theorem Th23:
  A |^ (n + 1) = (A |^ n) ^^ A
proof
  consider concat being sequence of  bool (E^omega) such that
A1: A |^ n = concat.n and
A2: concat.0 = {<%>E} and
A3: for i holds concat.(i + 1) = concat.i ^^ A by Def2;
  concat.(n + 1) = (A |^ n) ^^ A by A1,A3;
  hence thesis by A2,A3,Def2;
end;
