 reserve n,s for Nat;

theorem Th10:
  Triangle (n + 1) = (Triangle n) + (n + 1)
  proof
    defpred P[Nat] means
      (Triangle $1) + ($1 + 1) = Triangle($1+1);
A1: P[0] by FINSEQ_2:50,RVSUM_1:73;
A2: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat such that P[k];
      reconsider k1 = k as Element of NAT by ORDINAL1:def 12;
      (Triangle (k+1)) + k + 1 + 1
           = Sum ((idseq k1) ^ <*k1 + 1*>) + k + 1 + 1 by FINSEQ_2:51
          .= Sum ((idseq k1) ^ <*k1 + 1*>) + (k + 1 + 1)
          .= Sum (idseq (k1 + 1)) + (k1 + 1 + 1) by FINSEQ_2:51
          .= Sum ((idseq (k1 + 1)) ^ <*k1 + 1 + 1*>) by RVSUM_1:74
          .= Triangle (k + 1 + 1) by FINSEQ_2:51;
      hence thesis;
    end;
    for n being Nat holds P[n] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
