Lambda Calculus AND Implementation in CLISP

Question

I'm very new with functional programming, lisp and lambda calculus. Im trying to implement the AND operator with Common Lisp Lambda Calc style.

From Wikipedia:

AND := λp.λq.p q p

So far this is my code:

(defvar TRUE #'(lambda(x)#'(lambda(y)x)))
(defvar FALSE #'(lambda(x)#'(lambda(y)y)))

(defun OPAND (p q)
    #'(lambda(f) 
        #'(lambda(p) #'(lambda(q) (funcall p (funcall q(funcall p))))))
)

I found this 2 conversion functions:

(defun church2int(numchurch)
    (funcall (funcall numchurch #'(lambda (x) (+ x 1))) 0)
)

(defun int2church(n)
    (cond
        ((= n 0) #'(lambda(f) #'(lambda(x)x)))
        (t #'(lambda(f) #'(lambda(x) (funcall f
            (funcall(funcall(int2church (- n 1))f)x))))))

)

If I do:

(church2int FALSE)

I've got 0. If I do this:

(church2int TRUE)

I have

#<FUNCTION :LAMBDA (X) (+ X 1)>

Which I think it's ok. But if I do this:

 (church2int (OPAND FALSE FALSE))

I've got:

#<FUNCTION :LAMBDA (Q) (FUNCALL P (FUNCALL Q (FUNCALL P)))>

Where I should have 0. Is there something wrong with my code? Or am I missing something?

Thanks

Answer 1

If you want to define opand as a function with 2 parameters, like you are trying to, you need to do this:

(defun OPAND (p q)
    (funcall (funcall p q) p) )

and then:

(opand false false)
#<FUNCTION :LAMBDA (X) #'(LAMBDA (Y) Y)> ;; which is FALSE

(opand true true)
#<FUNCTION :LAMBDA (X) #'(LAMBDA (Y) X)> ;; which is TRUE

This is my implementation, based on the original paper http://www.utdallas.edu/~gupta/courses/apl/lambda.pdf, of the and operator λxy.xyF

(defvar OPAND 
    #'(lambda(x) 
        #'(lambda(y) 
            (funcall (funcall x y) FALSE) ) ) )

And if you do

(funcall (funcall opand false) false)
#<FUNCTION :LAMBDA (X) #'(LAMBDA (Y) Y)> ;; which is FALSE

(funcall (funcall opand true) true)
#<FUNCTION :LAMBDA (X) #'(LAMBDA (Y) X)> ;; which is TRUE