let f be Function; for x, y being object st x <> y holds
not x in rng (f +~ (x,y))
let x, y be object ; ( x <> y implies not x in rng (f +~ (x,y)) )
assume A1:
x <> y
; not x in rng (f +~ (x,y))
assume
x in rng (f +~ (x,y))
; contradiction
then consider z being object such that
A2:
z in dom (f +~ (x,y))
and
A3:
(f +~ (x,y)) . z = x
by FUNCT_1:def 3;
A4:
z in dom f
by A2, Th99;
(x .--> y) . (f . z) =
((x .--> y) * f) . z
by A4, FUNCT_1:13
.=
x
by A3, A5, Th13
;
then
f . z <> x
by A1, FUNCOP_1:72;
then
not f . z in dom (x .--> y)
by FUNCOP_1:75;
hence
contradiction
by A5, FUNCT_1:11; verum