set R = r (#) f;
assume not r (#) f is non-empty ; :: thesis: contradiction
then consider x being object such that
A1: x in dom (r (#) f) and
A2: (r (#) f) . x = 0 by FUNCT_1:def 3;
dom (r (#) f) = dom f by VALUED_1:def 5;
then f . x in rng f by A1, FUNCT_1:def 3;
then reconsider a = f . x as non zero Real ;
not r * a is zero ;
hence contradiction by A1, A2, VALUED_1:def 5; :: thesis: verum