theorem ThB25:
  a * b * (-b) = a & a * (-b) * b = a
proof
  thus a * b * (-b) = a * (b + (-b)) by Th24
    .= a * 0_G by Def5
    .= a by Th19;
  thus a * (-b) * b = a * ((-b) + b) by Th24
    .= a * 0_G by Def5
    .= a by Th19;
end;
