
theorem PCZ:
  for p be Prime, n be odd Nat st n < p holds
    ((p - 1) choose n) mod p = p - 1 & ((p - 1) choose (n - 1)) mod p = 1
  proof
    let p be Prime, n be odd Nat such that
    A1: n < p;
    per cases;
    suppose p is even; then
      B1: p = 2 by LAGRA4SQ:13; then
      n = 1 by A1,NAT_1:23; then
      ((p - 1) choose (n - 1)) mod 2 = 1 mod (1+1) &
        ((p - 1) choose n) mod 2 = 1 mod (1 + 1) by B1,NEWTON:19;
      hence thesis by B1;
    end;
    suppose p is odd; then
      reconsider p as odd Nat;
      reconsider m = (n - 1)/2 as Nat;
      2*m + 1 < p by A1;
      hence thesis by LmPCZ;
    end;
  end;
