let Y be set ; for g, f being Function holds
( g = Y | f iff ( ( for x being set holds
( x in dom g iff ( x in dom f & f . x in Y ) ) ) & ( for x being set st x in dom g holds
g . x = f . x ) ) )
let g, f be Function; ( g = Y | f iff ( ( for x being set holds
( x in dom g iff ( x in dom f & f . x in Y ) ) ) & ( for x being set st x in dom g holds
g . x = f . x ) ) )
hereby ( ( for x being set holds
( x in dom g iff ( x in dom f & f . x in Y ) ) ) & ( for x being set st x in dom g holds
g . x = f . x ) implies g = Y | f )
end;
assume that
A6:
for x being set holds
( x in dom g iff ( x in dom f & f . x in Y ) )
and
A7:
for x being set st x in dom g holds
g . x = f . x
; g = Y | f
now let x,
y be
set ;
( ( [x,y] in g implies ( y in Y & [x,y] in f ) ) & ( y in Y & [x,y] in f implies [x,y] in g ) )hereby ( y in Y & [x,y] in f implies [x,y] in g )
end; assume A11:
y in Y
;
( [x,y] in f implies [x,y] in g )assume A12:
[x,y] in f
;
[x,y] in gthen A13:
y = f . x
by Th8;
x in dom f
by A12, RELAT_1:def 4;
then A14:
x in dom g
by A6, A11, A13;
then
y = g . x
by A7, A13;
hence
[x,y] in g
by A14, Def4;
verum end;
hence
g = Y | f
by RELAT_1:def 12; verum