assume not c (#) f is non-empty ; :: thesis: contradiction
then consider x being object such that
A2: x in dom (c (#) f) and
A3: (c (#) f) . x = 0 by FUNCT_1:def 3;
A4: (c (#) f) . x = c * (f . x) by VALUED_1:6;
dom (c (#) f) = dom f by VALUED_1:def 5;
then f . x in rng f by A2, FUNCT_1:def 3;
hence contradiction by A3, A4; :: thesis: verum