theorem
  A |^ n \/ A |^.. (n + 1) = A |^.. n
proof
A1: n <= n + 1 by NAT_1:13;
  thus A |^ n \/ A |^.. (n + 1) = A |^ (n, n) \/ A |^.. (n + 1) by FLANG_2:22
    .= A |^.. n by A1,Th7;
end;
