,A9,Th1,NAT_1:11;
end;
suppose
A10: d1 = TRUE;
then d2 = FALSE by A7,XBOOLEAN:def 3;
then
A11: IFEQ(d2,FALSE,0 qua Real,2 to_power k) = 0 by FUNCOP_1:def 8;
IFEQ(d1,FALSE,0 qua Real,2 to_power k) = 2 to_power k by A10,
FUNCOP_1:def 8;
hence contradiction by A6,A11,Th1,NAT_1:11;
end;
end;
hence thesis by A2,A3,A5,A6,XCMPLX_1:2;
end;
A12: P[1]
proof
let F1,F2 be Tuple of 1,BOOLEAN;
consider d1 be Element of BOOLEAN such that
A13: F1 = <*d1*> by FINSEQ_2:97;
assume
A14: Absval F1 = Absval F2;
assume
A15: F1 <> F2;
consider d2 be Element of BOOLEAN such that
A16: F2 = <*d2*> by FINSEQ_2:97;
per cases by XBOOLEAN:def 3;
suppose
A17: d1 = FALSE;
then
A18: Absval F1 = 0 by A13,BINARITH:15;
d2 = TRUE by A13,A16,A15,A17,XBOOLEAN:def 3;
hence contradiction by A16,A14,A18,BINARITH:16;
end;
suppose
A19: d1 = TRUE;
then
A20: Absval F1 = 1 by A13,BINARITH:16;
d2 = FALSE by A13,A16,A15,A19,XBOOLEAN:def 3;
hence contradiction by A16,A14,A20,BINARITH:15;
end;
end;
thus for n being non zero Nat holds P[n] from NAT_1:sch 10(A12,A1);
end;
theorem
for t1,t2 be FinSequence st Rev t1 = Rev t2 holds t1 = t2
proof
let t1,t2 be FinSequence;
assume
A1: Rev t1 = Rev t2;
thus t1 = Rev Rev t1
.= t2 by A1;
end;
theorem Th4:
for n be Nat holds 0*n in BOOLEAN*
proof
let n be Nat;
n |-> FALSE is FinSequence of BOOLEAN;
hence thesis by FINSEQ_1:def 11;
end;
theorem Th5:
for n be Nat for y be Tuple of n,BOOLEAN st y = 0*n holds 'not' y = n |-> 1
proof
let n be Nat;
let y be Tuple of n,BOOLEAN;
assume
A1: y = 0*n;
A2: now
A3: len y = n by CARD_1:def 7;
let i be Nat;
assume that
A4: 1 <= i and
A5: i <= len 'not' y;
A6: len 'not' y = n by CARD_1:def 7;
then
A7: i in Seg n by A4,A5,FINSEQ_1:1;
then
A8: y.i = FALSE by A1,FUNCOP_1:7;
thus ('not' y).i = ('not' y)/.i by A4,A5,FINSEQ_4:15
.= 'not' (y/.i) by A7,BINARITH:def 1
.= 'not' FALSE by A4,A5,A6,A3,A8,FINSEQ_4:15
.= (n |-> 1).i by A7,FUNCOP_1:7;
end;
len 'not' y = n by CARD_1:def 7
.= len (n |-> 1) by CARD_1:def 7;
hence thesis by A2,FINSEQ_1:14;
end;
theorem Th6:
for n be non zero Nat for F be Tuple of n,BOOLEAN st F = 0*n
holds Absval F = 0
proof
defpred P[Nat] means for F be Tuple of $1,BOOLEAN st F = 0*$1 holds Absval F
= 0;
A1: for n be non zero Nat st P[n] holds P[n+1]
proof
let n be non zero Nat;
assume
A2: for F be Tuple of n,BOOLEAN st F = 0*n holds Absval F = 0;
let F be Tuple of n+1,BOOLEAN;
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider F1 = 0*n as Tuple of n,BOOLEAN;
assume F = 0*(n+1);
hence Absval F = Absval(F1 ^ <*FALSE*>) by FINSEQ_2:60
.= Absval(F1) + IFEQ(FALSE,FALSE,0,2 to_power n) by BINARITH:20
.= 0 + IFEQ(FALSE,FALSE,0,2 to_power n) by A2
.= 0 by FUNCOP_1:def 8;
end;
A3: P[1] by BINARITH:15,FINSEQ_2:59;
thus for n being non zero Nat holds P[n] from NAT_1:sch 10(A3,A1);
end;
theorem
for n be non zero Nat for F be Tuple of n,BOOLEAN st F = 0*n holds
Absval 'not' F = 2 to_power n - 1
proof
let n be non zero Nat;
let F be Tuple of n,BOOLEAN;
assume
A1: F = 0*n;
thus Absval 'not' F = - Absval F + 2 to_power n - 1 by BINARI_2:13
.= - 0 + 2 to_power n - 1 by A1,Th6
.= 2 to_power n - 1;
end;
theorem
for n be Nat holds Rev (0*n) = 0*n
proof
let n be Nat;
A1: now
let k be Nat;
assume
A2: k in dom 0*n;
then k in Seg len 0*n by FINSEQ_1:def 3;
then
A3: k in Seg n by CARD_1:def 7;
then n - k + 1 in Seg n by FINSEQ_5:2;
then
A4: len 0*n - k + 1 in Seg n by CARD_1:def 7;
thus (Rev 0*n).k = (0*n).(len 0*n - k + 1) by A2,FINSEQ_5:58
.= 0 by A4,FUNCOP_1:7
.= (0*n).k by A3,FUNCOP_1:7;
end;
dom Rev (0*n) = Seg len Rev (0*n) by FINSEQ_1:def 3
.= Seg len 0*n by FINSEQ_5:def 3
.= dom 0*n by FINSEQ_1:def 3;
hence thesis by A1,FINSEQ_1:13;
end;
theorem
for n be Nat for y be Tuple of n,BOOLEAN st y = 0*n holds Rev 'not' y
= 'not' y
proof
let n be Nat;
let y be Tuple of n,BOOLEAN;
assume
A1: y = 0*n;
A2: now
let k be Nat;
assume
A3: k in dom 'not' y;
then k in Seg len 'not' y by FINSEQ_1:def 3;
then
A4: k in Seg n by CARD_1:def 7;
then n - k + 1 in Seg n by FINSEQ_5:2;
then
A5: len 'not' y - k + 1 in Seg n by CARD_1:def 7;
thus (Rev 'not' y).k = ('not' y).(len 'not' y - k + 1) by A3,FINSEQ_5:58
.= (n |-> 1).(len 'not' y - k + 1) by A1,Th5
.= 1 by A5,FUNCOP_1:7
.= (n |-> 1).k by A4,FUNCOP_1:7
.= ('not' y).k by A1,Th5;
end;
dom Rev 'not' y = Seg len Rev 'not' y by FINSEQ_1:def 3
.= Seg len 'not' y by FINSEQ_5:def 3
.= dom 'not' y by FINSEQ_1:def 3;
hence thesis by A2,FINSEQ_1:13;
end;
theorem Th10:
Bin1 1 = <*TRUE*>
proof
1 in Seg 1 by FINSEQ_1:3;
then
A1: (Bin1 1)/.1 = TRUE by BINARI_2:5;
ex d be Element of BOOLEAN st Bin1 1 = <*d*> by FINSEQ_2:97;
hence thesis by A1,FINSEQ_4:16;
end;
theorem Th11:
for n be non zero Nat holds Absval (Bin1 n) = 1
proof
defpred P[Nat] means Absval (Bin1 $1) = 1;
A1: for n be non zero Nat st P[n] holds P[n+1]
proof
let n be non zero Nat;
assume
A2: Absval (Bin1 n) = 1;
thus Absval (Bin1 (n+1)) = Absval (Bin1 n ^ <*FALSE*>) by BINARI_2:7
.= Absval Bin1 n + IFEQ(FALSE,FALSE,0,2 to_power n) by BINARITH:20
.= Absval Bin1 n + 0 by FUNCOP_1:def 8
.= 1 by A2;
end;
A3: P[1] by Th10,BINARITH:16;
thus for n being non zero Nat holds P[n] from NAT_1:sch 10(A3,A1);
end;
theorem Th12:
for x,y be Element of BOOLEAN holds (x 'or' y = TRUE iff x =
TRUE or y = TRUE) & (x 'or' y = FALSE iff x = FALSE & y = FALSE)
proof
let x,y be Element of BOOLEAN;
thus x 'or' y = TRUE implies x = TRUE or y = TRUE
proof
assume x 'or' y = TRUE;
then 'not' x = FALSE or 'not' y = FALSE by MARGREL1:12;
