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My solution based on the following procedure.

```
(define (reverse seq)
(if (null? seq)
'()
(append (reverse (cdr seq)) (list (car seq)))))
```

So, there will be two rules:

--1st: an empty list will be got if reverse an empty list.

--2nd: for x y v z, we will get a reversed seq 'z' of (cons x y) only if 'v' is a reversed seq of 'y' and 'z' is (append v '(x))

(assert! (rule (reverse () ()))) (assert! (rule (reverse (?x . ?y) ?z) (and (reverse ?y ?v) (append-to-form ?v (?x) ?z)))))

the above one will work for (reverse (1 2 3) ?x), but end up with an infinite loop for (reverse ?x (1 2 3)). But this situation will be opposite if change the order of sequence like:

(assert! (rule (reverse () ()))) (assert! (rule (reverse (?x . ?y) ?z) (and (append-to-form ?v (?x) ?z) ;; changed (reverse ?y ?v))))) ;;

meteorgan