Am 01.02.11 14:05, schrieb Houdini:
that is really.....complicated...hmm
Maybe you find assign :: (Atom,Bool)->Formula->Formula assign _ [] = [] assign (a,b) (c:cs) | (b,a) `elem` c = cs' | otherwise = filter ((/= a).snd) c : cs' where cs' = assign (a,b) cs easier to read. With example :: Formula example = [[p,q,r],[n p,q,n r],[p,n q]] where p = l "P" q = l "Q" r = l "R" l a = (True, a) n (b,a) = (not b, a) we get *Algorithm> example [[(True,"P"),(True,"Q"),(True,"R")],[(False,"P"),(True,"Q"),(False,"R")],[(True,"P"),(False,"Q")]] *Algorithm> assign ("P", True) example [[(True,"Q"),(False,"R")]] *Algorithm> assign ("P", False) example [[(True,"Q"),(True,"R")],[(False,"Q")]] But note that we also get *Algorithm> assign ("P", False) [[(True,"P"),(True,"Q"),(True,"R")],[(False,"P"),(True,"Q"),(False,"R")],[(True,"P")]] [[(True,"Q"),(True,"R")],[]] So this is not reduced to [[]]. hth C.