
theorem
  for K be Function holds
    rng K is empty-membered iff (for x be object holds K.x = {})
proof
   let K be Function;
   hereby assume A1: rng K is empty-membered;
    let x be object;
    per cases;
    suppose x in dom K;
     hence K.x = {} by A1,FUNCT_1:3;
    end;
    suppose not x in dom K;
     hence K.x = {} by FUNCT_1:def 2;
    end;
   end;
   assume A2: for x be object holds K.x = {};
   now assume ex y be non empty set st y in rng K; then
    consider y be non empty set such that
A3:  y in rng K;
    ex a be object st a in dom K & y = K.a by A3,FUNCT_1:def 3;
    hence contradiction by A2;
   end;
   hence rng K is empty-membered;
end;
