Exercise 2.59 Union-set for sets represented as unordered lists
grevtsevalex

Code Review

Compare your solutions
    #| BEGIN (Введите свое решение) |#
(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set? x (cdr set)))))

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

#| END |#

    #| BEGIN (Введите свое решение) |#
(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set? x (cdr set)))))

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

#| END |#

    #| BEGIN (Введите свое решение) |#
(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set? x (cdr set)))))

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

#| END |#

    #| BEGIN (Введите свое решение) |#
(define (element-of-set? x set)
  (cond ((null? set) #f)
        ((equal? x (car set)) #t)
        (else (element-of-set? x (cdr set)))))

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

#| END |#