theorem Th36:
  (A |^ (m, n)) ^^ A = A ^^ (A |^ (m, n))
proof
  thus (A |^ (m, n)) ^^ A = (A |^ (m, n)) ^^ (A |^ 1) by FLANG_1:25
    .= (A |^ 1) ^^ (A |^ (m, n)) by Th35
    .= A ^^ (A |^ (m, n)) by FLANG_1:25;
end;
