let a, b be Element of NAT ; ( a <> b implies not (((prop a) => (prop b)) => (prop a)) => (prop a) is pseudo-canonical )
assume A1:
a <> b
; not (((prop a) => (prop b)) => (prop a)) => (prop a) is pseudo-canonical
set A = { ((0,1) --> (i,j)) where i, j is Element of INT : ( i < j or ( i is even & i = j ) ) } ;
set X = product ((0,1) --> (INT,INT));
{ ((0,1) --> (i,j)) where i, j is Element of INT : ( i < j or ( i is even & i = j ) ) } c= product ((0,1) --> (INT,INT))
proof
let x be
object ;
TARSKI:def 3 ( not x in { ((0,1) --> (i,j)) where i, j is Element of INT : ( i < j or ( i is even & i = j ) ) } or x in product ((0,1) --> (INT,INT)) )
assume
x in { ((0,1) --> (i,j)) where i, j is Element of INT : ( i < j or ( i is even & i = j ) ) }
;
x in product ((0,1) --> (INT,INT))
then
ex
i,
j being
Element of
INT st
(
x = (
0,1)
--> (
i,
j) & (
i < j or (
i is
even &
i = j ) ) )
;
hence
x in product ((0,1) --> (INT,INT))
by Th10;
verum
end;
then reconsider A = { ((0,1) --> (i,j)) where i, j is Element of INT : ( i < j or ( i is even & i = j ) ) } as Subset of (product ((0,1) --> (INT,INT))) ;
set p1 = (0,1) --> (1,0);
A2:
not b in dom (a .--> 2)
by A1, FUNCOP_1:75;
reconsider p1 = (0,1) --> (1,0) as Permutation of 2 by Th14;
defpred S1[ set , set ] means ex i being Integer st
( $1 = i & $2 = i + 1 );
set V = (NAT --> INT) +* (a .--> (Segm 2));
reconsider V = (NAT --> INT) +* (a .--> (Segm 2)) as SetValuation ;
A3:
a in dom (a .--> 2)
by FUNCOP_1:74;
A4: 2 =
(a .--> 2) . a
by FUNCOP_1:72
.=
V . a
by A3, FUNCT_4:13
.=
SetVal (V,(prop a))
by Def2
;
A5:
for e being Element of INT ex u being Element of INT st S1[e,u]
consider p0 being Function of INT,INT such that
A6:
for e being Element of INT holds S1[e,p0 . e]
from FUNCT_2:sch 3(A5);
A7:
dom p0 = INT
by FUNCT_2:def 1;
for y being object holds
( y in INT iff ex x being object st
( x in dom p0 & y = p0 . x ) )
then A10:
rng p0 = INT
by FUNCT_1:def 3;
A11:
p0 is one-to-one
A13: SetVal (V,(prop b)) =
V . b
by Def2
.=
(NAT --> INT) . b
by A2, FUNCT_4:11
.=
INT
;
A14: product ((0,1) --> (INT,INT)) =
product (2 --> INT)
by CARD_1:50, FUNCT_4:65
.=
Funcs (2,INT)
by CARD_3:11
.=
SetVal (V,((prop a) => (prop b)))
by A4, A13, Def2
;
then reconsider f = chi (A,(product ((0,1) --> (INT,INT)))) as Function of (SetVal (V,((prop a) => (prop b)))),(SetVal (V,(prop a))) by A4, CARD_1:50;
A15:
a in dom (a .--> p1)
by FUNCOP_1:74;
reconsider p0 = p0 as Permutation of INT by A11, A10, FUNCT_2:57;
set P = (NAT --> p0) +* (a .--> p1);
A16: dom ((NAT --> p0) +* (a .--> p1)) =
(dom (NAT --> p0)) \/ (dom (a .--> p1))
by FUNCT_4:def 1
.=
(dom (NAT --> p0)) \/ {a}
.=
NAT \/ {a}
