Рекурсия. Определить предикат, проверяющий пересечение двух множеств

(defun _intersection (w v)
  (cond ((null w) nil)
        ((member (car w) v) (cons (car w) (_intersection (cdr w) v)))
        ((_intersection (cdr w) v))))
 
> (_intersection '(a s d f g h ) '(q a w s e d r))
(A S D)

(define (by-value l1 l2 (acc null))
  (if (null? l1)
      acc
      (by-value (cdr l1) l2 (if (member (car l1) l2) (cons (car l1) acc) acc))))
 
(define (by-arg l1 l2)
  (cond ((null? l1) null)
        ((member (car l1) l2) (cons (car l1) (by-arg (cdr l1) l2)))
        ((by-arg (cdr l1) l2))))
 
(by-value '(1 2 3) '(2 3 4))
(by-arg '(1 2 3) '(2 3 4))
;'(3 2)
;'(2 3)

(defun inters-1 (L1 L2 &optional (r nil))
  (cond ((null L1) r)
        ((member (car L1) L2) (inters-1 (cdr L1) L2 (cons (car L1) r)))
        (t  (inters-1 (cdr L1) L2 r))))
 
==> inters-1
 
(inters-1 '(1 2 3 4 5) '( 7 6 5 4 3))
 
==> (5 4 3)
 
(defun inters-2 (L1 L2)
   (cond ((null L1) nil)
         ((member (car L1) L2) (cons (car L1) (inters-2 (cdr L1) L2)))
         (t (inters-2 (cdr L1) L2))))
 
==> inters-2
 
(inters-2 '(1 2 3 4 5) '( 7 6 5 4 3))
 
==> (3 4 5)

(defun _intersection (w v &optional acc &aux (a (car w)))
  (cond ((null w) (nreverse acc))
        ((member a v) (_intersection (cdr w) v (cons a acc )))
        ((_intersection (cdr w) v acc))))
 
> (_intersection '(a s d f g h ) '(q a w s e d r))
(A S D)

(defun _intersection (w v &optional acc &aux (a (car w)))
  (if w
      (_intersection (cdr w) v (if (member a v)
                                   (cons a acc )
                                   acc))
      (nreverse acc)))
 
> (_intersection '(a s d f g h ) '(q a w s e d r))
(A S D)