
theorem
  for p be Prime holds ((2*p) choose p) mod p = 2 mod p
  proof
    let p be Prime;
    reconsider n = p - 1 as Nat;
    n + 0 < n + 1 by XREAL_1:6; then
    ((1*p + n) choose n) mod p = 1 &
      ((1*p + n) choose (1*p)) mod p = 1 by NPK,KPN; then
    2 mod p = ((((p + n) choose n) mod p) + (((p + n) choose p) mod p)) mod p
    .=  (((p + n) choose n) + ((p + n) choose p)) mod p by NAT_D:66
    .= (((p + n) + 1) choose (n + 1)) mod p by NEWTON:22;
    hence thesis;
  end;
