theorem Th29:
  RAS has_property_of_zero_in m & RAS is_additive_in m implies RAS
  is_semi_additive_in m
proof
  assume that
A1: RAS has_property_of_zero_in m and
A2: RAS is_additive_in m;
  let a,pm,p;
  assume p.m = pm;
  then *'(a,(p+*(m,a@pm))) = *'(a,p)@*'(a,(p+*(m,a))) by A2
    .= a@*'(a,p) by A1;
  hence thesis;
end;
