theorem Th54:
  A * a * (-a) = A & A * (-a) * a = A
proof
  thus A * a * (-a) = A * (a + (-a)) by Th47
    .= A * 0_G by Def5
    .= A by ThB52;
  thus A * (-a) * a = A * ((-a) + a) by Th47
    .= A * 0_G by Def5
    .= A by ThB52;
end;
