theorem Th140:
  r ++ (F \+\ G) = (r++F) \+\ (r++G)
proof
  thus r ++ (F \+\ G) = (r++(F\G)) \/ (r++(G\F)) by Th41
    .= ((r++F)\(r++G)) \/ (r++(G\F)) by Th139
    .= (r++F) \+\ (r++G) by Th139;
end;
