ami_5.miz
begin
reserve x,y for
set;
theorem ::
AMI_5:1
Th1: for dl be
Data-Location holds ex i be
Nat st dl
= (
dl. i)
proof
let dl be
Data-Location;
dl
in (
Data-Locations
SCM ) by
AMI_2:def 16,
AMI_3: 27;
then
consider x,y be
object such that
A1: x
in
{1} and
A2: y
in
NAT and
A3: dl
=
[x, y] by
AMI_3: 27,
ZFMISC_1: 84;
reconsider k = y as
Nat by
A2;
take k;
thus thesis by
A1,
A3,
TARSKI:def 1;
end;
theorem ::
AMI_5:2
Th2: for dl be
Data-Location holds dl
<> (
IC
SCM ) by
Th1,
AMI_3: 13;
theorem ::
AMI_5:3
for il be
Nat, dl be
Data-Location holds il
<> dl
proof
let il be
Nat, dl be
Data-Location;
ex j be
Nat st dl
= (
dl. j) by
Th1;
hence thesis;
end;
reserve i,j,k for
Nat;
theorem ::
AMI_5:4
for s be
State of
SCM , d be
Data-Location holds d
in (
dom s)
proof
let s be
State of
SCM , d be
Data-Location;
A1: (
dom s)
= the
carrier of
SCM by
PARTFUN1:def 2;
thus d
in (
dom s) by
A1;
end;
registration
cluster (
Data-Locations
SCM ) ->
infinite;
coherence by
AMI_3: 27;
end
reserve I,J,K for
Element of (
Segm 9),
a,a1 for
Nat,
b,b1,c for
Element of (
Data-Locations
SCM );
Lm1: b is
Data-Location
proof
b
in (
Data-Locations
SCM );
then
reconsider b as
Object of
SCM ;
b is
Data-Location by
AMI_2:def 16,
AMI_3: 27;
hence thesis;
end;
theorem ::
AMI_5:5
for l be
Instruction of
SCM holds (
InsCode l)
<= 8
proof
let l be
Instruction of
SCM ;
l
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) or l
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
A1: l
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) or l
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} } or l
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
XBOOLE_0:def 3;
per cases by
A1,
XBOOLE_0:def 3;
suppose l
in
{
[
SCM-Halt ,
{} ,
{} ]};
then l
=
[
SCM-Halt ,
{} ,
{} ] by
TARSKI:def 1;
then (l
`1_3 )
=
0 ;
hence thesis;
end;
suppose l
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st l
=
[J,
<*a*>,
{} ] & J
= 6;
then (l
`1_3 )
= 6;
hence thesis;
end;
suppose l
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then ex K, a1, b1 st l
=
[K,
<*a1*>,
<*b1*>] & K
in
{7, 8};
then (l
`1_3 )
in
{7, 8};
then (l
`1_3 )
= 7 or (l
`1_3 )
= 8 by
TARSKI:def 2;
hence thesis;
end;
suppose l
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} };
then ex I, b, c st l
=
[I,
{} ,
<*b, c*>] & I
in
{1, 2, 3, 4, 5};
then (l
`1_3 )
in
{1, 2, 3, 4, 5};
then (l
`1_3 )
= 1 or (l
`1_3 )
= 2 or (l
`1_3 )
= 3 or (l
`1_3 )
= 4 or (l
`1_3 )
= 5 by
ENUMSET1:def 3;
hence thesis;
end;
end;
reserve a,b for
Data-Location,
loc for
Nat;
reserve I,J,K for
Element of (
Segm 9),
a,a1 for
Nat,
b,b1,c for
Element of (
Data-Locations
SCM ),
da,db for
Data-Location;
::$Canceled
theorem ::
AMI_5:7
for ins be
Instruction of
SCM st (
InsCode ins)
=
0 holds ins
= (
halt
SCM )
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
=
0 ;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
then (
InsCode ins)
= 6;
hence contradiction by
A1;
end;
now
assume ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} };
then ex I, b, c st ins
=
[I,
{} ,
<*b, c*>] & I
in
{1, 2, 3, 4, 5};
then (
InsCode ins)
in
{1, 2, 3, 4, 5};
hence contradiction by
A1,
ENUMSET1:def 3;
end;
then
A3: ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
AMI_3: 27,
XBOOLE_0:def 3;
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then ex K, a1, b1 st ins
=
[K,
<*a1*>,
<*b1*>] & K
in
{7, 8};
then (
InsCode ins)
in
{7, 8};
hence contradiction by
A1,
TARSKI:def 2;
end;
then ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A3,
XBOOLE_0:def 3;
then ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A2,
XBOOLE_0:def 3;
then ins
=
[
SCM-Halt ,
{} ,
{} ] by
TARSKI:def 1;
hence thesis by
AMI_3: 26;
end;
theorem ::
AMI_5:8
for ins be
Instruction of
SCM st (
InsCode ins)
= 1 holds ex da, db st ins
= (da
:= db)
