theorem Th50:
  A * g = (-g) + A + g
proof
  A * g c= -{g} + A + {g} by ThB33;
  hence A * g c= (-g) + A + g by ThB3;
  let x be object;
  assume x in (-g) + A + g;
  then consider a such that
A1: x = a + g and
A2: a in (-g) + A by Th28;
  consider b such that
A3: a = (-g) + b and
A4: b in A by A2,Th27;
  x = b * g by A1,A3;
  hence thesis by A4,Th41;
end;
