anchor.diff:  
   	Breaking this down you realize we simply take a list, split it up into two and take the first
   	list and second list and mix it back together so no doubt we are bound to get the same list back.
  -
   |#
   
   
  
positive.code: 	(defun subsetsum (L S)
  		(cond 
  			((null L) nil)
  			((> (car L) S)
  				(subsetsum (cdr L) S)
  			)
  			((= (car L) S)
  				(cons (car L) ()))
  				;(cons (car L))
  			((< (car L) S)
  				(if (null (subsetsum (cdr L) (- s (car L))))
  					(subsetsum (cdr L) S)
  					(cons (car L) (subsetsum (cdr L) (- S (car L))))
  				)
  
negative.code: #|	ASSIGNMENT ONE
  	NAME: SONOLA SOWEMIMO
  	CLASS: CMPUT 325
  	ID: 1293430
  |#
  
  ;QUESTION ONE
  #| 
   The function returns T if argument X is a member of the argument list Y and NIL otherwise. This should
   also test for lists being members of lists. Both the argument X and the list Y may be NIL or lists containing NIL.
  	Examples: 
  	(xmember '1 '(1)) --> T
  	(xmember '1 '( (1) 2 3)) --> NIL
  	(xmember '(1) '((1) 2 3)) --> NIL
  	(xmember nil nil) --> NIL 
  |#
  
  	(defun xmember (X Y)
  		(cond
  