
theorem CPK:
  for p be Prime, n,k be Nat st k <> p holds
    (((n mod p) + 1) choose k) mod p = (((n + 1) mod p) choose k) mod p
  proof
    let p be Prime, n,k be Nat such that
    A0: k <> p;
    A1: (n + 1) mod p = ((n mod p) + 1) mod p by NAT_D:22;
    per cases;
    suppose k = 0;
      hence thesis by NEWTON:19;
    end;
    suppose k > 0; then
      reconsider k as non zero Nat;
      (n mod p) + 1 <= p by NAT_D:1,INT_1:7; then
      per cases by XXREAL_0:1;
      suppose
        C1: (n mod p) + 1 = p;
        (((n mod p) + 1) choose k) mod p = (0 choose k) mod p by C1,A0,PCK;
        hence thesis by A1,C1;
      end;
      suppose (n mod p) + 1 < p;
        hence thesis by A1,NAT_D:24;
      end;
    end;
  end;
