
theorem
  for p be Prime, k,n be Nat holds k < n <= p implies
     ((p + n) choose k) mod p = ((n choose k) mod p)
  proof
    let p be Prime, k,n be Nat;
    assume
    A1: k < n <= p;
    per cases;
    suppose k is non zero; then
      reconsider k as non zero Nat;
      per cases by A1,XXREAL_0:1;
      suppose
        B1: n = p;
        ((2*p) choose k) mod p = 0 by A1,B1,PCN;
        hence thesis by A1,B1,PCK;
      end;
      suppose n < p;
        hence thesis by A1,PCM;
      end;
    end;
    suppose k is zero; then
      n choose k = 1 & (p + n) choose k = 1 by NEWTON:19;
      hence thesis;
    end;
  end;
