Prolog 误解链/2？

如果我有（加载clpfd）：

正如我所料，这给出了一个答案，但是：

test2(Ps):-
    permutation(Ps,[(a,1,1),(b,2,2),(c,3,1),(d,4,2)]), 
    Ps =[(L1,W1,X1),(L2,W2,X2),(L3,W3,Y1),(L4,W4,Y2)], 
    chain([X1,X2],#<).

我不希望得到这样的答案：

P = [(c,3,1), (b,2,2), (d,4,2), (a,1,1)]
P = [(c,3,1), (d,4,2), (b,2,2), (a,1,1)]

其中as的条款在bs的条款之后。

---

我误解了什么？

我现在可以看出我做错了什么

答案如下：

P = [(c,3,1), (b,2,2), (d,4,2), (a,1,1)]
P = [(c,3,1), (d,4,2), (b,2,2), (a,1,1)]

显然，前两个元素的第三个参数中有1和2，因此满足链约束

?-test2(P):-
P = [(a,1,1), (b,2,2), (c,3,1), (d,4,2)]
P = [(a,1,1), (b,2,2), (d,4,2), (c,3,1)]
P = [(a,1,1), (d,4,2), (b,2,2), (c,3,1)]
P = [(a,1,1), (d,4,2), (c,3,1), (b,2,2)]
P = [(c,3,1), (b,2,2), (a,1,1), (d,4,2)]
P = [(c,3,1), (d,4,2), (a,1,1), (b,2,2)]
P = [(c,3,1), (b,2,2), (d,4,2), (a,1,1)]
P = [(c,3,1), (d,4,2), (b,2,2), (a,1,1)]
false

P = [(c,3,1), (b,2,2), (d,4,2), (a,1,1)]
P = [(c,3,1), (d,4,2), (b,2,2), (a,1,1)]

P = [(c,3,1), (b,2,2), (d,4,2), (a,1,1)]
P = [(c,3,1), (d,4,2), (b,2,2), (a,1,1)]