theorem
  X <> {} implies rng uncurry (X --> f) = rng f & rng uncurry' (X --> f)
  = rng f
proof
  set x = the Element of X;
  assume
A1: X <> {};
  then dom (X --> f) = X & (X --> f).x = f by FUNCOP_1:7;
  hence rng uncurry (X --> f) c= rng f & rng f c= rng uncurry (X --> f) & rng
  uncurry' (X --> f) c= rng f & rng f c= rng uncurry' (X --> f) by A1,Th7,Th8;
end;
