set R = r (#) f;

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

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

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

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

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

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