theorem Th25:
  A |^ 1 = A
proof
  consider concat being sequence of  bool (E^omega) such that
A1: A |^ 1 = concat.1 and
A2: concat.0 = {<%>E} & for i holds concat.(i + 1) = concat.i ^^ A by Def2;
  thus A |^ 1 = concat.(0 + 1) by A1
    .= {<%>E} ^^ A by A2
    .= A by Th13;
end;
