問題2.60

(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set? x (cdr set)))))

(element-of-set? 4 '(1 2 3 2 3 1 5 4))
;#t

(element-of-set? 6 '(1 2 3 2 3 1 5 4))
;#f

(define adjoin-set cons)

(adjoin-set 4 '(1 2 4))
;(4 1 2 4)

(define union-set append)

(union-set '(1 2 3 5 9) '(1 2 4))
;(1 2 3 5 9 1 2 4)

(define (intersection-set set1 set2)
  (cond ((or (null? set1) (null? set2)) '())
        ((element-of-set? (car set1) set2)
         (cons (car set1)
               (intersection-set (cdr set1) set2)))
        (else (intersection-set (cdr set1) set2))))

(intersection-set '(1 2 5 6 8 4 3) '(1 2 3 4))
;(1 2 4 3)

(intersection-set '(1 8) '(7 2))
;()