theorem Th25:
  a |^ b |^ b" = a & a |^ b" |^ b = a
proof
  thus a |^ b |^ b" = a |^ (b * b") by Th24
    .= a |^ 1_G by GROUP_1:def 5
    .= a by Th19;
  thus a |^ b" |^ b = a |^ (b" * b) by Th24
    .= a |^ 1_G by GROUP_1:def 5
    .= a by Th19;
end;
