set R = Inv f;
assume not Inv f is non-empty ; :: thesis: contradiction
then consider x being object such that
A1: x in dom (Inv f) and
A2: (Inv f) . x = 0 by FUNCT_1:def 3;
dom (Inv f) = dom f by VALUED_1:def 7;
then f . x in rng f by A1, FUNCT_1:def 3;
then reconsider a = f . x as non zero Real ;
not a " is zero ;
hence contradiction by A2, VALUED_1:10; :: thesis: verum