hence thesis;
end;
thus x = TRUE or y = TRUE implies x 'or' y = TRUE;
thus x 'or' y = FALSE implies x = FALSE & y = FALSE
proof
assume
A1: x 'or' y = FALSE;
then 'not' x = TRUE by MARGREL1:12;
hence thesis by A1;
end;
thus thesis;
end;
theorem Th13:
for x,y be Element of BOOLEAN holds add_ovfl(<*x*>,<*y*>) = TRUE
iff x = TRUE & y = TRUE
proof
let x,y be Element of BOOLEAN;
A1: (<*TRUE*>/.1) '&' (<*TRUE*>/.1) = TRUE by FINSEQ_4:16;
thus add_ovfl(<*x*>,<*y*>) = TRUE implies x = TRUE & y = TRUE
proof
assume add_ovfl(<*x*>,<*y*>) = TRUE;
then
(<*x*>/.1) '&' (<*y*>/.1) 'or' (<*x*>/.1) '&' ((carry(<*x*>,<*y*>))/.1
) 'or' (<*y*>/.1) '&' ((carry(<*x*>,<*y*>))/.1) = TRUE by BINARITH:def 6;
then
A2: (<*x*>/.1) '&' (<*y*>/.1) 'or' (<*x*>/.1) '&' ((carry(<*x*>,<*y*>))/.1
) = TRUE or (<*y*>/.1) '&' ((carry(<*x*>,<*y*>))/.1) = TRUE by Th12;
now
per cases by A2,Th12;
suppose
A3: (<*x*>/.1) '&' (<*y*>/.1) = TRUE;
then (<*x*>/.1) = TRUE by MARGREL1:12;
hence thesis by A3,FINSEQ_4:16;
end;
suppose
(<*x*>/.1) '&' ((carry(<*x*>,<*y*>))/.1) = TRUE;
then (carry(<*x*>,<*y*>))/.1 = TRUE by MARGREL1:12;
hence thesis by BINARITH:def 2;
end;
suppose
(<*y*>/.1) '&' ((carry(<*x*>,<*y*>))/.1) = TRUE;
then (carry(<*x*>,<*y*>))/.1 = TRUE by MARGREL1:12;
hence thesis by BINARITH:def 2;
end;
end;
hence thesis;
end;
assume that
A4: x = TRUE and
A5: y = TRUE;
thus add_ovfl(<*x*>,<*y*>) = (<*TRUE*>/.1) '&' (<*TRUE*>/.1) 'or' (<*TRUE*>
/.1) '&' ((carry(<*TRUE*>,<*TRUE*>))/.1) 'or' (<*TRUE*>/.1) '&' (((carry(<*TRUE
*>,<*TRUE*>))/.1)) by A4,A5,BINARITH:def 6
.= TRUE by A1;
end;
theorem Th14:
'not' <*FALSE*> = <*TRUE*>
proof
now
let i be Nat;
assume
A1: i in Seg 1;
then
A2: i = 1 by FINSEQ_1:2,TARSKI:def 1;
len <*FALSE*> = 1 by CARD_1:def 7;
then
A3: (<*FALSE*>/.i) = <*FALSE*>.1 by A2,FINSEQ_4:15;
len 'not' <*FALSE*> = 1 by CARD_1:def 7;
hence ('not' <*FALSE*>).i = ('not' <*FALSE*>)/.i by A2,FINSEQ_4:15
.= 'not' (<*FALSE*>/.i) by A1,BINARITH:def 1
.= 'not' FALSE by A3
.= <*TRUE*>.i by A2;
end;
hence thesis by FINSEQ_2:119;
end;
theorem
'not' <*TRUE*> = <*FALSE*>
proof
now
let i be Nat;
assume
A1: i in Seg 1;
then
A2: i = 1 by FINSEQ_1:2,TARSKI:def 1;
len <*TRUE*> = 1 by CARD_1:def 7;
then
A3: (<*TRUE*>/.i) = <*TRUE*>.1 by A2,FINSEQ_4:15;
len 'not' <*TRUE*> = 1 by CARD_1:def 7;
hence ('not' <*TRUE*>).i = ('not' <*TRUE*>)/.i by A2,FINSEQ_4:15
.= 'not' (<*TRUE*>/.i) by A1,BINARITH:def 1
.= 'not' TRUE by A3
.= <*FALSE*>.i by A2;
end;
hence thesis by FINSEQ_2:119;
end;
theorem
<*FALSE*> + <*FALSE*> = <*FALSE*>
proof
add_ovfl(<*FALSE*>,<*FALSE*>) <> TRUE by Th13;
then add_ovfl(<*FALSE*>,<*FALSE*>) = FALSE by XBOOLEAN:def 3;
then <*FALSE*>,<*FALSE*> are_summable by BINARITH:def 7;
then Absval(<*FALSE*> + <*FALSE*>) = Absval <*FALSE*> + Absval <*FALSE*> by
BINARITH:22
.= Absval <*FALSE*> + 0 by BINARITH:15
.= Absval <*FALSE*>;
hence thesis by Th2;
end;
theorem Th17:
<*FALSE*> + <*TRUE*> = <*TRUE*> & <*TRUE*> + <*FALSE*> = <*TRUE *>
proof
add_ovfl(<*FALSE*>,<*TRUE*>) <> TRUE by Th13;
then add_ovfl(<*FALSE*>,<*TRUE*>) = FALSE by XBOOLEAN:def 3;
then <*FALSE*>,<*TRUE*> are_summable by BINARITH:def 7;
then Absval(<*FALSE*> + <*TRUE*>) = Absval <*FALSE*> + Absval <*TRUE*> by
BINARITH:22
.= Absval <*FALSE*> + 1 by BINARITH:16
.= 0 + 1 by BINARITH:15
.= Absval <*TRUE*> by BINARITH:16;
hence <*FALSE*> + <*TRUE*> = <*TRUE*> by Th2;
add_ovfl(<*TRUE*>,<*FALSE*>) <> TRUE by Th13;
then add_ovfl(<*TRUE*>,<*FALSE*>) = FALSE by XBOOLEAN:def 3;
then <*TRUE*>,<*FALSE*> are_summable by BINARITH:def 7;
then Absval(<*TRUE*> + <*FALSE*>) = Absval <*TRUE*> + Absval <*FALSE*> by
BINARITH:22
.= Absval <*TRUE*> + 0 by BINARITH:15
.= Absval <*TRUE*>;
hence thesis by Th2;
end;
theorem
<*TRUE*> + <*TRUE*> = <*FALSE*>
proof
A1: add_ovfl(<*TRUE*>,<*TRUE*>) = TRUE by Th13;
Absval(<*TRUE*> + <*TRUE*>) = Absval(<*TRUE*> + <*TRUE*>) + 2 - 2
.= Absval(<*TRUE*> + <*TRUE*>) + 2 to_power 1 - 2 by POWER:25
.= Absval(<*TRUE*> + <*TRUE*>) + IFEQ(add_ovfl(<*TRUE*>,<*TRUE*>), FALSE
,0,2 to_power (1)) - 2 by A1,FUNCOP_1:def 8
.= Absval <*TRUE*> + Absval <*TRUE*> - 2 by BINARITH:21
.= Absval <*TRUE*> + 1 - 2 by BINARITH:16
.= 1 + 1 - 2 by BINARITH:16
.= Absval <*FALSE*> by BINARITH:15;
hence thesis by Th2;
end;
theorem Th19:
for n be non zero Nat for x,y be Tuple of n,BOOLEAN st x/.n =
TRUE & (carry(x,Bin1 n))/.n = TRUE holds for k be non zero Nat st k <> 1 & k
<= n holds x/.k = TRUE & (carry(x,Bin1 n))/.k = TRUE
proof
let n be non zero Nat;
let x,y be Tuple of n,BOOLEAN;
assume that
A1: x/.n = TRUE and
A2: (carry(x,Bin1 n))/.n = TRUE;
defpred P[Nat] means ($1 in Seg (n -' 1) implies x/.(n -' $1 + 1) = TRUE & (
carry(x,Bin1 n))/.(n -' $1 + 1) = TRUE);
let k be non zero Nat;
assume that
A3: k <> 1 and
A4: k <= n;
set i = n -' k + 1;
1 < k by A3,NAT_2:19;
then 1 + 1 <= k by NAT_1:13;
then
A5: 1 <= k - 1 by XREAL_1:19;
A6: for j be non zero Nat st P[j] holds P[j+1]
proof
let j be non zero Nat;
assume that
A7: P[j] and
A8: j + 1 in Seg (n -' 1);
A9: j + 1 <= n -' 1 by A8,FINSEQ_1:1;
then
A10: j < n -' 1 by NAT_1:13;
then j < n - 1 by NAT_1:14,XREAL_1:233;
then j + 1 < n by XREAL_1:20;
then
A11: j < n by NAT_1:13;
j + 1 <= n - 1 by A9,NAT_1:14,XREAL_1:233;
then 1 <= n - 1 - j by XREAL_1:19;
then 1 <= n - j - 1;
then 1 + 1 <= n - j by XREAL_1:19;
then 1 + 1 <= n -' j by A11,XREAL_1:233;
then
A12: n -' j > 1 by NAT_1:13;
A13: 1 <= j by NAT_1:14;
A14: n -' j < n by NAT_2:9;
then n -' j in Seg n by A12,FINSEQ_1:1;
