theorem Th70:
  A+^succ n = succ A +^ n & A +^ (n+1) = succ A +^ n
proof
  defpred P[Nat] means A+^succ $1 = succ A +^ $1;
  A+^succ 0 = succ A by ORDINAL2:31
    .= succ A +^ 0 by ORDINAL2:27;
  then
A1: P[0];
A2: P[k] implies P[k+1]
  proof
    assume
A3: P[k];
A4: Segm(k+1) = succ Segm k by NAT_1:38;
    hence A+^succ (k+1) = succ (succ A +^ k) by A3,ORDINAL2:28
      .= succ A +^ k +^ 1 by ORDINAL2:31
      .= succ A +^ ( k +^ 1) by ORDINAL3:30
      .= succ A +^ (k+1) by A4,ORDINAL2:31;
  end;
A5: P[k] from NAT_1:sch 2(A1,A2);
  thus
A6:  A+^succ n = succ A +^ n by A5;
   Segm(n+1) = succ Segm n by NAT_1:38;
  hence thesis by A6;
end;