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 1;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
A3:
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A4: ins
=
[K,
<*a1*>,
<*b1*>] and
A5: K
in
{7, 8};
(
InsCode ins)
= K by
A4;
hence contradiction by
A1,
A5,
TARSKI:def 2;
end;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then not ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
A3,
XBOOLE_0:def 3;
then ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
consider I, b, c such that
A6: ins
=
[I,
{} ,
<*b, c*>] and I
in
{1, 2, 3, 4, 5};
reconsider da = b, db = c as
Data-Location by
Lm1;
take da, db;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:9
for ins be
Instruction of
SCM st (
InsCode ins)
= 2 holds ex da, db st ins
= (
AddTo (da,db))
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 2;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
A3:
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A4: ins
=
[K,
<*a1*>,
<*b1*>] and
A5: K
in
{7, 8};
(
InsCode ins)
= K by
A4;
hence contradiction by
A1,
A5,
TARSKI:def 2;
end;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then not ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
A3,
XBOOLE_0:def 3;
then ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
consider I, b, c such that
A6: ins
=
[I,
{} ,
<*b, c*>] and I
in
{1, 2, 3, 4, 5};
reconsider da = b, db = c as
Data-Location by
Lm1;
take da, db;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:10
for ins be
Instruction of
SCM st (
InsCode ins)
= 3 holds ex da, db st ins
= (
SubFrom (da,db))
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 3;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
A3:
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A4: ins
=
[K,
<*a1*>,
<*b1*>] and
A5: K
in
{7, 8};
(
InsCode ins)
= K by
A4;
hence contradiction by
A1,
A5,
TARSKI:def 2;
end;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then not ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
A3,
XBOOLE_0:def 3;
then ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
consider I, b, c such that
A6: ins
=
[I,
{} ,
<*b, c*>] and I
in
{1, 2, 3, 4, 5};
reconsider da = b, db = c as
Data-Location by
Lm1;
take da, db;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:11
for ins be
Instruction of
SCM st (
InsCode ins)
= 4 holds ex da, db st ins
= (
MultBy (da,db))
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 4;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
A3:
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A4: ins
=
[K,
<*a1*>,
<*b1*>] and
A5: K
in
{7, 8};
(
InsCode ins)
= K by
A4;
hence contradiction by
A1,
A5,
TARSKI:def 2;
end;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then not ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
A3,
XBOOLE_0:def 3;
then ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
consider I, b, c such that
A6: ins
=
[I,
{} ,
<*b, c*>] and I
in
{1, 2, 3, 4, 5};
reconsider da = b, db = c as
Data-Location by
Lm1;
take da, db;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:12
for ins be
Instruction of
SCM st (
InsCode ins)
= 5 holds ex da, db st ins
= (
Divide (da,db))
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 5;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
A3:
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A4: ins
=
[K,
<*a1*>,
<*b1*>] and
A5: K
in
{7, 8};
(
InsCode ins)
= K by
A4;
hence contradiction by
A1,
A5,
TARSKI:def 2;
end;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then not ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
A3,
XBOOLE_0:def 3;
then ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} } by
AMI_3: 27,
XBOOLE_0:def 3;
then
consider I, b, c such that
A6: ins
=
[I,
{} ,
<*b, c*>] and I
in
{1, 2, 3, 4, 5};
reconsider da = b, db = c as
Data-Location by
Lm1;
take da, db;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:13