then
A15: (Bin1 n)/.(n -' j) = FALSE by A12,BINARI_2:6;
then ((Bin1 n)/.(n -' j)) '&' ((carry(x,Bin1 n))/.(n -' j)) = FALSE;
then
A16: TRUE = (x/.(n -' j)) '&' ((Bin1 n)/.(n -' j)) 'or' (x/.(n -' j)) '&'
((carry(x,Bin1 n))/.(n -' j)) by A7,A13,A10,A14,A12,A15,BINARITH:def 2
,FINSEQ_1:1
.= (x/.(n -' j)) '&' ((carry(x,Bin1 n))/.(n -' j)) by A15;
then x/.(n -' j) = TRUE by MARGREL1:12;
hence x/.(n -' (j+1) + 1) = TRUE by A11,NAT_2:7;
(carry(x,Bin1 n))/.(n -' j) = TRUE by A16,MARGREL1:12;
hence thesis by A11,NAT_2:7;
end;
A17: 1 <= i by NAT_1:11;
i = n - k + 1 by A4,XREAL_1:233
.= n - (k - 1);
then
A18: i <= n - 1 by A5,XREAL_1:13;
then i + 1 <= n by XREAL_1:19;
then i < n by NAT_1:13;
then
A19: k = n -' i + 1 by A4,NAT_2:5;
i <= n -' 1 by A18,NAT_1:14,XREAL_1:233;
then
A20: i in Seg (n -' 1) by A17,FINSEQ_1:1;
A21: P[1] by A1,A2,NAT_1:14,XREAL_1:235;
for j be non zero Nat holds P[j] from NAT_1:sch 10(A21,A6);
hence thesis by A19,A20;
end;
registration
cluster zero -> non positive for Nat;
coherence;
end;
theorem Th20:
for n be non zero Nat for x be Tuple of n,BOOLEAN st x/.n =
TRUE & (carry(x,Bin1 n))/.n = TRUE holds carry(x,Bin1 n) = 'not' Bin1 n
proof
let n be non zero Nat;
let x be Tuple of n,BOOLEAN;
assume that
A1: x/.n = TRUE and
A2: (carry(x,Bin1 n))/.n = TRUE;
now
A3: len 'not' Bin1 n = n by CARD_1:def 7;
let i be Nat;
reconsider z = i as Nat;
A4: len carry(x,Bin1 n) = n by CARD_1:def 7;
assume
A5: i in Seg n;
then
A6: 1 <= i by FINSEQ_1:1;
A7: i <= n by A5,FINSEQ_1:1;
per cases;
suppose
A8: i = 1;
thus carry(x,Bin1 n).i = (carry(x,Bin1 n))/.z by A6,A7,A4,FINSEQ_4:15
.= 'not' TRUE by A8,BINARITH:def 2
.= 'not' ((Bin1 n)/.i) by A5,A8,BINARI_2:5
.= ('not' Bin1 n)/.z by A5,BINARITH:def 1
.= ('not' Bin1 n).i by A6,A7,A3,FINSEQ_4:15;
end;
suppose
A9: i <> 1;
1 <= i by A5,FINSEQ_1:1;
then 0 < i;
then
A10: i is non zero;
thus carry(x,Bin1 n).i = (carry(x,Bin1 n))/.z by A6,A7,A4,FINSEQ_4:15
.= 'not' FALSE by A1,A2,A7,A9,A10,Th19
.= 'not' ((Bin1 n)/.i) by A5,A9,BINARI_2:6
.= ('not' Bin1 n)/.z by A5,BINARITH:def 1
.= ('not' Bin1 n).i by A6,A7,A3,FINSEQ_4:15;
end;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th21:
for n be non zero Nat for x,y be Tuple of n,BOOLEAN st y = 0*n
& x/.n = TRUE & (carry(x,Bin1 n))/.n = TRUE holds x = 'not' y
proof
let n be non zero Nat;
let x,y be Tuple of n,BOOLEAN;
assume that
A1: y = 0*n and
A2: x/.n = TRUE and
A3: (carry(x,Bin1 n))/.n = TRUE;
A4: len x = n by CARD_1:def 7;
A5: len 'not' y = n by CARD_1:def 7;
A6: len y = n by CARD_1:def 7;
A7: len carry(x,Bin1 n) = n by CARD_1:def 7;
now
let i be Nat;
reconsider z = i as Nat;
assume
A8: i in Seg n;
then
A9: 1 <= i by FINSEQ_1:1;
A10: i <= n by A8,FINSEQ_1:1;
A11: y.i = FALSE by A1,A8,FUNCOP_1:7;
now
per cases;
suppose
A12: i = 1;
A13: n >= 1 by NAT_1:14;
now
per cases by A13,XXREAL_0:1;
suppose
n = 1;
hence x .i = ('not' y).i by A3,BINARITH:def 2;
end;
suppose
A14: n > 1;
A15: len 'not' Bin1 n = n by CARD_1:def 7;
A16: (carry(x,Bin1 n))/.i = FALSE by A12,BINARITH:def 2;
then
A17: ((Bin1 n)/.i) '&' ((carry(x,Bin1 n))/.i) = FALSE;
A18: (1+1) <= n by A14,NAT_1:13;
then
A19: 2 in Seg n by FINSEQ_1:1;
carry(x,Bin1 n).(i+1) = ('not' Bin1 n).2 by A2,A3,A12,Th20
.= ('not' Bin1 n)/.2 by A18,A15,FINSEQ_4:15
.= 'not' ((Bin1 n)/.2) by A19,BINARITH:def 1
.= 'not' FALSE by A19,BINARI_2:6
.= TRUE;
then
A20: TRUE = (carry(x,Bin1 n))/.(i+1) by A7,A12,A18,FINSEQ_4:15
.= (x/.i) '&' ((Bin1 n)/.i) 'or' (x/.i) '&' ((carry(x,Bin1 n))
/.i) by A12,A14,A16,A17,BINARITH:def 2
.= (x/.i) '&' ((Bin1 n)/.i) by A16;
thus x .i = x/.z by A4,A9,A10,FINSEQ_4:15
.= 'not' FALSE by A20,MARGREL1:12
.= 'not' (y/.z) by A6,A9,A10,A11,FINSEQ_4:15
.= ('not' y)/.z by A8,BINARITH:def 1
.= ('not' y).i by A5,A9,A10,FINSEQ_4:15;
end;
end;
hence x .i = ('not' y).i;
end;
suppose
A21: i <> 1;
1 <= i by A8,FINSEQ_1:1;
then 0 < i;
then
A22: i is non empty;
thus x .i = x/.z by A4,A9,A10,FINSEQ_4:15
.= 'not' FALSE by A2,A3,A10,A21,A22,Th19
.= 'not' (y/.z) by A6,A9,A10,A11,FINSEQ_4:15
.= ('not' y)/.z by A8,BINARITH:def 1
.= ('not' y).i by A5,A9,A10,FINSEQ_4:15;
end;
end;
hence x .i = ('not' y).i;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th22:
for n be non zero Nat for y be Tuple of n,BOOLEAN st y = 0*n
holds carry('not' y,Bin1 n) = 'not' Bin1 n
proof
let n be non zero Nat;
let y be Tuple of n,BOOLEAN;
A1: n >= 1 by NAT_1:14;
A2: len y = n by CARD_1:def 7;
assume
A3: y = 0*n;
then
A4: y.n = 0 by FINSEQ_1:3,FUNCOP_1:7;
n in Seg n by FINSEQ_1:3;
then
A5: ('not' y)/.n = 'not' (y/.n) by BINARITH:def 1
.= 'not' FALSE by A1,A4,A2,FINSEQ_4:15
.= TRUE;
per cases;
suppose
A6: n = 1;
now
let i be Nat;
A7: len 'not' Bin1 n = n by CARD_1:def 7;
assume
A8: i in Seg n;
then
A9: i = 1 by A6,FINSEQ_1:2,TARSKI:def 1;
len carry('not' y,Bin1 n) = n by CARD_1:def 7;
hence carry('not' y,Bin1 n).i = (carry('not' y,Bin1 n))/.i by A6,A9,
FINSEQ_4:15
.= 'not' TRUE by A9,BINARITH:def 2
.= 'not' ((Bin1 n)/.i) by A8,A9,BINARI_2:5
.= ('not' Bin1 n)/.i by A8,BINARITH:def 1
.= ('not' Bin1 n).i by A6,A9,A7,FINSEQ_4:15;
end;
hence thesis by FINSEQ_2:119;
end;
suppose
n <> 1;
then
A10: n is non trivial by NAT_2:def 1;
defpred P[Nat] means $1 <= n implies (carry('not' y,Bin1 n))/.$1 = TRUE;
A11: for i be non trivial Nat st P[i] holds P[i+1]
proof
let i be non trivial Nat;
assume that
A12: i <= n implies (carry('not' y,Bin1 n))/.i = TRUE and
A13: i + 1 <= n;
A14: 1 <= i by NAT_1:14;
A15: i < n by A13,NAT_1:13;
