
theorem PCN:
  for p be Prime, n be non zero Nat st n < p holds ((2*p) choose n) mod p = 0
  proof
    let p be Prime, n be non zero Nat such that
    A1: n < p;
    reconsider m = p - 1 as non zero Nat;
    reconsider k = n - 1 as Nat;
    A2: m + 0 < m + 1 by XREAL_1:6;
    k + 1 < m + 1 by A1; then
    A3: k + 1 <= m by INT_1:7;
    k + 1 < m + 1 by A1; then
    A4: k < m by XREAL_1:6;
    A5: ((m choose (k + 1))) mod p = ((p + m) choose (k + 1)) mod p &
      ((p + m) choose k) mod p = (m choose k)mod p by PCM,A2,A3,A4;
    ((p + p) choose n) mod p = (((p + m) + 1) choose (k + 1)) mod p
    .= (((p + m) choose (k + 1))+ ((p + m) choose k)) mod p by NEWTON:22
    .= ((((p + m) choose (k + 1)) mod p) +
         ((((p + m) choose k))mod p)) mod p by NAT_D:66
    .= ((m choose (k + 1)) + (m choose k)) mod p by NAT_D:66,A5
    .= ((m + 1) choose (k + 1)) mod p by NEWTON:22
    .= 0 by A1,PCK;
    hence thesis;
  end;