for ins be
Instruction of
SCM st (
InsCode ins)
= 6 holds ex loc st ins
= (
SCM-goto loc)
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 6;
now
assume ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} };
then
consider I, b, c such that
A2: ins
=
[I,
{} ,
<*b, c*>] and
A3: I
in
{1, 2, 3, 4, 5};
(
InsCode ins)
= I by
A2;
hence contradiction by
A1,
A3,
ENUMSET1:def 3;
end;
then
A4: ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
AMI_3: 27,
XBOOLE_0:def 3;
now
assume ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} };
then
consider K, a1, b1 such that
A5: ins
=
[K,
<*a1*>,
<*b1*>] and
A6: K
in
{7, 8};
(
InsCode ins)
= K by
A5;
hence contradiction by
A1,
A6,
TARSKI:def 2;
end;
then
A7: ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A4,
XBOOLE_0:def 3;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then ins
in {
[J,
<*a*>,
{} ] : J
= 6 } by
A7,
XBOOLE_0:def 3;
then
consider J, a such that
A8: ins
=
[J,
<*a*>,
{} ] & J
= 6;
reconsider loc = a as
Nat;
take loc;
thus thesis by
A8;
end;
theorem ::
AMI_5:14
for ins be
Instruction of
SCM st (
InsCode ins)
= 7 holds ex loc, da st ins
= (da
=0_goto loc)
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 7;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
now
assume ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} };
then
consider I, b, c such that
A3: ins
=
[I,
{} ,
<*b, c*>] and
A4: I
in
{1, 2, 3, 4, 5};
(
InsCode ins)
= I by
A3;
hence contradiction by
A1,
A4,
ENUMSET1:def 3;
end;
then
A5: ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
AMI_3: 27,
XBOOLE_0:def 3;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} } by
A5,
XBOOLE_0:def 3;
then
consider K, a1, b1 such that
A6: ins
=
[K,
<*a1*>,
<*b1*>] and K
in
{7, 8};
reconsider da = b1 as
Data-Location by
Lm1;
reconsider loc = a1 as
Nat;
take loc, da;
thus thesis by
A1,
A6;
end;
theorem ::
AMI_5:15
for ins be
Instruction of
SCM st (
InsCode ins)
= 8 holds ex loc, da st ins
= (da
>0_goto loc)
proof
let ins be
Instruction of
SCM such that
A1: (
InsCode ins)
= 8;
A2:
now
assume ins
in {
[J,
<*a*>,
{} ] : J
= 6 };
then ex J, a st ins
=
[J,
<*a*>,
{} ] & J
= 6;
hence contradiction by
A1;
end;
now
assume ins
in {
[I,
{} ,
<*b, c*>] : I
in
{1, 2, 3, 4, 5} };
then
consider I, b, c such that
A3: ins
=
[I,
{} ,
<*b, c*>] and
A4: I
in
{1, 2, 3, 4, 5};
(
InsCode ins)
= I by
A3;
hence contradiction by
A1,
A4,
ENUMSET1:def 3;
end;
then
A5: ins
in ((
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 })
\/ {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} }) by
AMI_3: 27,
XBOOLE_0:def 3;
(
InsCode (
halt
SCM ))
=
0 by
COMPOS_1: 70;
then not ins
in
{
[
SCM-Halt ,
{} ,
{} ]} by
A1,
AMI_3: 26,
TARSKI:def 1;
then not ins
in (
{
[
SCM-Halt ,
{} ,
{} ]}
\/ {
[J,
<*a*>,
{} ] : J
= 6 }) by
A2,
XBOOLE_0:def 3;
then ins
in {
[K,
<*a1*>,
<*b1*>] : K
in
{7, 8} } by
A5,
XBOOLE_0:def 3;
then
consider K, a1, b1 such that
A6: ins
=
[K,
<*a1*>,
<*b1*>] and K
in
{7, 8};
reconsider da = b1 as
Data-Location by
Lm1;
reconsider loc = a1 as
Nat;
take loc, da;
thus thesis by
A1,
A6;
end;
begin
theorem ::
AMI_5:16
for s be
State of
SCM , iloc be
Nat, a be
Data-Location holds (s
. a)
= ((s
+* (
Start-At (iloc,
SCM )))
. a)
proof
let s be
State of
SCM , iloc be
Nat, a be
Data-Location;
a
in the
carrier of
SCM ;
then a
in (
dom s) by
PARTFUN1:def 2;
then
A1: (
dom (
Start-At (iloc,
SCM )))
=
{(
IC
SCM )} & a
in ((
dom s)
\/ (
dom (
Start-At (iloc,
SCM )))) by
XBOOLE_0:def 3;
a
<> (
IC
SCM ) by
Th2;
then not a
in
{(
IC
SCM )} by
TARSKI:def 1;
hence thesis by
A1,
FUNCT_4:def 1;
end;
begin
registration
cluster
SCM ->
IC-recognized;
coherence
proof
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic
FinPartState of
SCM st (
DataPart p)
<>
{} holds (
IC
SCM )
in (
dom p)
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic
FinPartState of
SCM ;
assume (
DataPart p)
<>
{} ;
then
A1: (
dom (
DataPart p))
<>
{} ;
assume
A2: not (
IC
SCM )
in (
dom p);
then (
dom p)
misses
{(
IC
SCM )} by
ZFMISC_1: 50;
then
A3: ((
dom p)
/\
{(
IC
SCM )})
=
{} by
XBOOLE_0:def 7;
not p is q
-autonomic
proof
set il = the
Element of (
NAT
\ (
dom q));
set d2 = the
Element of ((
Data-Locations
SCM )
\ (
dom p));
set d1 = the
Element of (
dom (
DataPart p));
A4: d1
in (
dom (
DataPart p)) by
A1;
(
DataPart p)
c= p by
MEMSTR_0: 12;
then
A5: (
dom (
DataPart p))
c= (
dom p) by
RELAT_1: 11;
(
dom (
DataPart p))
c= the
carrier of
SCM by
RELAT_1:def 18;
then
reconsider d1 as
Element of
SCM by
A4;
not (
Data-Locations
SCM )
c= (
dom p);
then
A6: ((
Data-Locations
SCM )
\ (
dom p))
<>
{} by
XBOOLE_1: 37;
then d2
in (
Data-Locations
SCM ) by
XBOOLE_0:def 5;
then
reconsider d2 as
Data-Location by
AMI_2:def 16,
AMI_3: 27;
A7: not d2
in (
dom p) by
A6,
XBOOLE_0:def 5;
then
A8: (
dom p)
misses
{d2} by
ZFMISC_1: 50;
not
NAT
c= (
dom q);
then
A9: (
NAT
\ (
dom q))
<>
{} by
XBOOLE_1: 37;
then
reconsider il as
Element of
NAT by
XBOOLE_0:def 5;
A10: not il
in (
dom q) by
A9,
XBOOLE_0:def 5;
(
dom (
DataPart p))
c= (
Data-Locations
SCM ) by
RELAT_1: 58;
then
reconsider d1 as
Data-Location by
A4,
AMI_2:def 16,
AMI_3: 27;
set p2 = (p
+* ((d2
.--> 1)
+* (
Start-At (il,
SCM ))));
set p1 = (p
+* ((d2
.-->
0 )
+* (
Start-At (il,
SCM ))));
set q2 = (q
+* (il
.--> (d1
:= d2)));
set q1 = (q
+* (il
.--> (d1
:= d2)));
consider s1 be
State of
SCM such that
A11: p1
c= s1 by
PBOOLE: 141;
consider S1 be
Instruction-Sequence of
SCM such that
A12: q1
c= S1 by
PBOOLE: 145;
A13: (
dom p)
misses
{d2} by
A7,
ZFMISC_1: 50;
A14: (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM ))))
= ((
dom (d2
.--> 1))
\/ (
dom (
Start-At (il,
SCM )))) by
FUNCT_4:def 1;
consider s2 be
State of
SCM such that
A15: p2
c= s2 by
PBOOLE: 141;
consider S2 be
Instruction-Sequence of
SCM such that
A16: q2
c= S2 by
PBOOLE: 145;
A17: (
dom p)
c= the
carrier of
SCM by
RELAT_1:def 18;
(
dom (
Comput (S2,s2,1)))
= the
carrier of
SCM by
PARTFUN1:def 2;
then
A18: (
dom ((
Comput (S2,s2,1))
| (
dom p)))
= (
dom p) by
A17,
RELAT_1: 62;
A19: (
dom (
Comput (S1,s1,1)))
= the
carrier of
SCM by
PARTFUN1:def 2;
A20: (
dom ((
Comput (S1,s1,1))
| (
dom p)))
= (
dom p) by
A17,
A19,
RELAT_1: 62;
A21: (
dom p2)
= ((
dom p)
\/ (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM ))))) by
FUNCT_4:def 1;
A22: (
dom q2)
= ((
dom q)
\/ (
dom (il
.--> (d1
:= d2)))) by
FUNCT_4:def 1;
A24: (
IC
SCM )
in (
dom (
Start-At (il,
SCM ))) by
TARSKI:def 1;
then
A25: (
IC
SCM )
in (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM )))) by
A14,
XBOOLE_0:def 3;
then (
IC
SCM )
in (
dom p2) by
A21,
XBOOLE_0:def 3;
then
A26: (
IC s2)
= (p2
. (
IC
SCM )) by
A15,
GRFUNC_1: 2
.= (((d2
.--> 1)
+* (
Start-At (il,
SCM )))
. (
IC
SCM )) by
A25,
FUNCT_4: 13
.= ((
Start-At (il,
SCM ))
. (
IC
SCM )) by
A24,
FUNCT_4: 13
.= il by
FUNCOP_1: 72;
d2
<> (
IC
SCM ) by
Th2;
then
A27: not d2
in (
dom (
Start-At (il,
SCM ))) by
TARSKI:def 1;
d2
in (
dom (d2
.--> 1)) by
TARSKI:def 1;
then
A28: d2
in (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM )))) by
A14,
XBOOLE_0:def 3;
then d2
in (
dom p2) by
A21,
XBOOLE_0:def 3;
then
A29: (s2
. d2)
= (p2
. d2) by
A15,
GRFUNC_1: 2
.= (((d2
.--> 1)
+* (
Start-At (il,
SCM )))
. d2) by
A28,
FUNCT_4: 13
.= ((d2
.--> 1)
. d2) by
A27,
FUNCT_4: 11
.= 1 by
FUNCOP_1: 72;
A31: il
in (
dom (il
.--> (d1
:= d2))) by
TARSKI:def 1;
then il
in (
dom q2) by
A22,
XBOOLE_0:def 3;
then
A32: (S2
. il)
= (q2
. il) by
A16,
GRFUNC_1: 2
.= ((il
.--> (d1
:= d2))
. il) by
A31,
FUNCT_4: 13
.= (d1
:= d2) by
FUNCOP_1: 72;
A33: (S2
/. (
IC s2))
= (S2
. (
IC s2)) by
PBOOLE: 143;
A34: ((
Comput (S2,s2,(
0
+ 1)))
. d1)
= ((
Following (S2,(
Comput (S2,s2,
0 ))))
. d1) by
EXTPRO_1: 3
.= ((
