let a, b, c, d be set ; not ( ( a = b implies c = d ) & ( c = d implies a = b ) & not ((a,b) --> (c,d)) " = (c,d) --> (a,b) )
assume A1:
( a = b iff c = d )
; ((a,b) --> (c,d)) " = (c,d) --> (a,b)
per cases
( ( a = b & c = d ) or ( a <> b & c <> d ) )
by A1;
suppose A3:
(
a <> b &
c <> d )
;
((a,b) --> (c,d)) " = (c,d) --> (a,b)set f = (
a,
b)
--> (
c,
d);
set g = (
c,
d)
--> (
a,
b);
A4:
for
y,
x being
object holds
(
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y iff (
x in dom ((a,b) --> (c,d)) &
y = ((a,b) --> (c,d)) . x ) )
proof
let y,
x be
object ;
( y in rng ((a,b) --> (c,d)) & x = ((c,d) --> (a,b)) . y iff ( x in dom ((a,b) --> (c,d)) & y = ((a,b) --> (c,d)) . x ) )
A5:
(
x in dom ((a,b) --> (c,d)) &
y = ((a,b) --> (c,d)) . x implies (
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y ) )
proof
assume that A6:
x in dom ((a,b) --> (c,d))
and A7:
y = ((a,b) --> (c,d)) . x
;
( y in rng ((a,b) --> (c,d)) & x = ((c,d) --> (a,b)) . y )
A8:
x in {a,b}
by A6, FUNCT_4:62;
per cases
( x = a or x = b )
by A8, TARSKI:def 2;
suppose A9:
x = a
;
( y in rng ((a,b) --> (c,d)) & x = ((c,d) --> (a,b)) . y )then A10:
y = c
by A3, A7, FUNCT_4:63;
then
y in {c,d}
by TARSKI:def 2;
hence
(
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y )
by A3, A9, A10, FUNCT_4:63, FUNCT_4:64;
verum end; suppose A11:
x = b
;
( y in rng ((a,b) --> (c,d)) & x = ((c,d) --> (a,b)) . y )then A12:
y = d
by A7, FUNCT_4:63;
then
y in {c,d}
by TARSKI:def 2;
hence
(
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y )
by A3, A11, A12, FUNCT_4:63, FUNCT_4:64;
verum end; end;
end;
(
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y implies (
x in dom ((a,b) --> (c,d)) &
y = ((a,b) --> (c,d)) . x ) )
proof
assume that A13:
y in rng ((a,b) --> (c,d))
and A14:
x = ((c,d) --> (a,b)) . y
;
( x in dom ((a,b) --> (c,d)) & y = ((a,b) --> (c,d)) . x )
A15:
y in {c,d}
by A3, A13, FUNCT_4:64;
per cases
( y = c or y = d )
by A15, TARSKI:def 2;
suppose A16:
y = c
;
( x in dom ((a,b) --> (c,d)) & y = ((a,b) --> (c,d)) . x )then A17:
x = a
by A3, A14, FUNCT_4:63;
then
x in {a,b}
by TARSKI:def 2;
hence
(
x in dom ((a,b) --> (c,d)) &
y = ((a,b) --> (c,d)) . x )
by A3, A16, A17, FUNCT_4:62, FUNCT_4:63;
verum end; end;
end;
hence
(
y in rng ((a,b) --> (c,d)) &
x = ((c,d) --> (a,b)) . y iff (
x in dom ((a,b) --> (c,d)) &
y = ((a,b) --> (c,d)) . x ) )
by A5;
verum
end; A20:
(
a,
b)
--> (
c,
d) is
one-to-one
proof
A21:
dom ((a,b) --> (c,d)) = {a,b}
by FUNCT_4:62;
let x1,
x2 be
object ;
FUNCT_1:def 4 ( not x1 in dom ((a,b) --> (c,d)) or not x2 in dom ((a,b) --> (c,d)) or not ((a,b) --> (c,d)) . x1 = ((a,b) --> (c,d)) . x2 or x1 = x2 )
assume that A22:
(
x1 in dom ((a,b) --> (c,d)) &
x2 in dom ((a,b) --> (c,d)) )
and A23:
((a,b) --> (c,d)) . x1 = ((a,b) --> (c,d)) . x2
;
x1 = x2
end; dom ((c,d) --> (a,b)) =
{c,d}
by FUNCT_4:62
.=
rng ((a,b) --> (c,d))
by A3, FUNCT_4:64
;
hence
((a,b) --> (c,d)) " = (
c,
d)
--> (
a,
b)
by A20, A4, FUNCT_1:32;
verum end; end;