theorem
  m <= n & n > 0 & <%>E in A implies (A |^ (m, n))* = A*
proof
  assume that
A1: m <= n and
A2: n > 0 and
A3: <%>E in A;
  thus (A |^ (m, n))* = (A |^ n)* by A1,A3,Th34
    .= A* by A2,A3,Th16;
end;
