在没有CLP（set）的情况下在core.logic中设置成员资格-定义行为

Question

由于CLP（set）似乎在“立即”路线图中已经停留了7年之久，而且还在不断增加，所以我试图做一个不完整，效率低下的替代方案，以至少能够反映集合成员资格的语义。主要思想是检查类成员函数的成员资格，并使用CLP（fd）强制执行整数列表中的顺序，因此“集合”列表的排列不会使可能的解决方案数量呈指数级增长。我提出了以下建议：

(defne set-membero [elt set both]
  ([elt set both]
   (== set both)
   (membero elt set))
  ([elt [] [elt]])
  ([elt [x . smallrest] [x . bigrest]]
   (fd/in elt x Integer/MIN_VALUE Integer/MAX_VALUE)
   (fd/< x elt)
   (set-membero elt smallrest bigrest))
  ([elt [x . _] [elt . set]]
   (fd/in elt x Integer/MIN_VALUE Integer/MAX_VALUE)
   (fd/< elt x)))

预期用途类似于以下内容：

(run 2 [q] (set-membero 1 [2 3] q)) "=> ((1 2 3))"   ; adds non existing members as membero                                                                                                                      
(run 2 [q] (set-membero 1 [1 2] q)) "=> ((1 2))"     ; but does not add existing elements                                                                                                                        
(run 2 [q] (set-membero q [1 2] [1 2 3])) "=> (3)"   ; substracts sets as expected                                                                                                                               
(run 2 [q] (set-membero q [1] [1 2 3])) "=> ()"      ; remember it is not set-difference!                                                                                                                        
(run 2 [q] (set-membero 1 q [1 2])) "=> ((1 2))"     ; extracts elements as expected
(run 2 [q] (set-membero 2 [3 1] q)) "=> ()"          ; fails with non-compliant "sets" (unordered)

但是我一直坚持为什么会发生这种情况：

(run 2 [q] (set-membero 4 [3] q))
=> () ; expected ([3 4])

这种简单的情况有效：

(run 2 [q] (set-membero 4 [] q))
=> ([4]) ; as expected

不是应该通过第三条defne减少规则将(set-membero 4 [3] q)减少到(set-membero 4 [] q)吗？