Найти все слова, которые есть в первом предложении и отсутствуют во втором

(defun MemberOf (obj lst)
 ((null lst) nil)
 ((equal obj (car lst)) T)
 (Member obj (cdr lst))
)
 
(defun Find (lst1 lst2)
 ((null lst) nil)
 ((not (MemberOf (car lst1) lst2))
  (Find (cdr lst1) lst2)
)
(
 (cons (car lst1) (Find (cdr lst1) lst2))
)
)

(defun member-of (lst elem)
  (cond ((null lst)
     nil)
    ((equal elem (car lst))
     t)
    (t 
     (member-of (cdr lst) elem))))
 
(defun in-1st-not-2nd (list1 list2)
  (defun iter (resulting-list testing-list)
    (cond ((null testing-list)
       resulting-list)
      ((not (member-of list2 (car testing-list)))
       (iter (cons (car testing-list) resulting-list)
         (cdr testing-list)))
      (t
       (iter resulting-list
         (cdr testing-list)))))
  (iter nil list1))

(defun our-set-difference (w v)
  (cond ((null w) w)
        ((member (car w) v) (our-set-difference (cdr w) v))
        ((cons (car w) (our-set-difference (cdr w) v)))))
 
> (our-set-difference  '(a b c d e f) '(c d f))
(A B E)
> (set-difference  '(a b c d e f) '(c d f))
(A B E)

(defun task (s1 s2)
   (let ((r nil))
    (dolist (i s1 r) (when (not (member i s2)) (setq r (cons i r))))))
 
==> task
 
(task '(a b c) '(a d e f))
 
==> (c b)
 
;; HomeLisp
 
(defun task (s1 s2)
  (iter (for i in s1) (when (not (member i s2)) (collecting i))))
 
==> task
 
(task '(a b c) '(a d e f))
 
==> (b c)

(defun task (s1 s2)
 (let ((r nil))
  (mapcar #'(lambda (x) (when (not (member x s2)) (setq r (cons x r)))) s1) r ))
 
==> task
 
(task '(a b c) '(a d e f))
 
==> (c b)