Following (S2,s2))
. d1)
.= 1 by
A26,
A32,
A29,
A33,
AMI_3: 2;
(
dom p)
misses
{(
IC
SCM )} by
A2,
ZFMISC_1: 50;
then
A35: ((
dom p)
/\
{(
IC
SCM )})
=
{} by
XBOOLE_0:def 7;
take P = S1, Q = S2;
(
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM ))))
= ((
dom (d2
.-->
0 ))
\/ (
dom (
Start-At (il,
SCM )))) by
FUNCT_4:def 1
.= ((
dom (d2
.-->
0 ))
\/
{(
IC
SCM )})
.= (
{d2}
\/
{(
IC
SCM )});
then ((
dom p)
/\ (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM )))))
= (((
dom p)
/\
{d2})
\/
{} ) by
A35,
XBOOLE_1: 23
.=
{} by
A8,
XBOOLE_0:def 7;
then (
dom p)
misses (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM )))) by
XBOOLE_0:def 7;
then p
c= p1 by
FUNCT_4: 32;
then
A36: p
c= s1 by
A11,
XBOOLE_1: 1;
(
dom q)
misses (
dom (il
.--> (d1
:= d2))) by
A10,
ZFMISC_1: 50;
then q
c= q1 by
FUNCT_4: 32;
hence q
c= P by
A12,
XBOOLE_1: 1;
A37: (
dom p1)
= ((
dom p)
\/ (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM ))))) by
FUNCT_4:def 1;
(
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM ))))
= ((
dom (d2
.--> 1))
\/ (
dom (
Start-At (il,
SCM )))) by
FUNCT_4:def 1
.= ((
dom (d2
.--> 1))
\/
{(
IC
SCM )})
.= (
{d2}
\/
{(
IC
SCM )});
then ((
dom p)
/\ (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM )))))
= (((
dom p)
/\
{d2})
\/
{} ) by
A3,
XBOOLE_1: 23
.=
{} by
A13,
XBOOLE_0:def 7;
then (
dom p)
misses (
dom ((d2
.--> 1)
+* (
Start-At (il,
SCM )))) by
XBOOLE_0:def 7;
then p
c= p2 by
FUNCT_4: 32;
then
A38: p
c= s2 by
A15,
XBOOLE_1: 1;
(
dom q)
misses (
dom (il
.--> (d1
:= d2))) by
A10,
ZFMISC_1: 50;
then q
c= q2 by
FUNCT_4: 32;
hence q
c= Q by
A16,
XBOOLE_1: 1;
take s1, s2;
thus p
c= s1 by
A36;
thus p
c= s2 by
A38;
take 1;
A39: (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM ))))
= ((
dom (d2
.-->
0 ))
\/ (
dom (
Start-At (il,
SCM )))) by
FUNCT_4:def 1;
A41: (
IC
SCM )
in (
dom (
Start-At (il,
SCM ))) by
TARSKI:def 1;
then
A42: (
IC
SCM )
in (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM )))) by
A39,
XBOOLE_0:def 3;
then (
IC
SCM )
in (
dom p1) by
A37,
XBOOLE_0:def 3;
then
A43: (
IC s1)
= (p1
. (
IC
SCM )) by
A11,
GRFUNC_1: 2
.= (((d2
.-->
0 )
+* (
Start-At (il,
SCM )))
. (
IC
SCM )) by
A42,
FUNCT_4: 13
.= ((
Start-At (il,
SCM ))
. (
IC
SCM )) by
A41,
FUNCT_4: 13
.= il by
FUNCOP_1: 72;
d2
<> (
IC
SCM ) by
Th2;
then
A44: not d2
in (
dom (
Start-At (il,
SCM ))) by
TARSKI:def 1;
d2
in (
dom (d2
.-->
0 )) by
TARSKI:def 1;
then
A45: d2
in (
dom ((d2
.-->
0 )
+* (
Start-At (il,
SCM )))) by
A39,
XBOOLE_0:def 3;
then d2
in (
dom p1) by
A37,
XBOOLE_0:def 3;
then
A46: (s1
. d2)
= (p1
. d2) by
A11,
GRFUNC_1: 2
.= (((d2
.-->
0 )
+* (
Start-At (il,
SCM )))
. d2) by
A45,
FUNCT_4: 13
.= ((d2
.-->
0 )
. d2) by
A44,
FUNCT_4: 11
.=
0 by
FUNCOP_1: 72;
A47: il
in (
dom (il
.--> (d1
:= d2))) by
TARSKI:def 1;
(
dom q1)
= ((
dom q)
\/ (
dom (il
.--> (d1
:= d2)))) by
FUNCT_4:def 1;
then il
in (
dom q1) by
A47,
XBOOLE_0:def 3;
then
A48: (S1
. il)
= (q1
. il) by
A12,
GRFUNC_1: 2
.= ((il
.--> (d1
:= d2))
. il) by
A47,
FUNCT_4: 13
.= (d1
:= d2) by
FUNCOP_1: 72;
A49: (S1
/. (
IC s1))
= (S1
. (
IC s1)) by
PBOOLE: 143;
((
Comput (S1,s1,(
0
+ 1)))
. d1)
= ((
Following (S1,(
Comput (S1,s1,
0 ))))
. d1) by
EXTPRO_1: 3
.=
0 by
A43,
A48,
A46,
A49,
AMI_3: 2;
then (((
Comput (P,s1,1))
| (
dom p))
. d1)
=
0 by
A4,
A5,
A20,
FUNCT_1: 47;
hence ((
Comput (P,s1,1))
| (
dom p))
<> ((
Comput (Q,s2,1))
| (
dom p)) by
A18,
A34,
A4,
A5,
FUNCT_1: 47;
end;
hence contradiction;
end;
hence thesis by
AMISTD_5: 3;
end;
end
registration
cluster
SCM ->
CurIns-recognized;
coherence
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s be
State of
SCM such that
A1: p
c= s;
let P be
Instruction-Sequence of
SCM such that
A2: q
c= P;
let i be
Nat;
set Csi = (
Comput (P,s,i));
set loc = (
IC Csi);
assume
A3: not (
IC (
Comput (P,s,i)))
in (
dom q);
set I = ((
dl.
0 )
:= (
dl.
