theorem
  A |^.. k ^^ (A?) = A? ^^ (A |^.. k)
proof
  thus A |^.. k ^^ (A?) = A |^.. k ^^ (A |^ (0, 1)) by FLANG_2:79
    .= A |^ (0, 1) ^^ (A |^.. k) by Th25
    .= A? ^^ (A |^.. k) by FLANG_2:79;
end;
