let f, g, h be Function; :: thesis: ( f +* h = g +* h iff f,g equal_outside dom h )
thus
( f +* h = g +* h implies f,g equal_outside dom h )
:: thesis: ( f,g equal_outside dom h implies f +* h = g +* h )
assume A3:
f,g equal_outside dom h
; :: thesis: f +* h = g +* h
then A4:
(dom f) \ (dom h) = (dom g) \ (dom h)
by Th30;
A5: dom (f +* h) =
(dom f) \/ (dom h)
by FUNCT_4:def 1
.=
((dom g) \ (dom h)) \/ (dom h)
by A4, XBOOLE_1:39
.=
(dom g) \/ (dom h)
by XBOOLE_1:39
.=
dom (g +* h)
by FUNCT_4:def 1
;
for x being set st x in dom (f +* h) holds
(f +* h) . x = (g +* h) . x
proof
let x be
set ;
:: thesis: ( x in dom (f +* h) implies (f +* h) . x = (g +* h) . x )
assume A6:
x in dom (f +* h)
;
:: thesis: (f +* h) . x = (g +* h) . x
per cases
( x in dom h or not x in dom h )
;
suppose A8:
not
x in dom h
;
:: thesis: (f +* h) . x = (g +* h) . xA9:
f | ((dom f) \ (dom h)) = g | ((dom g) \ (dom h))
by A3, Def2;
dom (f +* h) = (dom f) \/ (dom h)
by FUNCT_4:def 1;
then
x in dom f
by A6, A8, XBOOLE_0:def 3;
then A10:
x in (dom f) \ (dom h)
by A8, XBOOLE_0:def 5;
thus (f +* h) . x =
f . x
by A8, FUNCT_4:12
.=
(g | ((dom g) \ (dom h))) . x
by A9, A10, FUNCT_1:72
.=
g . x
by A4, A10, FUNCT_1:72
.=
(g +* h) . x
by A8, FUNCT_4:12
;
:: thesis: verum end; end;
end;
hence
f +* h = g +* h
by A5, FUNCT_1:9; :: thesis: verum