;given definitions
(define zero (lambda (f) (lambda (x) x)))
(define (add-1 n)
  (lambda (f) (lambda (x) (f ((n f) x)))))
; exercise 2.6: define one and two directly -
; not in terms of zero or add-1
(define one
  (lambda (f) (lambda (x) (f x))))
 
(define two
  (lambda (f) (lambda (x) (f (f x)))))
 
(define (church-plus a b)
  ((a (lambda (n) (lambda (f) (lambda (x) (f ((n f) x)))))) b))
 
(define (church->integer a)
  ((a add1) 0))
 
(display (church->integer zero))

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