then
A16: i in Seg n by A14,FINSEQ_1:1;
then
A17: y.i = FALSE by A3,FUNCOP_1:7;
A18: ('not' y)/.i = 'not' (y/.i) by A16,BINARITH:def 1
.= 'not' FALSE by A2,A14,A15,A17,FINSEQ_4:15
.= TRUE;
i <> 1 by NAT_2:def 1;
then
A19: ((Bin1 n)/.i) = FALSE by A16,BINARI_2:6;
then ((Bin1 n)/.i) '&' ((carry('not' y,Bin1 n))/.i) = FALSE;
hence
(carry('not' y,Bin1 n))/.(i + 1) = (('not' y)/.i) '&' ((Bin1 n)/.i)
'or' (('not' y)/.i) '&' ((carry('not' y,Bin1 n))/.i) by A14,A15,A19,
BINARITH:def 2
.= TRUE by A12,A13,A18,NAT_1:13;
end;
A20: P[2]
proof
assume 2 <= n;
then 1 + 1 <= n;
then
A21: 1 < n by NAT_1:13;
then
A22: 1 in Seg n by FINSEQ_1:1;
then
A23: y.1 = FALSE by A3,FUNCOP_1:7;
('not' y)/.1 = 'not' (y/.1) by A22,BINARITH:def 1
.= 'not' FALSE by A2,A21,A23,FINSEQ_4:15
.= TRUE;
then
A24: (('not' y)/.1) '&' ((Bin1 n)/.1) = TRUE by A22,BINARI_2:5;
thus (carry('not' y,Bin1 n))/.2 = (carry('not' y,Bin1 n))/.(1 + 1)
.= (('not' y)/.1) '&' ((Bin1 n)/.1) 'or' (('not' y)/.1) '&' ((carry(
'not' y,Bin1 n))/.1) 'or' ((Bin1 n)/.1) '&' ( (carry('not' y,Bin1 n))/.1) by
A21,BINARITH:def 2
.= TRUE by A24;
end;
for i be non trivial Nat holds P[i] from NAT_2:sch 2(A20,A11);
then (carry('not' y,Bin1 n))/.n = TRUE by A10;
hence thesis by A5,Th20;
end;
end;
theorem Th23:
for n be non zero Nat for x,y be Tuple of n,BOOLEAN st y = 0*n
holds add_ovfl(x,Bin1 n) = TRUE iff x = 'not' y
proof
let n be non zero Nat;
let x,y be Tuple of n,BOOLEAN;
assume
A1: y = 0*n;
A2: n in Seg n by FINSEQ_1:3;
A3: 1 in Seg 1 by FINSEQ_1:3;
thus add_ovfl(x,Bin1 n) = TRUE implies x = 'not' y
proof
assume
A4: add_ovfl(x,Bin1 n) = TRUE;
then
A5: (x/.n) '&' ((Bin1 n)/.n) 'or' (x/.n) '&' ((carry(x,Bin1 n))/.n) 'or' (
(Bin1 n)/.n) '&' ((carry(x,Bin1 n))/.n) = TRUE by BINARITH:def 6;
per cases;
suppose
A6: n <> 1;
now
per cases by A5,Th12;
suppose
A7: (x/.n) '&' ((Bin1 n)/.n) 'or' (x/.n) '&' ((carry(x,Bin1 n))
/.n) = TRUE;
now
per cases by A7,Th12;
suppose
A8: (x/.n) '&' ((Bin1 n)/.n) = TRUE;
assume x <> 'not' y;
((Bin1 n)/.n) = TRUE by A8,MARGREL1:12;
hence contradiction by A2,A6,BINARI_2:6;
end;
suppose
A9: (x/.n) '&' ((carry(x,Bin1 n))/.n) = TRUE;
then (x/.n) = TRUE by MARGREL1:12;
hence thesis by A1,A9,Th21;
end;
end;
hence thesis;
end;
suppose
A10: ((Bin1 n)/.n) '&' ((carry(x,Bin1 n))/.n) = TRUE;
assume x <> 'not' y;
((Bin1 n)/.n) = TRUE by A10,MARGREL1:12;
hence contradiction by A2,A6,BINARI_2:6;
end;
end;
hence thesis;
end;
suppose
A11: n = 1;
then len y = 1 by CARD_1:def 7;
then 1 in dom y by A3,FINSEQ_1:def 3;
then
A12: (y/.1) = y.1 by PARTFUN1:def 6
.= 0 by A1,A11,FINSEQ_1:3,FUNCOP_1:7;
consider d be Element of BOOLEAN such that
A13: x = <*d*> by A11,FINSEQ_2:97;
A14: d = TRUE by A4,A11,A13,Th10,Th13;
now
let i be Nat;
assume i in Seg n;
then i = 1 by A11,FINSEQ_1:2,TARSKI:def 1;
hence x/.i = 'not' (y/.i) by A13,A14,A12,FINSEQ_4:16;
end;
hence thesis by BINARITH:def 1;
end;
end;
assume
A15: x = 'not' y;
per cases;
suppose
A16: n <> 1;
A17: (carry(x,Bin1 n))/.n = ('not' Bin1 n)/.n by A1,A15,Th22
.= 'not' ((Bin1 n)/.n) by A2,BINARITH:def 1
.= 'not' FALSE by A2,A16,BINARI_2:6
.= TRUE;
len y = n by CARD_1:def 7;
then n in dom y by A2,FINSEQ_1:def 3;
then
A18: (y/.n) = y.n by PARTFUN1:def 6
.= 0 by A1,FINSEQ_1:3,FUNCOP_1:7;
A19: x/.n = 'not' (y/.n) by A2,A15,BINARITH:def 1
.= TRUE by A18;
thus add_ovfl(x,Bin1 n) = (x/.n) '&' ((Bin1 n)/.n) 'or' (x/.n) '&' ((carry
(x,Bin1 n))/.n) 'or' ((Bin1 n)/.n) '&' ((carry(x,Bin1 n))/.n) by BINARITH:def 6
.= TRUE by A19,A17;
end;
suppose
A20: n = 1;
then len y = 1 by CARD_1:def 7;
then 1 in dom y by A3,FINSEQ_1:def 3;
then
A21: (y/.1) = y.1 by PARTFUN1:def 6
.= 0 by A1,A20,FINSEQ_1:3,FUNCOP_1:7;
consider d be Element of BOOLEAN such that
A22: x = <*d*> by A20,FINSEQ_2:97;
d = ('not' y)/.1 by A15,A22,FINSEQ_4:16
.= 'not' (y/.1) by A3,A20,BINARITH:def 1
.= TRUE by A21;
hence thesis by A20,A22,Th10,Th13;
end;
end;
theorem Th24:
for n be non zero Nat for z be Tuple of n,BOOLEAN st z = 0*n
holds 'not' z + Bin1 n = z
proof
let n be non zero Nat;
let z be Tuple of n,BOOLEAN;
assume
A1: z = 0*n;
then
A2: add_ovfl('not' z,Bin1 n) = TRUE by Th23;
Absval ('not' z + Bin1 n) = Absval ('not' z + Bin1 n) + 2 to_power n - 2
to_power n
.= Absval ('not' z + Bin1 n) + IFEQ(add_ovfl('not' z,Bin1 n),FALSE, 0,2
to_power (n)) - 2 to_power n by A2,FUNCOP_1:def 8
.= Absval 'not' z + Absval Bin1 n - 2 to_power n by BINARITH:21
.= - Absval(z) + 2 to_power n - 1 + Absval Bin1 n - 2 to_power n by
BINARI_2:13
.= - Absval(z) + 2 to_power n - 1 + 1 - 2 to_power n by Th11
.= -0 by A1,Th6
.= Absval z by A1,Th6;
hence thesis by Th2;
end;
begin :: Binary Sequences
definition
let n,k be Nat;
func n-BinarySequence k -> Tuple of n,BOOLEAN means
:Def1:
for i be Nat st i
in Seg n holds it/.i = IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE);
existence
proof
reconsider n9 = n as Nat;
deffunc F(Nat) = IFEQ((k div 2 to_power ($1-'1)) mod 2,0,FALSE,TRUE);
consider p be FinSequence of BOOLEAN such that
A1: len p = n9 and
A2: for i be Nat st i in dom p holds p.i = F(i) from FINSEQ_2:sch 1;
A3: dom p = Seg n9 by A1,FINSEQ_1:def 3;
reconsider p as Element of n-tuples_on BOOLEAN by A1,FINSEQ_2:92;
take p;
let i be Nat;
assume
A4: i in Seg n;
then i in dom p by A1,FINSEQ_1:def 3;
hence p/.i = p.i by PARTFUN1:def 6
.= IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A2,A3,A4;
end;
uniqueness
proof
let p1,p2 be Tuple of n,BOOLEAN such that
