
theorem MPO:
  for a,b be odd Integer st |.a.| <> |.b.| holds
    min (Parity(a-b),Parity(a+b)) = 2
  proof
    let a,b be odd Integer such that
    A1: |.a.| <> |.b.|;
    A0: |.a.| in NAT & |.b.| in NAT by INT_1:3; then
    reconsider k = |.a.| as odd Nat;
    reconsider l = |.b.| as odd Nat by A0;
    k-l <> 0 & k+l <> 0 by A1; then
    A2: Parity(k-l) = 2|^(2|-count (k-l)) &
    Parity(k+l) = 2|^(2|-count (k+l)) by Def1;
    A3: Parity (k-l) = Parity |.k-l.| &
    Parity (a+b) = Parity |.a+b.| &
    Parity (a-b) = Parity |.a-b.| by PMP;
    min (2|-count (k-l), 2|-count (k+l)) = 1 by A1,NEWTON03:82; then
    A4: min (Parity (k-l),Parity (k+l)) = 2|^1 by A2,MIN1;
    per cases by ABS;
    suppose
      |.a + b.| = |.a.| + |.b.|; then
      |.a -(-b).| = |.a.| + |.-b.| by COMPLEX1:52; then
      |.a -(-b).| = (|.a.|+|.-b.|) & |.a + (-b).| = |.|.a.|-|.-b.|.|
        by LmABS; then
      Parity |.a + b.| = Parity(|.a.|+|.b.|) &
      Parity|.a - b.| = Parity|.|.a.|-|.b.|.| by COMPLEX1:52;
      hence thesis by A4,A3;
    end;
    suppose
      |.a - b.| = |.a.| + |. b.|;
      hence thesis by A4,A3,LmABS;
    end;
  end;
