A25: dom g = X by FUNCT_2:def 1;
A26: dom (eq f,g) = (dom f) /\ (dom g) by Def7;
A27: dom f = X by FUNCT_2:def 1;
rng (eq f,g) c= INT
proof
let y be set ; :: according to TARSKI:def 3 :: thesis: ( not y in rng (eq f,g) or y in INT )
assume y in rng (eq f,g) ; :: thesis: y in INT
then consider a being set such that
A28: a in dom (eq f,g) and
A29: y = (eq f,g) . a by FUNCT_1:def 5;
A30: g . a in rng g by A26, A25, A28, FUNCT_1:12;
f . a in rng f by A26, A27, A28, FUNCT_1:12;
then reconsider i = f . a, j = g . a as Element of INT by A30;
thus y in INT by A29, INT_1:def 2; :: thesis: verum
end;
hence eq f,g is Function of X,INT by A26, A27, A25, FUNCT_2:4; :: thesis: verum