.=
NAT
by ZFMISC_1:40
;
for n being Element of NAT holds ((NAT --> p0) +* (a .--> p1)) . n is Permutation of (V . n)
then reconsider P = (NAT --> p0) +* (a .--> p1) as Permutation of V by A16, Def4;
A21: Perm (P,(prop a)) =
P . a
by Def5
.=
(a .--> p1) . a
by A15, FUNCT_4:13
.=
p1
by FUNCOP_1:72
;
A22:
f is_a_fixpoint_of Perm (P,(((prop a) => (prop b)) => (prop a)))
proof
set Q =
Perm (
P,
(((prop a) => (prop b)) => (prop a)));
f in Funcs (
(SetVal (V,((prop a) => (prop b)))),
(SetVal (V,(prop a))))
by FUNCT_2:8;
then
f in SetVal (
V,
(((prop a) => (prop b)) => (prop a)))
by Def2;
hence
f in dom (Perm (P,(((prop a) => (prop b)) => (prop a))))
by FUNCT_2:def 1;
ABIAN:def 3 f = (Perm (P,(((prop a) => (prop b)) => (prop a)))) . f
rng ((Perm (P,((prop a) => (prop b)))) ") =
dom (((Perm (P,((prop a) => (prop b)))) ") ")
by FUNCT_1:32
.=
product ((0,1) --> (INT,INT))
by A14, FUNCT_2:def 1
.=
dom f
by FUNCT_2:def 1
;
then A23:
dom (f * ((Perm (P,((prop a) => (prop b)))) ")) =
dom ((Perm (P,((prop a) => (prop b)))) ")
by RELAT_1:27
.=
rng (Perm (P,((prop a) => (prop b))))
by FUNCT_1:32
.=
product ((0,1) --> (INT,INT))
by A14, FUNCT_2:def 3
.=
dom (chi ((A `),(product ((0,1) --> (INT,INT)))))
by FUNCT_3:def 3
;
for
x being
object st
x in dom (chi ((A `),(product ((0,1) --> (INT,INT))))) holds
(f * ((Perm (P,((prop a) => (prop b)))) ")) . x = (chi ((A `),(product ((0,1) --> (INT,INT))))) . x
proof
A24:
dom (p0 ") = INT
by A10, FUNCT_1:32;
let x be
object ;
( x in dom (chi ((A `),(product ((0,1) --> (INT,INT))))) implies (f * ((Perm (P,((prop a) => (prop b)))) ")) . x = (chi ((A `),(product ((0,1) --> (INT,INT))))) . x )
A25:
not
b in dom (a .--> p1)
by A1, FUNCOP_1:75;
assume A26:
x in dom (chi ((A `),(product ((0,1) --> (INT,INT)))))
;
(f * ((Perm (P,((prop a) => (prop b)))) ")) . x = (chi ((A `),(product ((0,1) --> (INT,INT))))) . x
then
x in product ((0,1) --> (INT,INT))
by FUNCT_3:def 3;
then
x in Funcs (
(SetVal (V,(prop a))),
(SetVal (V,(prop b))))
by A14, Def2;
then reconsider g =
x as
Function of
(SetVal (V,(prop a))),
(SetVal (V,(prop b))) by FUNCT_2:66;
consider i0,
j0 being
Element of
INT such that A27:
g = (
0,1)
--> (
i0,
j0)
by A4, A13, Th11;
reconsider i0 =
i0,
j0 =
j0 as
Integer ;
A28:
j0 - 1
in dom p0
by A7, INT_1:def 2;
then
ex
i being
Integer st
(
j0 - 1
= i &
p0 . (j0 - 1) = i + 1 )
by A6, A7;
then A29:
p0 . (j0 - 1) = j0
by XCMPLX_1:27;
A30:
i0 - 1
in dom p0
by A7, INT_1:def 2;
then
ex
i being
Integer st
(
i0 - 1
= i &
p0 . (i0 - 1) = i + 1 )
by A6, A7;
then
p0 . (i0 - 1) = i0
by XCMPLX_1:27;
then A31:
(p0 ") . i0 = i0 - 1
by A30, FUNCT_1:34;
Perm (
P,
(prop b)) =
P . b