A5: for i be Nat st i in Seg n holds p1/.i = IFEQ((k div 2 to_power (i
-'1)) mod 2,0,FALSE,TRUE) and
A6: for i be Nat st i in Seg n holds p2/.i = IFEQ((k div 2 to_power (i
-'1)) mod 2,0,FALSE,TRUE);
A7: len p1 = n by CARD_1:def 7;
then
A8: dom p1 = Seg n by FINSEQ_1:def 3;
A9: len p2 = n by CARD_1:def 7;
now
let i be Nat;
assume
A10: i in dom p1;
then
A11: i in dom p2 by A9,A8,FINSEQ_1:def 3;
thus p1.i = p1/.i by A10,PARTFUN1:def 6
.= IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A5,A8,A10
.= p2/.i by A6,A8,A10
.= p2.i by A11,PARTFUN1:def 6;
end;
hence thesis by A7,A9,FINSEQ_2:9;
end;
end;
theorem Th25:
for n be Nat holds n-BinarySequence 0 = 0*n
proof
let n be Nat;
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider F = 0*n as Tuple of n,BOOLEAN;
now
let i be Nat;
assume
A1: i in Seg n;
len (n-BinarySequence 0) = n by CARD_1:def 7;
then i in dom (n-BinarySequence 0) by A1,FINSEQ_1:def 3;
hence (n-BinarySequence 0).i = (n-BinarySequence 0)/.i by PARTFUN1:def 6
.= IFEQ((0 div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A1,Def1
.= IFEQ(0 mod 2,0,FALSE,TRUE)
.= IFEQ(0,0,FALSE,TRUE) by NAT_D:26
.= 0 by FUNCOP_1:def 8
.= F.i by A1,FUNCOP_1:7;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th26:
for n,k be Nat st k < 2 to_power n holds ((n+1)-BinarySequence k
).(n+1) = FALSE
proof
let n,k be Nat;
A1: n+1-'1 = n+1-1 by NAT_D:37
.= n;
assume k < 2 to_power n;
then
A2: (k div 2 to_power (n+1-'1)) mod 2 = 0 mod 2 by A1,NAT_D:27
.= 0 by NAT_D:26;
A3: n + 1 in Seg (n+1) by FINSEQ_1:4;
then n + 1 in Seg len ((n+1)-BinarySequence k) by CARD_1:def 7;
then n + 1 in dom ((n+1)-BinarySequence k) by FINSEQ_1:def 3;
hence ((n+1)-BinarySequence k).(n+1) = ((n+1)-BinarySequence k)/.(n+1) by
PARTFUN1:def 6
.= IFEQ((k div 2 to_power (n+1-'1)) mod 2,0,FALSE,TRUE) by A3,Def1
.= FALSE by A2,FUNCOP_1:def 8;
end;
theorem Th27:
for n be non zero Nat for k be Nat st k < 2 to_power n holds (n
+1)-BinarySequence k = (n-BinarySequence k)^<*FALSE*>
proof
let n be non zero Nat;
let k be Nat;
assume
A1: k < 2 to_power n;
now
let i be Nat;
assume
A2: i in Seg (n + 1);
then i in Seg len ((n+1)-BinarySequence k) by CARD_1:def 7;
then
A3: i in dom ((n+1)-BinarySequence k) by FINSEQ_1:def 3;
now
per cases by A2,FINSEQ_2:7;
suppose
A4: i in Seg n;
then i in Seg len (n-BinarySequence k) by CARD_1:def 7;
then
A5: i in dom (n-BinarySequence k) by FINSEQ_1:def 3;
thus ((n+1)-BinarySequence k).i = ((n+1)-BinarySequence k)/.i by A3,
PARTFUN1:def 6
.= IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A2,Def1
.= (n-BinarySequence k)/.i by A4,Def1
.= (n-BinarySequence k).i by A5,PARTFUN1:def 6
.= ((n-BinarySequence k)^<*FALSE*>).i by A5,FINSEQ_1:def 7;
end;
suppose
A6: i = n + 1;
hence ((n+1)-BinarySequence k).i = FALSE by A1,Th26
.= ((n-BinarySequence k)^<*FALSE*>).i by A6,FINSEQ_2:116;
end;
end;
hence ((n+1)-BinarySequence k).i = ((n-BinarySequence k)^<*FALSE*>).i;
end;
hence thesis by FINSEQ_2:119;
end;
Lm1: for n be non zero Nat holds (n+1)-BinarySequence 2 to_power n = 0*n^<*
TRUE*>
proof
let n be non zero Nat;
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider 0n = 0*n as Tuple of n,BOOLEAN;
now
let i be Nat;
assume
A1: i in Seg (n+1);
now
per cases by A1,FINSEQ_2:7;
suppose
A2: i in Seg n;
then
A3: i >= 1 by FINSEQ_1:1;
i <= n + 1 by A1,FINSEQ_1:1;
then i - 1 <= n + 1 - 1 by XREAL_1:9;
then
A4: i-'1 <= n + 1 - 1 by A3,XREAL_1:233;
n = n-(i-'1)+(i-'1) .= n-'(i-'1)+(i-'1) by A4,XREAL_1:233;
then
A5: 2 to_power n = 2 to_power (n-'(i-'1)) * 2 to_power (i-'1) by POWER:27;
i in Seg len 0n by A2,CARD_1:def 7;
then
A6: i in dom 0n by FINSEQ_1:def 3;
n >= i by A2,FINSEQ_1:1;
then n + 1 > i by NAT_1:13;
then n > i - 1 by XREAL_1:19;
then n - (i - 1) > 0 by XREAL_1:50;
then n - (i-'1) > 0 by A3,XREAL_1:233;
then n-'(i-'1) > 0 by A4,XREAL_1:233;
then consider n1 be Nat such that
A7: n-'(i-'1) = n1 + 1 by NAT_1:6;
reconsider n1 as Nat;
A8: 2 to_power (n -' (i-'1)) = 2 to_power n1 * 2 to_power 1 by A7,POWER:27
.= 2 to_power n1 * 2 by POWER:25;
2 to_power (i-'1) > 0 by POWER:34;
then
A9: ((2 to_power n) div 2 to_power (i-'1)) mod 2 = (2 to_power (n-'(i
-'1))) mod 2 by A5,NAT_D:18
.= 0 by A8;
i in Seg len ((n+1)-BinarySequence 2 to_power n) by A1,CARD_1:def 7;
then i in dom ((n+1)-BinarySequence 2 to_power n) by FINSEQ_1:def 3;
hence ((n+1)-BinarySequence 2 to_power n).i = ((n+1)-BinarySequence 2
to_power n)/.i by PARTFUN1:def 6
.= IFEQ(((2 to_power n) div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE)
by A1,Def1
.= 0 by A9,FUNCOP_1:def 8
.= 0n.i by A2,FUNCOP_1:7
.= (0n^<*TRUE*>).i by A6,FINSEQ_1:def 7;
end;
suppose
A10: i = n + 1;
A11: 2 to_power n > 0 by POWER:34;
i-'1 = n+1-1 by A10,NAT_D:37
.= n;
then
A12: ((2 to_power n) div 2 to_power (i-'1)) mod 2 = 1 mod 2 by A11,NAT_2:3
.= 1 by NAT_D:14;
n + 1 in Seg (n+1) by FINSEQ_1:4;
then n + 1 in Seg len ((n+1)-BinarySequence 2 to_power n) by
CARD_1:def 7;
then n + 1 in dom ((n+1)-BinarySequence 2 to_power n) by FINSEQ_1:def 3
;
hence ((n+1)-BinarySequence 2 to_power n).i = ((n+1)-BinarySequence 2
to_power n)/.i by A10,PARTFUN1:def 6
.= IFEQ(((2 to_power n) div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE)
by A1,Def1
.= TRUE by A12,FUNCOP_1:def 8
.= (0n^<*TRUE*>).i by A10,FINSEQ_2:116;
end;
end;
hence ((n+1)-BinarySequence 2 to_power n).i = (0n^<*TRUE*>).i;
end;
hence thesis by FINSEQ_2:119;
end;
Lm2: for n be non zero Nat for k be Nat st 2 to_power n <= k & k < 2 to_power
