theorem Th33:
  g.(a |^ b) = (g.a) |^ (g.b)
proof
  thus g.(a |^ b) = g.(b" * a * b) by GROUP_3:def 2
    .= g.(b" * a) * g.b by Def6
    .= g.(b") * g.a * g.b by Def6
    .= (g.b)" * g.a * g.b by Th32
    .= (g.a) |^ (g.b) by GROUP_3:def 2;
end;
