theorem
  |{ p2,p1,p2 }| = 0
proof
A1:(p1<X>p2).1 = (p1.2*p2.3)-(p1.3*p2.2);
A2:(p1<X>p2).2 = (p1.3*p2.1)-(p1.1*p2.3);
(p1<X>p2).3 = (p1.1*p2.2)-(p1.2*p2.1);
   then |{ p2,p1,p2 }| = p2.1*((p1.2*p2.3)-(p1.3*p2.2))
  +p2.2*((p1.3*p2.1)-(p1.1*p2.3))
  +p2.3*((p1.1*p2.2)-(p1.2*p2.1)) by A2,A1,Lm5
.= 0;
  hence thesis;
end;