0 ));
set q1 = (q
+* (loc
.--> I));
set q2 = (q
+* (loc
.--> (
halt
SCM )));
reconsider P1 = (P
+* (loc
.--> I)) as
Instruction-Sequence of
SCM ;
reconsider P2 = (P
+* (loc
.--> (
halt
SCM ))) as
Instruction-Sequence of
SCM ;
A5: loc
in (
dom (loc
.--> (
halt
SCM ))) by
TARSKI:def 1;
A7: loc
in (
dom (loc
.--> I)) by
TARSKI:def 1;
A8: (
dom q)
misses (
dom (loc
.--> (
halt
SCM ))) by
A3,
ZFMISC_1: 50;
A9: (
dom q)
misses (
dom (loc
.--> I)) by
A3,
ZFMISC_1: 50;
A10: q1
c= P1 by
A2,
FUNCT_4: 123;
A11: q2
c= P2 by
A2,
FUNCT_4: 123;
set Cs2i = (
Comput (P2,s,i)), Cs1i = (
Comput (P1,s,i));
not p is q
-autonomic
proof
((loc
.--> (
halt
SCM ))
. loc)
= (
halt
SCM ) by
FUNCOP_1: 72;
then
A12: (P2
. loc)
= (
halt
SCM ) by
A5,
FUNCT_4: 13;
A13: ((loc
.--> I)
. loc)
= I by
FUNCOP_1: 72;
take P1, P2;
q
c= q1 by
A9,
FUNCT_4: 32;
hence
A14: q
c= P1 by
A10,
XBOOLE_1: 1;
q
c= q2 by
A8,
FUNCT_4: 32;
hence
A15: q
c= P2 by
A11,
XBOOLE_1: 1;
take s, s;
thus p
c= s by
A1;
A16: (Cs1i
| (
dom p))
= (Csi
| (
dom p)) by
A14,
A2,
A1,
EXTPRO_1:def 10;
thus p
c= s by
A1;
A17: (Cs1i
| (
dom p))
= (Cs2i
| (
dom p)) by
A14,
A15,
A1,
EXTPRO_1:def 10;
take k = (i
+ 1);
set Cs1k = (
Comput (P1,s,k));
A18: (
IC
SCM )
in (
dom p) by
AMISTD_5: 6;
(
IC Csi)
= (
IC (Csi
| (
dom p))) by
A18,
FUNCT_1: 49;
then (
IC Cs1i)
= loc by
A16,
A18,
FUNCT_1: 49;
then
A19: (
CurInstr (P1,Cs1i))
= (P1
. loc) by
PBOOLE: 143
.= I by
A13,
A7,
FUNCT_4: 13;
A20: Cs1k
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec (I,Cs1i)) by
A19;
A21: (
IC (
Exec (I,Cs1i)))
= ((
IC Cs1i)
+ 1) by
AMI_3: 2;
A22: (
IC
SCM )
in (
dom p) by
AMISTD_5: 6;
A23: (
IC Csi)
= (
IC (Csi
| (
dom p))) by
A22,
FUNCT_1: 49;
then
A24: (
IC Cs1k)
= (loc
+ 1) by
A20,
A21,
A16,
A22,
FUNCT_1: 49;
set Cs2k = (
Comput (P2,s,k));
A25: Cs2k
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
A26: (P2
/. (
IC Cs2i))
= (P2
. (
IC Cs2i)) by
PBOOLE: 143;
(
IC Cs2i)
= loc by
A16,
A23,
A17,
A22,
FUNCT_1: 49;
then
A27: (
IC Cs2k)
= loc by
A25,
A12,
A26,
EXTPRO_1:def 3;
(
IC (Cs1k
| (
dom p)))
= (
IC Cs1k) & (
IC (Cs2k
| (
dom p)))
= (
IC Cs2k) by
A22,
FUNCT_1: 49;
hence thesis by
A24,
A27;
end;
hence contradiction;
end;
end
theorem ::
AMI_5:17
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (da
:= db) & da
in (
dom p) implies ((
Comput (P1,s1,i))
. db)
= ((
Comput (P2,s2,i))
. db)
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (da
:= db) and
A6: da
in (
dom p) & ((
Comput (P1,s1,i))
. db)
<> ((
Comput (P2,s2,i))
. db);
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A7: (Cs2i1
. da)
= (Cs2i
. db) by
A4,
A5,
AMI_3: 2;
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
A8: da
in (
dom p) implies ((Cs1i1
| (
dom p))
. da)
= (Cs1i1
. da) & ((Cs2i1
| (
dom p))
. da)
= (Cs2i1
. da) by
FUNCT_1: 49;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. da)
= (Cs1i
. db) by
A3,
A5,
AMI_3: 2;
hence contradiction by
A8,
A6,
A7,
A2,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:18
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (
AddTo (da,db)) & da
in (
dom p) implies (((
Comput (P1,s1,i))
. da)
+ ((
Comput (P1,s1,i))
. db))
= (((
Comput (P2,s2,i))
. da)
+ ((
Comput (P2,s2,i))
. db))
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (
AddTo (da,db)) and
A6: da
in (
dom p) & (((
Comput (P1,s1,i))
. da)
+ ((
Comput (P1,s1,i))
. db))
<> (((
Comput (P2,s2,i))
. da)
+ ((
Comput (P2,s2,i))
. db));
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A7: (Cs2i1
. da)
= ((Cs2i
. da)
+ (Cs2i
. db)) by
A4,
A5,
AMI_3: 3;
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
A8: da
in (
dom p) implies ((Cs1i1
| (
dom p))
. da)
= (Cs1i1
. da) & ((Cs2i1
| (
dom p))
. da)
= (Cs2i1
. da) by
FUNCT_1: 49;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. da)
= ((Cs1i
. da)
+ (Cs1i
. db)) by
A3,
A5,
AMI_3: 3;
hence contradiction by
A8,
A6,
A7,
A2,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:19
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (
SubFrom (da,db)) & da
in (
dom p) implies (((
Comput (P1,s1,i))
. da)
- ((
Comput (P1,s1,i))
. db))
= (((
Comput (P2,s2,i))
. da)
- ((
Comput (P2,s2,i))
. db))
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (
