set R = f (#) g;

assume not f (#) g is non-empty ; :: thesis: contradiction

then 0 in rng (f (#) g) ;

then consider x being object such that

A1: x in dom (f (#) g) and

A2: (f (#) g) . x = 0 by FUNCT_1:def 3;

A3: dom (f (#) g) = (dom f) /\ (dom g) by VALUED_1:def 4;

then x in dom g by A1, XBOOLE_0:def 4;

then A4: g . x in rng g by FUNCT_1:def 3;

x in dom f by A1, A3, XBOOLE_0:def 4;

then f . x in rng f by FUNCT_1:def 3;

then reconsider a = f . x, b = g . x as non zero Real by A4;

not a * b is zero ;

hence contradiction by A2, VALUED_1:5; :: thesis: verum

assume not f (#) g is non-empty ; :: thesis: contradiction

then 0 in rng (f (#) g) ;

then consider x being object such that

A1: x in dom (f (#) g) and

A2: (f (#) g) . x = 0 by FUNCT_1:def 3;

A3: dom (f (#) g) = (dom f) /\ (dom g) by VALUED_1:def 4;

then x in dom g by A1, XBOOLE_0:def 4;

then A4: g . x in rng g by FUNCT_1:def 3;

x in dom f by A1, A3, XBOOLE_0:def 4;

then f . x in rng f by FUNCT_1:def 3;

then reconsider a = f . x, b = g . x as non zero Real by A4;

not a * b is zero ;

hence contradiction by A2, VALUED_1:5; :: thesis: verum