theorem Th2:
  y in rng f iff ex c st c in dom f & y = f/.c
proof
  thus y in rng f implies ex c st c in dom f & y = f/.c
  proof
    assume y in rng f;
    then consider x being object such that
A1: x in dom f and
A2: y = (f qua Function).x by FUNCT_1:def 3;
    reconsider x as Element of C by A1;
    take x;
    thus thesis by A1,A2,PARTFUN1:def 6;
  end;
  given c such that
A3: c in dom f and
A4: y = f/.c;
  (f qua Function).c in rng f by A3,FUNCT_1:def 3;
  hence thesis by A3,A4,PARTFUN1:def 6;
end;
