 reserve n,s for Nat;

theorem
  3 * (Triangle n) + Triangle (n + 1) = Triangle (2 * n + 1)
  proof
A1: Triangle n = n * (n + 1) / 2 by Th19;
A2: Triangle (n + 1) = (n + 1) * (n + 1 + 1) / 2 by Th19
                    .= (n + 1) * (n + 2) / 2;
    Triangle (2 * n + 1) = (2 * n + 1) * (2 * n + 1 + 1) / 2 by Th19
                        .= (2 * n + 1) * (2 * n + 2) / 2;
    hence thesis by A1,A2;
  end;
