A17: dom g = X by FUNCT_2:def 1;
A18: dom (eq (f,g)) = (dom f) /\ (dom g) by Def7;
A19: 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
A20: y = (eq (f,g)) . a by FUNCT_1:def 3;
thus y in INT by A20, INT_1:def 2; :: thesis: verum

end;

hence eq (f,g) is Function of X,INT by A18, A19, A17, FUNCT_2:2; :: thesis: verum