theorem
  (F\G) -- r = (F--r) \ (G--r)
proof
  thus (F\G) -- r = --(r--(F\G)) by Th60
    .= --((r--F) \ (r--G)) by Th158
    .= (--(r--F)) \ (--(r--G)) by Th7
    .= (--(r--F)) \ (G--r) by Th60
    .= (F--r) \ (G--r) by Th60;
end;