SubFrom (da,db)) and
A6: da
in (
dom p) & (((
Comput (P1,s1,i))
. da)
- ((
Comput (P1,s1,i))
. db))
<> (((
Comput (P2,s2,i))
. da)
- ((
Comput (P2,s2,i))
. db));
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A7: (Cs2i1
. da)
= ((Cs2i
. da)
- (Cs2i
. db)) by
A4,
A5,
AMI_3: 4;
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
A8: da
in (
dom p) implies ((Cs1i1
| (
dom p))
. da)
= (Cs1i1
. da) & ((Cs2i1
| (
dom p))
. da)
= (Cs2i1
. da) by
FUNCT_1: 49;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. da)
= ((Cs1i
. da)
- (Cs1i
. db)) by
A3,
A5,
AMI_3: 4;
hence contradiction by
A8,
A6,
A7,
A2,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:20
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (
MultBy (da,db)) & da
in (
dom p) implies (((
Comput (P1,s1,i))
. da)
* ((
Comput (P1,s1,i))
. db))
= (((
Comput (P2,s2,i))
. da)
* ((
Comput (P2,s2,i))
. db))
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (
MultBy (da,db)) and
A6: da
in (
dom p) & (((
Comput (P1,s1,i))
. da)
* ((
Comput (P1,s1,i))
. db))
<> (((
Comput (P2,s2,i))
. da)
* ((
Comput (P2,s2,i))
. db));
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A7: (Cs2i1
. da)
= ((Cs2i
. da)
* (Cs2i
. db)) by
A4,
A5,
AMI_3: 5;
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
A8: da
in (
dom p) implies ((Cs1i1
| (
dom p))
. da)
= (Cs1i1
. da) & ((Cs2i1
| (
dom p))
. da)
= (Cs2i1
. da) by
FUNCT_1: 49;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. da)
= ((Cs1i
. da)
* (Cs1i
. db)) by
A3,
A5,
AMI_3: 5;
hence contradiction by
A8,
A6,
A7,
A2,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:21
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (
Divide (da,db)) & da
in (
dom p) & da
<> db implies (((
Comput (P1,s1,i))
. da)
div ((
Comput (P1,s1,i))
. db))
= (((
Comput (P2,s2,i))
. da)
div ((
Comput (P2,s2,i))
. db))
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (
Divide (da,db)) and
A6: da
in (
dom p) and
A7: da
<> db and
A8: (((
Comput (P1,s1,i))
. da)
div ((
Comput (P1,s1,i))
. db))
<> (((
Comput (P2,s2,i))
. da)
div ((
Comput (P2,s2,i))
. db));
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A9: (Cs2i1
. da)
= ((Cs2i
. da)
div (Cs2i
. db)) by
A4,
A5,
A7,
AMI_3: 6;
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
A10: da
in (
dom p) implies ((Cs1i1
| (
dom p))
. da)
= (Cs1i1
. da) & ((Cs2i1
| (
dom p))
. da)
= (Cs2i1
. da) by
FUNCT_1: 49;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. da)
= ((Cs1i
. da)
div (Cs1i
. db)) by
A3,
A5,
A7,
AMI_3: 6;
hence contradiction by
A10,
A8,
A9,
A2,
A6,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:22
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (
Divide (da,db)) & db
in (
dom p) implies (((
Comput (P1,s1,i))
. da)
mod ((
Comput (P1,s1,i))
. db))
= (((
Comput (P2,s2,i))
. da)
mod ((
Comput (P2,s2,i))
. db))
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da,db be
Data-Location, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs1i = (
Comput (P1,s1,i));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
A4: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A5: I
= (
Divide (da,db)) and
A6: db
in (
dom p) and
A7: (((
Comput (P1,s1,i))
. da)
mod ((
Comput (P1,s1,i))
. db))
<> (((
Comput (P2,s2,i))
. da)
mod ((
Comput (P2,s2,i))
. db));
A8: ((Cs1i1
| (
dom p))
. db)
= (Cs1i1
. db) & ((Cs2i1
| (
dom p))
. db)
= (Cs2i1
. db) by
A6,
FUNCT_1: 49;
I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
then
A9: (Cs2i1
. db)
= ((Cs2i
. da)
mod (Cs2i
. db)) by
A4,
A5,
AMI_3: 6;
Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
then (Cs1i1
. db)
= ((Cs1i
. da)
mod (Cs1i
. db)) by
A3,
A5,
AMI_3: 6;
hence contradiction by
A7,
A8,
A9,
A2,
A1,
EXTPRO_1:def 10;
end;
theorem ::
AMI_5:23
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da be
Data-Location, loc be
Nat, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (da
=0_goto loc) & loc
<> ((
IC (
Comput (P1,s1,i)))
+ 1) implies (((
Comput (P1,s1,i))
. da)
=
0 iff ((
Comput (P2,s2,i))
. da)
=
0 )
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da be
Data-Location, loc be
Nat, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
set Cs2i = (
