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
a in dom (eq (f,g)) and
A29: y = (eq (f,g)) . a by FUNCT_1:def 5;
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