reserve a,b,c,d,x,j,k,l,m,n,o,xi,xj for Nat,
  p,q,t,z,u,v for Integer,
  a1,b1,c1,d1 for Complex;

theorem Odd:
  for a,b be odd Integer, n be Nat holds
  2 |-count (a|^(2*n+1) - b|^(2*n+1)) = 2 |-count (a-b)
  proof
    let a,b be odd Integer, n be Nat;
    per cases;
    suppose a = b;
      hence thesis;
    end;
    suppose
  A0: a <> b; then
      reconsider c = a - b as non zero Integer;
      defpred P[Nat] means
        2 |-count (a|^(2*$1+1) - b|^(2*$1+1)) = 2 |-count (a-b);
  A1: P[0];
  A2: P[k] implies P[k+1]
      proof
        assume
    B0: P[k];
        reconsider d = a|^(2*k+1) - b|^(2*k+1) as non zero Integer by A0,POW2;
    B1: 2|^(2 |-count c) divides c &
        2|^(2 |-count d) divides d by Def6;
        reconsider x = d/2|^(2|-count d) as odd Integer;
        consider y be Integer such that
    B3: c = 2|^(2|-count c)*y by B1;
    B5: a|^((2*k+1)+2) - b|^((2*k+1)+2) =
          a|^(2*k+1)*(a|^2 - b|^2) + b|^2*(a|^(2*k+1) - b|^(2*k+1)) by RI2
        .= a|^(2*k+1)*(a|^2 - b|^2) + b|^2* (2|^(2 |-count (a-b))*x)
          by B0,XCMPLX_1:87
        .= a|^(2*k+1)*((a+b)*(2|^(2 |-count (a-b))*y)) +
          b|^2* 2|^(2 |-count (a-b))*x by B3,NEWTON01:1
        .= (a|^(2*k+1)*(a+b)*y + b|^2*x)* 2|^(2 |-count (a-b));
    B6: 2 is non trivial;
        2 |-count (a|^((2*k+1)+2) - b|^((2*k+1)+2))
        = 2 |-count (a|^(2*k+1)*(a+b)*y + b|^2*x) +
        2 |-count (2|^(2 |-count (a-b))) by B5,NAT328,INT_2:28
        .= 2 |-count c by B6;
        hence thesis;
      end;
      for x holds P[x] from NAT_1:sch 2(A1,A2);
      hence thesis;
    end;
  end;