by Def5
.=
(NAT --> p0) . b
by A25, FUNCT_4:11
.=
p0
;
then A32:
((Perm (P,(prop b))) ") * g =
(
0,1)
--> (
((p0 ") . i0),
((p0 ") . j0))
by A27, A24, Th13
.=
(
0,1)
--> (
(i0 - 1),
(j0 - 1))
by A31, A28, A29, FUNCT_1:34
;
A33:
(f * ((Perm (P,((prop a) => (prop b)))) ")) . x =
f . (((Perm (P,((prop a) => (prop b)))) ") . x)
by A23, A26, FUNCT_1:12
.=
f . ((((Perm (P,(prop b))) ") * g) * (Perm (P,(prop a))))
by Th37
.=
f . ((0,1) --> ((j0 - 1),(i0 - 1)))
by A21, A32, Th12
;
per cases
( x in A or not x in A )
;
suppose A34:
x in A
;
(f * ((Perm (P,((prop a) => (prop b)))) ")) . x = (chi ((A `),(product ((0,1) --> (INT,INT))))) . xthen consider i,
j being
Element of
INT such that A35:
x = (
0,1)
--> (
i,
j)
and A36:
(
i < j or (
i is
even &
i = j ) )
;
A37:
(
i = i0 &
j = j0 )
by A27, A35, Th9;
A38:
now not (0,1) --> ((j0 - 1),(i0 - 1)) in Aassume
(
0,1)
--> (
(j0 - 1),
(i0 - 1))
in A
;
contradictionthen consider i,
j being
Element of
INT such that A39:
(
0,1)
--> (
(j0 - 1),
(i0 - 1))
= (
0,1)
--> (
i,
j)
and A40:
(
i < j or (
i is
even &
i = j ) )
;
A41:
(
i = j0 - 1 &
j = i0 - 1 )
by A39, Th9;
end; A42:
x in (A `) `
by A34;
(
j0 - 1
in INT &
i0 - 1
in INT )
by INT_1:def 2;
then
(
0,1)
--> (
(j0 - 1),
(i0 - 1))
in product ((0,1) --> (INT,INT))
by Th10;
then
(
0,1)
--> (
(j0 - 1),
(i0 - 1))
in (product ((0,1) --> (INT,INT))) \ A
by A38, XBOOLE_0:def 5;
hence (f * ((Perm (P,((prop a) => (prop b)))) ")) . x =
0
by A33, FUNCT_3:37
.=
(chi ((A `),(product ((0,1) --> (INT,INT))))) . x
by A42, FUNCT_3:37
;
verum end; suppose A43:
not
x in A
;
(f * ((Perm (P,((prop a) => (prop b)))) ")) . x = (chi ((A `),(product ((0,1) --> (INT,INT))))) . x
x in product ((0,1) --> (INT,INT))
by A27, Th10;
then A44:
x in (product ((0,1) --> (INT,INT))) \ A
by A43, XBOOLE_0:def 5;
A45:
j0 - 1 is
Element of
INT
by INT_1:def 2;
A46:
(
i0 is
odd or
i0 <> j0 )
by A27, A43;
A47:
i0 - 1 is
Element of
INT
by INT_1:def 2;
A48:
i0 >= j0
by A27, A43;
hence (f * ((Perm (P,((prop a) => (prop b)))) ")) . x =
1
by A33, FUNCT_3:def 3
.=
(chi ((A `),(product ((0,1) --> (INT,INT))))) . x
by A44, FUNCT_3:def 3
;
verum end; end;
end;
then
f * ((Perm (P,((prop a) => (prop b)))) ") = chi (
(A `),
(product ((0,1) --> (INT,INT))))
by A23;
hence f =
(Perm (P,(prop a))) * (f * ((Perm (P,((prop a) => (prop b)))) "))
by A21, Th8
.=
((Perm (P,(prop a))) * f) * ((Perm (P,((prop a) => (prop b)))) ")
by RELAT_1:36
.=
(Perm (P,(((prop a) => (prop b)) => (prop a)))) . f
by Th36
;
verum
end;
for x being set holds not x is_a_fixpoint_of Perm (P,(prop a))
hence
not (((prop a) => (prop b)) => (prop a)) => (prop a) is pseudo-canonical
by A22, Th43; verum