
theorem
  for a,b be Nat holds ((a*b + 1) choose 1) mod b = 1 mod b
  proof
    let a,b be Nat;
    per cases;
    suppose a is zero or b is zero;
      hence thesis;
    end;
    suppose
      B1: a is non zero & b is non zero; then
      reconsider a as non zero Nat;
      reconsider b as non zero Nat by B1;
      ((a*b + 1) choose (0 + 1)) mod b =
        (((a*b) choose 0) + ((a*b) choose (0+1))) mod b by NEWTON:22
      .= (((((a*b) choose 0) mod b)) +
        (((a*b) choose 1) mod b)) mod b by NAT_D:66
      .= (((((a*b) choose 0) mod b)) + 0) mod b by ILO
      .= (1 mod b) mod b by NEWTON:19;
      hence thesis;
    end;
  end;