(n+1) holds ((n+1)-BinarySequence k).(n+1) = TRUE
proof
let n be non zero Nat;
let k be Nat;
assume that
A1: 2 to_power n <= k and
A2: k < 2 to_power (n+1);
k < 2 to_power n * 2 to_power 1 by A2,POWER:27;
then k < 2 * 2 to_power n by POWER:25;
then
A3: k < 2 to_power n + 2 to_power n;
n+1-'1 = n+1-1 by NAT_D:37
.= n;
then
A4: (k div 2 to_power (n+1-'1)) mod 2 = 1 mod 2 by A1,A3,NAT_2:20
.= 1 by NAT_D:24;
A5: n + 1 in Seg (n+1) by FINSEQ_1:4;
then n + 1 in Seg len ((n+1)-BinarySequence k) by CARD_1:def 7;
then n + 1 in dom ((n+1)-BinarySequence k) by FINSEQ_1:def 3;
hence ((n+1)-BinarySequence k).(n+1) = ((n+1)-BinarySequence k)/.(n+1) by
PARTFUN1:def 6
.= IFEQ((k div 2 to_power (n+1-'1)) mod 2,0,FALSE,TRUE) by A5,Def1
.= TRUE by A4,FUNCOP_1:def 8;
end;
Lm3: for n be non zero Nat for k be Nat st 2 to_power n <= k & k < 2 to_power
(n+1) holds (n+1)-BinarySequence k = (n-BinarySequence (k -' 2 to_power n))^<*
TRUE*>
proof
let n be non zero Nat;
let k be Nat;
assume that
A1: 2 to_power n <= k and
A2: k < 2 to_power (n+1);
now
let i be Nat;
reconsider z = i as Nat;
assume
A3: i in Seg (n+1);
then i in Seg len ((n+1)-BinarySequence k) by CARD_1:def 7;
then
A4: i in dom ((n+1)-BinarySequence k) by FINSEQ_1:def 3;
now
per cases by A3,FINSEQ_2:7;
suppose
A5: i in Seg n;
then
A6: 1 <= i by FINSEQ_1:1;
A7: 2 * 2 to_power (i-'1) = 2 to_power (i-'1) * 2 to_power 1 by POWER:25
.= 2 to_power (i-'1+1) by POWER:27
.= 2 to_power (i-1+1) by A6,XREAL_1:233
.= 2 to_power i;
2 to_power (i-'1) > 0 by POWER:34;
then
A8: 0 + 1 <= 2 to_power (i-'1) by NAT_1:13;
i <= n by A5,FINSEQ_1:1;
then
A9: 2 * 2 to_power (z-'1) divides 2 to_power n by A7,NAT_2:11;
A10: now
per cases;
suppose
A11: k div 2 to_power (i-'1) is even;
then
A12: (k div 2 to_power (i-'1)) mod 2 = 0 by NAT_2:21;
(k-' 2 to_power n) div 2 to_power (i-'1) is even by A1,A8,A9,A11,
NAT_2:23;
hence (k div 2 to_power (i-'1)) mod 2 = ((k -' 2 to_power n) div 2
to_power (i-'1)) mod 2 by A12,NAT_2:21;
end;
suppose
A13: k div 2 to_power (i-'1) is odd;
then
A14: (k div 2 to_power (i-'1)) mod 2 = 1 by NAT_2:22;
(k -' 2 to_power n) div 2 to_power (i-'1) is odd by A1,A8,A9,A13,
NAT_2:23;
hence (k div 2 to_power (i-'1)) mod 2 = ((k -' 2 to_power n) div 2
to_power (i-'1)) mod 2 by A14,NAT_2:22;
end;
end;
i in Seg len (n-BinarySequence (k -' 2 to_power n)) by A5,CARD_1:def 7;
then
A15: i in dom (n-BinarySequence (k -' 2 to_power n)) by FINSEQ_1:def 3;
thus ((n+1)-BinarySequence k).i = ((n+1)-BinarySequence k)/.i by A4,
PARTFUN1:def 6
.= IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A3,Def1
.= (n-BinarySequence (k -' 2 to_power n))/.i by A5,A10,Def1
.= (n-BinarySequence (k -' 2 to_power n)).i by A15,PARTFUN1:def 6
.= ((n-BinarySequence (k -' 2 to_power n))^<*TRUE*>).i by A15,
FINSEQ_1:def 7;
end;
suppose
A16: i = n + 1;
hence ((n+1)-BinarySequence k).i = TRUE by A1,A2,Lm2
.= ((n-BinarySequence (k -' 2 to_power n))^<*TRUE*>).i by A16,
FINSEQ_2:116;
end;
end;
hence
((n+1)-BinarySequence k).i = ((n-BinarySequence (k -' 2 to_power n))^
<*TRUE*>).i;
end;
hence thesis by FINSEQ_2:119;
end;
Lm4: for n be non zero Nat for k be Nat st k < 2 to_power n for x be Tuple of
n,BOOLEAN st x = 0*n holds n-BinarySequence k = 'not' x iff k = 2 to_power n -
1
proof
let n be non zero Nat;
let k be Nat;
assume
A1: k < 2 to_power n;
let x be Tuple of n,BOOLEAN;
assume
A2: x = 0*n;
thus n-BinarySequence k = 'not' x implies k = 2 to_power n - 1
proof
defpred P[Nat] means for k be Nat holds (k < 2 to_power $1 implies for y
be Tuple of $1,BOOLEAN st y = 0*$1 holds $1-BinarySequence k = 'not' y implies
k = 2 to_power $1 - 1);
assume
A3: n-BinarySequence k = 'not' x;
A4: for m be non zero Nat st P[m] holds P[m+1]
proof
let m be non zero Nat;
assume
A5: P[m];
A6: m + 1 in Seg (m + 1) by FINSEQ_1:4;
A7: 0 + 1 <= m + 1 by XREAL_1:6;
0*m in BOOLEAN* by Th4;
then
0*m is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider y1 = 0*m as Tuple of m,BOOLEAN;
let k be Nat;
assume
A8: k < 2 to_power (m+1);
let y be Tuple of m+1,BOOLEAN;
assume that
A9: y = 0*(m+1) and
A10: (m+1)-BinarySequence k = 'not' y;
A11: y.(m + 1) = FALSE by A9,FINSEQ_1:4,FUNCOP_1:7;
A12: y = y1 ^ <* 0 *> by A9,FINSEQ_2:60;
A13: len y = m + 1 by CARD_1:def 7;
len 'not' y = m + 1 by CARD_1:def 7;
then
A14: ((m+1)-BinarySequence k).(m+1) = ('not' y)/.(m+1) by A10,A7,FINSEQ_4:15
.= 'not' (y/.(m+1)) by A6,BINARITH:def 1
.= 'not' FALSE by A13,A7,A11,FINSEQ_4:15
.= TRUE;
then (m+1)-BinarySequence k = (m-BinarySequence (k -' 2 to_power m))^<*
TRUE*> by A8,Lm3,Th26;
then (m-BinarySequence (k -' 2 to_power m))^<*TRUE*> = 'not' y1^<*'not'
FALSE*> by A10,A12,BINARI_2:9;
then
A15: m-BinarySequence (k -' 2 to_power m) = 'not' y1 by FINSEQ_2:17;
k < (2 to_power m) * (2 to_power 1) by A8,POWER:27;
then k < 2 * (2 to_power m) by POWER:25;
then k < 2 to_power m + 2 to_power m;
then k - 2 to_power m < 2 to_power m by XREAL_1:19;
then k -' 2 to_power m < 2 to_power m by A14,Th26,XREAL_1:233;
then k -' 2 to_power m = 2 to_power m - 1 by A5,A15;
then k - 2 to_power m = 2 to_power m - 1 by A14,Th26,XREAL_1:233;
hence k = (2 to_power m) * 2 - 1
.= (2 to_power m) * (2 to_power 1) - 1 by POWER:25
.= 2 to_power (m+1) - 1 by POWER:27;
end;
A16: P[1]
proof
let k be Nat;
A17: 1 <= len (1-BinarySequence k) by CARD_1:def 7;
assume k < 2 to_power 1;
then
A18: k < 2 by POWER:25;
let y be Tuple of 1,BOOLEAN;
assume y = 0*1;
then
A19: y = <*FALSE*> by FINSEQ_2:59;
assume
A20: 1-BinarySequence k = 'not' y;