Comput (P2,s2,i));
set Cs1i = (
Comput (P1,s1,i));
A4: Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
A5: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
(
IC
SCM )
in (
dom p) by
AMISTD_5: 6;
then
A6: ((Cs1i1
| (
dom p))
. (
IC
SCM ))
= (
IC Cs1i1) & ((Cs2i1
| (
dom p))
. (
IC
SCM ))
= (
IC Cs2i1) by
FUNCT_1: 49;
assume that
A7: I
= (da
=0_goto loc) and
A8: loc
<> ((
IC (
Comput (P1,s1,i)))
+ 1);
A9: I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
A10:
now
assume ((
Comput (P2,s2,i))
. da)
=
0 & ((
Comput (P1,s1,i))
. da)
<>
0 ;
then (Cs2i1
. (
IC
SCM ))
= loc & (Cs1i1
. (
IC
SCM ))
= ((
IC Cs1i)
+ 1) by
A3,
A9,
A4,
A5,
A7,
AMI_3: 8;
hence contradiction by
A6,
A8,
A2,
A1,
EXTPRO_1:def 10;
end;
A11: (Cs1i1
| (
dom p))
= (Cs2i1
| (
dom p)) by
A2,
A1,
EXTPRO_1:def 10;
now
assume ((
Comput (P1,s1,i))
. da)
=
0 & ((
Comput (P2,s2,i))
. da)
<>
0 ;
then (Cs1i1
. (
IC
SCM ))
= loc & (Cs2i1
. (
IC
SCM ))
= ((
IC Cs2i)
+ 1) by
A3,
A9,
A4,
A5,
A7,
AMI_3: 8;
hence contradiction by
A6,
A11,
A8,
A2,
A1,
AMISTD_5: 7;
end;
hence thesis by
A10;
end;
theorem ::
AMI_5:24
for q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function holds for p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM st p
c= s1 & p
c= s2 holds for P1,P2 be
Instruction-Sequence of
SCM st q
c= P1 & q
c= P2 holds for i be
Nat, da be
Data-Location, loc be
Nat, I be
Instruction of
SCM st I
= (
CurInstr (P1,(
Comput (P1,s1,i)))) holds I
= (da
>0_goto loc) & loc
<> ((
IC (
Comput (P1,s1,i)))
+ 1) implies (((
Comput (P1,s1,i))
. da)
>
0 iff ((
Comput (P2,s2,i))
. da)
>
0 )
proof
let q be non
halt-free
finitethe
InstructionsF of
SCM
-valued
NAT
-defined
Function;
let p be q
-autonomic non
empty
FinPartState of
SCM , s1,s2 be
State of
SCM such that
A1: p
c= s1 & p
c= s2;
let P1,P2 be
Instruction-Sequence of
SCM such that
A2: q
c= P1 & q
c= P2;
let i be
Nat, da be
Data-Location, loc be
Nat, I be
Instruction of
SCM such that
A3: I
= (
CurInstr (P1,(
Comput (P1,s1,i))));
set Cs2i1 = (
Comput (P2,s2,(i
+ 1)));
set Cs1i1 = (
Comput (P1,s1,(i
+ 1)));
A4: (Cs1i1
| (
dom p))
= (Cs2i1
| (
dom p)) by
A2,
A1,
EXTPRO_1:def 10;
set Cs2i = (
Comput (P2,s2,i));
set Cs1i = (
Comput (P1,s1,i));
A5: Cs1i1
= (
Following (P1,Cs1i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P1,Cs1i)),Cs1i));
(
IC
SCM )
in (
dom p) by
AMISTD_5: 6;
then
A6: ((Cs1i1
| (
dom p))
. (
IC
SCM ))
= (
IC Cs1i1) & ((Cs2i1
| (
dom p))
. (
IC
SCM ))
= (
IC Cs2i1) by
FUNCT_1: 49;
A7: Cs2i1
= (
Following (P2,Cs2i)) by
EXTPRO_1: 3
.= (
Exec ((
CurInstr (P2,Cs2i)),Cs2i));
assume that
A8: I
= (da
>0_goto loc) and
A9: loc
<> ((
IC (
Comput (P1,s1,i)))
+ 1);
A10: I
= (
CurInstr (P2,(
Comput (P2,s2,i)))) by
A3,
A2,
A1,
AMISTD_5: 7;
A11:
now
assume that
A12: ((
Comput (P2,s2,i))
. da)
>
0 and
A13: ((
Comput (P1,s1,i))
. da)
<=
0 ;
(Cs2i1
. (
IC
SCM ))
= loc by
A10,
A7,
A8,
A12,
AMI_3: 9;
hence contradiction by
A3,
A5,
A6,
A4,
A8,
A9,
A13,
AMI_3: 9;
end;
A14: (
IC Cs1i)
= (
IC Cs2i) by
A2,
A1,
AMISTD_5: 7;
now
assume that
A15: ((
Comput (P1,s1,i))
. da)
>
0 and
A16: ((
Comput (P2,s2,i))
. da)
<=
0 ;
(Cs1i1
. (
IC
SCM ))
= loc by
A3,
A5,
A8,
A15,
AMI_3: 9;
hence contradiction by
A14,
A10,
A7,
A6,
A4,
A8,
A9,
A16,
AMI_3: 9;
end;
hence thesis by
A11;
end;
theorem ::
AMI_5:25
for s1,s2 be
State of
SCM st (
IC s1)
= (
IC s2) & (for a be
Data-Location holds (s1
. a)
= (s2
. a)) holds s1
= s2
proof
let s1,s2 be
State of
SCM such that
A1: (
IC s1)
= (
IC s2);
(
IC
SCM )
in (
dom s1) & (
IC
SCM )
in (
dom s2) by
MEMSTR_0: 2;
then
A2: s1
= ((
DataPart s1)
+* (
Start-At ((
IC s1),
SCM ))) & s2
= ((
DataPart s2)
+* (
Start-At ((
IC s2),
SCM ))) by
MEMSTR_0: 26;
assume
A3: for a be
Data-Location holds (s1
. a)
= (s2
. a);
(
DataPart s1)
= (
DataPart s2)
proof
A4: (
dom (
DataPart s1))
= (
Data-Locations
SCM ) by
MEMSTR_0: 9;
hence (
dom (
DataPart s1))
= (
dom (
DataPart s2)) by
MEMSTR_0: 9;
let x be
object;
assume
A5: x
in (
dom (
DataPart s1));
then
A6: x is
Data-Location by
A4,
AMI_2:def 16,
AMI_3: 27;
thus ((
DataPart s1)
. x)
= (s1
. x) by
A5,
A4,
FUNCT_1: 49
.= (s2
. x) by
A6,
A3
.= ((
DataPart s2)
. x) by
A5,
A4,
FUNCT_1: 49;
end;
hence thesis by
A1,
A2;
end;