問題2.59

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

(define (adjoin-set x set)
  (if (element-of-set? x set) set
      (cons x set)))

(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))))

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

(union-set (list 1 2 3) (list 1 2 3))
;(1 2 3)

(union-set (list 1 2 3) (list 1 2 3 4))
;(1 2 3 4)

(union-set (list 1 2 3) (list 4 5 6))
;(1 2 3 4 5 6)