A21: 1 in Seg 1 by FINSEQ_1:3;
A22: 1 = <*1*>.1
.= (1-BinarySequence k)/.1 by A19,A20,A17,Th14,FINSEQ_4:15
.= IFEQ((k div 2 to_power (1-'1)) mod 2,0,FALSE,TRUE) by A21,Def1;
k = 1
proof
assume k <> 1;
then k = 0 by A18,NAT_1:23;
then k div 2 to_power (1-'1) = 0;
then (k div 2 to_power (1-'1)) mod 2 = 0 by NAT_D:26;
hence contradiction by A22,FUNCOP_1:def 8;
end;
hence k = 1 + 1 - 1 .= 2 to_power 1 - 1 by POWER:25;
end;
for m be non zero Nat holds P[m] from NAT_1:sch 10(A16,A4 );
hence thesis by A1,A2,A3;
end;
assume
A23: k = 2 to_power n - 1;
now
let i be Nat;
A24: len x = n by CARD_1:def 7;
2 to_power (i-'1) > 0 by POWER:34;
then
A25: 2 to_power (i-'1) >= 0 + 1 by NAT_1:13;
A26: len 'not' x = n by CARD_1:def 7;
A27: 2 to_power (n-'(i-'1)) > 0 by POWER:34;
then
A28: 2 to_power (n-'(i-'1)) >= 0 + 1 by NAT_1:13;
reconsider z = i as Nat;
assume
A29: i in Seg n;
then
A30: 1 <= i by FINSEQ_1:1;
i <= n by A29,FINSEQ_1:1;
then i < n + 1 by NAT_1:13;
then
A31: i - 1 < n + 1 - 1 by XREAL_1:9;
1 <= i by A29,FINSEQ_1:1;
then
A32: 0 + (i-'1) < n by A31,XREAL_1:233;
then n - (i-'1) > 0 by XREAL_1:20;
then n-'(i-'1) > 0 by A32,XREAL_1:233;
then
A33: 2 to_power (n-'(i-'1)) mod 2 = 0 by NAT_2:17;
2 to_power n > 0 by POWER:34;
then
A34: 2 to_power n >= 0 + 1 by NAT_1:13;
then k div 2 to_power (i-'1) = (2 to_power n -' 1) div 2 to_power (i-'1)
by A23,XREAL_1:233
.= (2 to_power n div 2 to_power (i-'1)) - 1 by A25,A32,A34,NAT_2:11,15
.= (2 to_power (n-'(i-'1))) - 1 by A32,NAT_2:16
.= (2 to_power (n-'(i-'1))) -' 1 by A28,XREAL_1:233;
then
A35: (k div 2 to_power (i-'1)) mod 2 = 1 by A33,A27,NAT_2:18;
A36: (x).i = FALSE by A2,A29,FUNCOP_1:7;
A37: i <= n by A29,FINSEQ_1:1;
i in Seg len (n-BinarySequence k) by A29,CARD_1:def 7;
then i in dom (n-BinarySequence k) by FINSEQ_1:def 3;
hence (n-BinarySequence k).i = (n-BinarySequence k)/.i by PARTFUN1:def 6
.= IFEQ((k div 2 to_power (i-'1)) mod 2,0,FALSE,TRUE) by A29,Def1
.= 'not' FALSE by A35,FUNCOP_1:def 8
.= 'not' (x/.z) by A24,A30,A37,A36,FINSEQ_4:15
.= ('not' x)/.z by A29,BINARITH:def 1
.= ('not' x).i by A26,A30,A37,FINSEQ_4:15;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th28:
for i being Nat holds (i+1)-BinarySequence 2 to_power i= 0*i^<*1 *>
proof
deffunc Bi(Nat) = ($1+1)-BinarySequence 2 to_power $1;
let i be Nat;
set Bi = Bi(i);
per cases;
suppose
A1: i = 0;
then
A2: 0*i = 0;
reconsider i1 = i+1 as non zero Nat;
A3: 0*i1 = <*FALSE*> by A1,FINSEQ_2:59;
then reconsider x = 0*i1 as Tuple of i1, BOOLEAN;
2 to_power i1 = 2 by A1,POWER:25;
then 1 = (2 to_power i1) - 1;
then i1-BinarySequence 1 = 'not' x by Lm4;
hence Bi = <*TRUE*> by A1,A3,Th14,POWER:24
.= 0*i^<*1*> by A2,FINSEQ_1:34;
end;
suppose
i > 0;
then reconsider i9 = i as non zero Nat;
Bi = 0*(i9)^<*1*> by Lm1;
hence thesis;
end;
end;
theorem
for n be non zero Nat for k be Nat st 2 to_power n <= k & k < 2
to_power (n+1) holds ((n+1)-BinarySequence k).(n+1) = TRUE by Lm2;
theorem
for n be non zero Nat for k be Nat st 2 to_power n <= k & k < 2
to_power (n+1) holds (n+1)-BinarySequence k = (n-BinarySequence (k -' 2
to_power n))^<*TRUE*> by Lm3;
theorem
for n be non zero Nat for k be Nat st k < 2 to_power n for x be Tuple
of n,BOOLEAN st x = 0*n holds n-BinarySequence k = 'not' x iff k = 2 to_power n
- 1 by Lm4;
theorem Th32:
for n be non zero Nat for k be Nat st k + 1 < 2 to_power n
holds add_ovfl(n-BinarySequence k,Bin1 n) = FALSE
proof
let n be non zero Nat;
let k be Nat;
assume
A1: k + 1 < 2 to_power n;
then
A2: k < 2 to_power n by NAT_1:13;
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider y = 0*n as Tuple of n,BOOLEAN;
k < 2 to_power n - 1 by A1,XREAL_1:20;
then n-BinarySequence k <> 'not' y by A2,Lm4;
then add_ovfl(n-BinarySequence k,Bin1 n) <> TRUE by Th23;
hence thesis by XBOOLEAN:def 3;
end;
theorem Th33:
for n be non zero Nat for k be Nat st k + 1 < 2 to_power n
holds n-BinarySequence (k+1) = n-BinarySequence k + Bin1 n
proof
defpred P[non zero Nat] means for k be Nat st k + 1 < 2 to_power $1 holds
$1-BinarySequence (k+1) = $1-BinarySequence k + Bin1 $1;
A1: for n be non zero Nat st P[n] holds P[n+1]
proof
let n be non zero Nat;
assume
A2: P[n];
let k be Nat;
assume
A3: k + 1 < 2 to_power (n+1);
then
A4: k < 2 to_power (n+1) by NAT_1:13;
now
per cases by XXREAL_0:1;
suppose
A5: k + 1 < 2 to_power n;
then
A6: k < 2 to_power n by NAT_1:13;
A7: add_ovfl(n-BinarySequence k,Bin1 n) = FALSE by A5,Th32;
thus (n+1)-BinarySequence (k+1) = (n-BinarySequence (k+1))^<*FALSE*>
by A5,Th27
.= (n-BinarySequence k + Bin1 n)^<*FALSE 'xor' FALSE 'xor'
add_ovfl(n-BinarySequence k,Bin1 n)*> by A2,A5,A7
.= ((n-BinarySequence k)^<*FALSE*>) + (Bin1 n^<*FALSE*>) by
BINARITH:19
.= ((n-BinarySequence k)^<*FALSE*>) + Bin1 (n+1) by BINARI_2:7
.= (n+1)-BinarySequence k + Bin1 (n+1) by A6,Th27;
end;
suppose
A8: k + 1 > 2 to_power n;
then
A9: k >= 2 to_power n by NAT_1:13;
k + 1 < (2 to_power n) * (2 to_power 1) by A3,POWER:27;
then k + 1 < (2 to_power n) * 2 by POWER:25;
then k + 1 < 2 to_power n + 2 to_power n;
then k + 1 - 2 to_power n < 2 to_power n by XREAL_1:19;
then k - 2 to_power n + 1 < 2 to_power n;
then
A10: k -' 2 to_power n + 1 < 2 to_power n by A9,XREAL_1:233;
then
A11: add_ovfl(n-BinarySequence (k -' 2 to_power n),Bin1 n) = FALSE by Th32;
thus (n+1)-BinarySequence (k+1) = (n-BinarySequence (k + 1 -' 2
to_power n))^<*TRUE*> by A3,A8,Lm3
.= (n-BinarySequence (k-'2 to_power n + 1))^<*TRUE*> by A9,NAT_D:38
