
theorem PEQ:
  for a,b be non zero Integer st a + b <> 0 holds
  Parity a = Parity b implies Parity (a+b) >= (Parity a) + (Parity b)
  proof
    let a,b be non zero Integer such that
    A0: a+b <> 0;
    reconsider k = 2|-count a as Nat;
    assume
    A1: Parity a = Parity b;
    reconsider l = a/(Parity a)+ b/(Parity b) as even Integer;
    A2: l*(Parity a) = a/(Parity a)*(Parity a) + b/(Parity b)*(Parity b) by A1
    .= a/(Parity a)*(Parity a) + b by XCMPLX_1:87
    .= a + b by XCMPLX_1:87;
    A3: 2*(Parity a) = 2*2|^k by Def1 .= 2|^(k+1) by NEWTON:6;
    2 divides l by ABIAN:def 1; then
    2 is non trivial & 2|^(k+1) divides (a+b) by NEWTON03:5,A2,A3; then
    2|-count (a+b) >= k+1 by A0,NEWTON03:59; then
    2|^(k+1) divides 2|^(2|-count (a+b)) by PEPIN:31; then
    2|^(k+1) divides Parity (a+b) by A0,Def1;
    hence thesis by A0,A1,A3,NAT_D:7;
  end;
