set R = f (#) g;
assume not f (#) g is non-empty ; :: thesis: contradiction
then 0 in rng (f (#) g) by RELAT_1:def 9;
then consider x being set such that
A1: x in dom (f (#) g) and
A2: (f (#) g) . x = 0 by FUNCT_1:def 5;
dom (f (#) g) = (dom f) /\ (dom g) by VALUED_1:def 4;
then ( x in dom f & x in dom g ) by A1, XBOOLE_0:def 4;
then ( f . x in rng f & g . x in rng g ) by FUNCT_1:def 5;
then reconsider a = f . x, b = g . x as non zero real number by RELAT_1:def 9;
not a * b is empty ;
hence contradiction by A2, VALUED_1:5; :: thesis: verum