
theorem PSD:
  for a,b be odd Integer holds Parity (a+b) <> Parity (a-b)
  proof
    let a,b be odd Integer;
    per cases;
    suppose
      B1: a+b is zero; then
      Parity (a-b) = Parity (2*a)
      .= (Parity 2)*(Parity a) by ILP;
      hence thesis by B1,Def1;
    end;
    suppose
      B1: a-b is zero; then
      Parity (a+b) = Parity (2*a)
      .= (Parity 2)*Parity a by ILP;
      hence thesis by B1,Def1;
    end;
    suppose
      B0: a+b <> 0  & a-b <> 0; then
      B1: 2|^(2|-count (a+b)) = Parity (a+b) &
        2|^(2|-count (a-b)) = Parity (a-b) by Def1;
      2*2 = 2|^2 by NEWTON:81; then
      B2: 2|^2 divides (a+b) iff not 2|^2 divides (a-b) by NEWTON03:73;
      2 is non trivial; then
      2|-count (a+b) >= 2 iff 2|-count (a-b) < 2 by B0,B2,NEWTON03:59; then
      per cases by XXREAL_0:1;
      suppose
        2|-count (a+b) > 2|-count (a-b);
        hence thesis by B1,PEPIN:66;
      end;
      suppose
        2|-count (a+b) < 2|-count (a-b);
        hence thesis by B1,PEPIN:66;
      end;
    end;
  end;
