theorem Th36:
  g (#) (f1/f2) = (g (#) f1)/f2
proof
  thus g (#) (f1/f2) = g (#) (f1 (#) (f2^)) by Th31
    .= g (#) f1 (#) (f2^) by Th9
    .= (g (#) f1)/f2 by Th31;
end;
