(define subst
  (lambda (new old ls)
    ;(writeln new old ls)
    (cond
      ((null? ls) '())
      ;None of the examples show what subst does when 
      ;new and old are lists. We could throw an error.
      ;In spite of the fact that this section is on flat
      ;recursion, I chose to use equal? which recursively 
      ;compares.
      (else 
        (cond 
          ((equal? (car ls) old) 
            ;(writeln "1: " new old ls)  
            (cons new (subst new old (cdr ls)))
          )
          (else
            ;(writeln "2: " new old ls) 
            (cons (car ls) (subst new old (cdr ls)))
          )
        )
      )
    )
  )
)
  
  
  
;============== Unit Tests ======================
(load "test.scm")
(load "writeln.scm")
(define msg "subst")
(test msg (subst 'z 'a '(a b a c a)) '(z b z c z)) 
(test msg (subst '0 '1 '(0 1 0 1)) '(0 0 0 0))
(test msg (subst 'dog 'cat '(my dog is fun)) 
                           '(my dog is fun))

(test msg (subst 'two 'one '()) '())

(test msg (subst '(a a) 'x '(x b x c x))
                           '((a a) b (a a) c (a a)))

(test msg (subst 'x '(a a) '((a a) b (a a) c (a a)))
                           '(x b x c x))

(test msg (subst 'x '(a b) '((a a) b (a a) c (a a)))
                           '((a a) b (a a) c (a a)))