theorem Th28:
  A |^ (n, n + 1) = A |^ n \/ A |^ (n + 1)
proof
  thus A |^ (n, n + 1) = A |^ (n, n) \/ A |^ (n + 1) by Th26,NAT_1:11
    .= A |^ n \/ A |^ (n + 1) by Th22;
end;
