
theorem Th15:
  for X being non empty set, i being object holds
  product ({i} --> X) = {{i} --> x where x is Element of X : not contradiction}
proof
  let X be non empty set, i be object;
  set S = {{i} --> x where x is Element of X : not contradiction};
  for z being object holds z in product ({i} --> X) iff z in S
  proof
    let z be object;
    hereby
      assume z in product ({i} --> X);
      then consider f being Function such that
        A1: z = f & dom f = dom ({i} --> X) and
        A2: for y being object st y in dom ({i} --> X)
          holds f.y in ({i} --> X).y by CARD_3:def 5;
      A3: dom f = {i} by A1;
      for y being object st y in dom f holds f.y = f.i
        by  A1,TARSKI:def 1;
      then A4: f = {i} --> f.i by A3, FUNCOP_1:11;
      i in dom ({i} --> X) by  TARSKI:def 1;
      then f.i in ({i} --> X).i by A2;
      then f.i in (i .--> X). i by FUNCOP_1:def 9;
      then f.i in X by FUNCOP_1:72;
      hence z in S by A1, A4;
    end;
    assume z in S;
    then ex x being Element of X st
       z = {i} --> x;
    hence thesis by Th13;
  end;
  hence thesis by TARSKI:2;
end;
