 reserve n,s for Nat;

theorem Th19:
  Triangle n = n * (n + 1) / 2
  proof
    defpred P[Nat] means
      Triangle $1 = $1 * ($1 + 1) / 2;
A1: P[0];
A2: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat such that
A3:   P[k];
      Triangle (k + 1) = (Triangle k) + (k + 1) by Th10
          .= (k + 1) * (k + 2) / 2 by A3;
      hence thesis;
    end;
    for n being Nat holds P[n] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
