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,bset f =
a,
b --> c,
d;
set g =
c,
d --> a,
b;
A4:
for
y,
x being
set 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
set ;
( 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_4:65;
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_4:66;
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_4:66, FUNCT_4:67;
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_4:66;
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_4:66, FUNCT_4:67;
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_4:67;
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_4:66;
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_4:65, FUNCT_4:66;
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_4:65;
let x1,
x2 be
set ;
FUNCT_1:def 8 ( not x1 in proj1 (a,b --> c,d) or not x2 in proj1 (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_4:65
.=
rng (a,b --> c,d)
by A3, FUNCT_4:67
;
hence
(a,b --> c,d) " = c,
d --> a,
b
by A20, A4, FUNCT_1:54;
verum end; end;