Invertible List Manipulation

(SETF DATABASE4 '(

;;; (APPEND L1 L2 L3) is true provided L3 is the

;;; result of appending L1 and L2.

((APPEND NILL L L))

((APPEND L NILL L))

((APPEND (CONS X L1) L2 (CONS X L3)) (APPEND L1 L2 L3))

))

(QUERY

'((APPEND (CONS X (CONS Y NILL))

(CONS Z (CONS W NILL))

L)) )

SOLUTION: L=(CONS X (CONS Y (CONS Z (CONS W NILL))))