let f, g be Function; ( ( for x being set st x in (dom f) /\ (dom g) holds
f . x = g . x ) implies ex h being Function st f \/ g = h )
assume A1:
for x being set st x in (dom f) /\ (dom g) holds
f . x = g . x
; ex h being Function st f \/ g = h
defpred S1[ set , set ] means [$1,$2] in f \/ g;
A2:
for x, y1, y2 being set st S1[x,y1] & S1[x,y2] holds
y1 = y2
proof
let x,
y1,
y2 be
set ;
( S1[x,y1] & S1[x,y2] implies y1 = y2 )
assume that A3:
[x,y1] in f \/ g
and A4:
[x,y2] in f \/ g
;
y1 = y2
hence
y1 = y2
;
verum
end;
consider h being Function such that
A8:
for x, y being set holds
( [x,y] in h iff ( x in (dom f) \/ (dom g) & S1[x,y] ) )
from FUNCT_1:sch 1(A2);
take
h
; f \/ g = h
let x be set ; RELAT_1:def 2 for b1 being set holds
( ( not [x,b1] in f \/ g or [x,b1] in h ) & ( not [x,b1] in h or [x,b1] in f \/ g ) )
let y be set ; ( ( not [x,y] in f \/ g or [x,y] in h ) & ( not [x,y] in h or [x,y] in f \/ g ) )
thus
( [x,y] in f \/ g implies [x,y] in h )
( not [x,y] in h or [x,y] in f \/ g )
thus
( not [x,y] in h or [x,y] in f \/ g )
by A8; verum