.= (n-BinarySequence (k -' 2 to_power n) + Bin1 n)^<*TRUE 'xor'
FALSE 'xor' add_ovfl(n-BinarySequence (k -' 2 to_power n),Bin1 n)*> by A2,A10
,A11
.= (n-BinarySequence (k -' 2 to_power n))^<*TRUE*> + Bin1 n^<*
FALSE*> by BINARITH:19
.= (n-BinarySequence (k -' 2 to_power n))^<*TRUE*> + Bin1 (n+1) by
BINARI_2:7
.= (n+1)-BinarySequence k + Bin1 (n+1) by A4,A9,Lm3;
end;
suppose
A12: k + 1 = 2 to_power n;
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider z = 0*n as Tuple of n,BOOLEAN;
A13: k < 2 to_power n by A12,NAT_1:13;
k = 2 to_power n - 1 by A12;
then
A14: n-BinarySequence k = 'not' z by A13,Lm4;
thus (n+1)-BinarySequence (k+1) = 0*n^<*1*> by A12,Th28
.= ('not' z + Bin1 n)^<*TRUE*> by Th24
.= (n-BinarySequence k + Bin1 n)^<*FALSE 'xor' FALSE 'xor'
add_ovfl(n-BinarySequence k,Bin1 n)*> by A14,Th23
.= (n-BinarySequence k)^<*FALSE*> + Bin1 n^<*FALSE*> by BINARITH:19
.= (n-BinarySequence k)^<*FALSE*> + Bin1 (n+1) by BINARI_2:7
.= (n+1)-BinarySequence k + Bin1 (n+1) by A13,Th27;
end;
end;
hence thesis;
end;
A15: P[1]
proof
0*1 in BOOLEAN* by Th4;
then
0*1 is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider x = 0*1 as Tuple of 1,BOOLEAN;
let k be Nat;
A16: 0*1 = <*FALSE*> by FINSEQ_2:59;
assume
A17: k + 1 < 2 to_power 1;
then k + 1 < 1 + 1 by POWER:25;
then k < 1 by XREAL_1:6;
then
A18: k = 0 by NAT_1:14;
then k + 1 = 2 - 1 .= 2 to_power 1 - 1 by POWER:25;
hence 1-BinarySequence (k+1) = 'not' x by A17,Lm4
.= 1-BinarySequence k + Bin1 1 by A18,A16,Th10,Th14,Th17,Th25;
end;
thus for n being non zero Nat holds P[n] from NAT_1:sch 10(A15,A1);
end;
theorem
for n,i be Nat holds (n+1)-BinarySequence i = <*i mod 2*> ^ (n
-BinarySequence (i div 2))
proof
let n,i be Nat;
A1: len ((n+1)-BinarySequence i) = n + 1 by CARD_1:def 7;
then
A2: dom ((n+1)-BinarySequence i) = Seg(n+1) by FINSEQ_1:def 3;
A3: len (<*i mod 2*> ^ (n-BinarySequence (i div 2))) = 1 + len (n
-BinarySequence (i div 2)) by FINSEQ_5:8
.= n + 1 by CARD_1:def 7;
now
let j be Nat;
reconsider z = j as Nat;
assume
A4: j in dom ((n+1)-BinarySequence i);
then
A5: 1 <= j by A2,FINSEQ_1:1;
A6: j <= n + 1 by A2,A4,FINSEQ_1:1;
A7: len <*i mod 2*> = 1 by FINSEQ_1:39;
now
per cases by A5,XXREAL_0:1;
suppose
A8: j > 1;
A9: 2 to_power (j-'1-'1 + 1) = (2 to_power (j-'1-'1)) * (2 to_power 1
) by POWER:27
.= 2 * 2 to_power (j-'1-'1) by POWER:25;
j - 1 > 1 - 1 by A8,XREAL_1:9;
then j-'1 > 0 by A5,XREAL_1:233;
then
A10: j-'1 >= 0 + 1 by NAT_1:13;
then
A11: i div 2 to_power (j-'1) = i div 2 to_power (j-'1-'1 + 1) by XREAL_1:235
.= (i div 2) div 2 to_power (j-'1-'1) by A9,NAT_2:27;
j - 1 <= n by A6,XREAL_1:20;
then
A12: j-'1 <= n by A5,XREAL_1:233;
then
A13: j-'1 <= len (n-BinarySequence (i div 2)) by CARD_1:def 7;
A14: j-'1 in Seg n by A10,A12,FINSEQ_1:1;
j <= len ((n+1)-BinarySequence i) by A6,CARD_1:def 7;
hence ((n+1)-BinarySequence i).j = ((n+1)-BinarySequence i)/.z by A5,
FINSEQ_4:15
.= IFEQ((i div 2 to_power (j-'1)) mod 2,0,FALSE,TRUE) by A2,A4,Def1
.= (n-BinarySequence (i div 2))/.(j-'1) by A14,A11,Def1
.= (n-BinarySequence (i div 2)).(j-'1) by A10,A13,FINSEQ_4:15
.= (n-BinarySequence (i div 2)).(j - 1) by A5,XREAL_1:233
.= (<*i mod 2*> ^ (n-BinarySequence (i div 2))).j by A3,A6,A7,A8,
FINSEQ_1:24;
end;
suppose
A15: j = 1;
A16: now
per cases;
suppose
i mod 2 = 0;
hence IFEQ(i mod 2,0,FALSE,TRUE) = i mod 2 by FUNCOP_1:def 8;
end;
suppose
A17: i mod 2 <> 0;
hence IFEQ(i mod 2,0,FALSE,TRUE) = 1 by FUNCOP_1:def 8
.= i mod 2 by A17,NAT_D:12;
end;
end;
A18: 2 to_power 0 = 1 by POWER:24;
thus ((n+1)-BinarySequence i).j = ((n+1)-BinarySequence i)/.z by A1,A5
,A6,FINSEQ_4:15
.= IFEQ((i div 2 to_power (1-'1)) mod 2,0,FALSE,TRUE) by A2,A4,A15
,Def1
.= IFEQ((i div 1) mod 2,0,FALSE,TRUE) by A18,XREAL_1:232
.= IFEQ(i mod 2,0,FALSE,TRUE)
.= (<*i mod 2*> ^ (n-BinarySequence (i div 2))).j by A15,A16,
FINSEQ_1:41;
end;
end;
hence ((n+1)-BinarySequence i).j = (<*i mod 2*> ^ (n-BinarySequence (i div
2))).j;
end;
hence thesis by A1,A3,FINSEQ_2:9;
end;
theorem Th35:
for n be non zero Nat for k be Nat holds k < 2 to_power n
implies Absval (n-BinarySequence k) = k
proof
let n be non zero Nat;
defpred P[Nat] means $1 < 2 to_power n implies Absval (n-BinarySequence $1)
= $1;
A1: for k be Nat st P[k] holds P[k+1]
proof
0*n in BOOLEAN* by Th4;
then
0*n is FinSequence of BOOLEAN by FINSEQ_1:def 11;
then reconsider 0n = 0*n as Tuple of n,BOOLEAN;
let k be Nat;
assume
A2: k < 2 to_power n implies Absval (n-BinarySequence k) = k;
assume
A3: k + 1 < 2 to_power n;
then
A4: k + 1 - 1 < 2 to_power n - 1 by XREAL_1:9;
k < 2 to_power n by A3,NAT_1:13;
then n-BinarySequence k <> 'not' 0n by A4,Lm4;
then add_ovfl(n-BinarySequence k,Bin1 n) <> TRUE by Th23;
then add_ovfl(n-BinarySequence k,Bin1 n) = FALSE by XBOOLEAN:def 3;
then
A5: n-BinarySequence k,Bin1 n are_summable by BINARITH:def 7;
thus Absval (n-BinarySequence (k+1)) = Absval (n-BinarySequence k + Bin1 n
) by A3,Th33
.= Absval (n-BinarySequence k) + Absval (Bin1 n) by A5,BINARITH:22
.= k + 1 by A2,A3,Th11,NAT_1:13;
end;
A6: P[0]
proof
assume 0 < 2 to_power n;
n-BinarySequence 0 = 0*n by Th25;
hence thesis by Th6;
end;
thus for n being Nat holds P[n] from NAT_1:sch 2(A6,A1);
end;
theorem
for n be non zero Nat for x be Tuple of n,BOOLEAN holds n
-BinarySequence (Absval x) = x
proof
let n be non zero Nat;
let x be Tuple of n,BOOLEAN;
Absval x < 2 to_power n by Th1;
then Absval (n-BinarySequence Absval x) = Absval x by Th35;
hence thesis by Th2;
end;