theorem
  (F\+\G) -- r = (F--r) \+\ (G--r)
proof
  thus (F\+\G) -- r = --(r--(F\+\G)) by Th60
    .= --((r--F) \+\ (r--G)) by Th159
    .= (--(r--F)) \+\ (--(r--G)) by Th8
    .= (--(r--F)) \+\ (G--r) by Th60
    .= (F--r) \+\ (G--r) by Th60;
end;
