
theorem
  for a,b be Nat holds
    (parity (a+b) = parity b implies parity a = 0) &
      (parity (a+b) <> parity b implies parity a = 1)
  proof
    let a,b be Nat;
    ((a+b) mod 2 = b mod 2 implies a mod 2 = 0) &
      ((a+b) mod 2 <> b mod 2 implies a mod 2 <> 0) by MAB;
    hence thesis by NAT_2:def 1;
  end;
