theorem
  r <> 0 implies r ** (F\+\G) = (r**F) \+\ (r**G)
proof
  assume
A1: r <> 0;
  thus r ** (F \+\ G) = (r**(F\G)) \/ (r**(G\F)) by Th82
    .= ((r**F)\(r**G)) \/ (r**(G\F)) by A1,Th191
    .= (r**F) \+\ (r**G) by A1,Th191;
end;
