theorem Th84:
  [+].<*i,j*> = i+j
proof
  reconsider i,j as Element of NAT by ORDINAL1:def 12;
  reconsider q = <*i*> as Element of 1-tuples_on NAT by FINSEQ_2:131;
  defpred p[Nat] means [+].<*i,$1*> = i+$1;
A1: now
    let j be Nat;
    reconsider jj=j as Element of NAT by ORDINAL1:def 12;
    reconsider r = <*i,jj,i+jj*> as Element of 3-tuples_on NAT by FINSEQ_2:104;
    assume p[j];
    then [+].<*i,jj+1*> = (3 succ 3).r by Th80
      .= (r.3)+1 by Def7
      .= i+(j+1);
    hence p[j+1];
  end;
  [+].<*i,0*> = (1 proj 1).q by Th78
    .= q.1 by Th37
    .= i+0;
  then
A2: p[0];
  for j being Nat holds p[j] from NAT_1:sch 2(A2,A1);
  hence thesis;
end;
