2.4|

$ cd ~

SICP Exercise 2.4

Using applicative-order evaluation on car:

(car (cons x y)) 
(car (lambda (m) (m x y))) 
((lambda (m) (m x y)) (lambda (p q) p)) 
((lambda (p q) p) x y) 
x 

Similarly, cdr is:

(define (cdr z)
  (z (lambda (p q) q)))