在haskell中构建重言式和可满足的公式

Question

我在试图了解如何构建重言式和可满足的公式时遇到一些困难。我正在研究需要我使用这种方法模拟NAND门和NOR门的问题。

问题：

通过扩展文件proplog.hs

A中的代码 - 模拟NAND，NOR中没有，或者与和条款。 NAND为真时的其输入中的至少一个为假

也不是真时它的两个输入均为假

乙 - 通过使用NAND，NOR，XOR，IMPL，T，F建立

一个）2重言

b）中令人满意的式

c）一种不可满足式

Proplog.hs：

-- definition of basic gates 
data Prop = T | F | 
    Not Prop | 
    And Prop Prop | 
Or Prop Prop 
deriving (Eq,Read,Show) 

-- truth tables of basic gates 

tt (Not F) = T 
tt (Not T) = F 
tt (And F F) = F 
tt (And F T) = F 
tt (And T F) = F 
tt (And T T) = T 

tt (Or F F) = F 
tt (Or F T) = T 
tt (Or T F) = T 
tt (Or T T) = T 

-- giving the tt of a derived gate 
xor' F F = F 
xor' F T = T 
xor' T F = T 
xor' T T = F 

-- building the derived gate from Not, And, Or 
xor x y = eval (And (Or x y) (Not (And x y))) 

-- evaluating expressions made of logic gates 

eval T = T 
eval F = F 
eval (Not x) = tt (Not (eval x)) 
eval (And x y) = tt (And (eval x) (eval y)) 
eval (Or x y) = tt (Or (eval x) (eval y)) 

ite c t e = eval (Or (And c t) (And (Not c) e)) 

truthTable1 f = [(x,f x)|x<-[F,T]] 

tt1 f = mapM_ print (truthTable1 f) 

truthTable2 f = [((x,y),f x y)|x<-[F,T],y<-[F,T]] 

tt2 f = mapM_ print (truthTable2 f) 

truthTable3 f = [((x,y),f x y z)|x<-[F,T],y<-[F,T],z<-[F,T]] 

tt3 f = mapM_ print (truthTable3 f) 

or' x y = eval (Or x y) 
and' x y = eval (And x y) 
not' x = eval (Not x) 

impl x y = eval (Or (Not x) y) 

eq x y = eval (And (impl x y) (impl y x)) 

deMorgan1 x y = eq (Not (And x y)) (Or (Not x) (Not y)) 
deMorgan2 x y = eq (Not (Or x y)) (And (Not x) (Not y)) 


-- tautologies, satisfiable and unsatisfiable formulas 

taut1 f = all (==T) [f x|x<-[F,T]] 

taut2 f = all (==T) [f x y|x<-[F,T],y<-[F,T]] 

sat1 f = any (==T) [f x|x<-[F,T]] 

sat2 f = any (==T) [f x y|x<-[F,T],y<-[F,T]] 

unsat1 f = not (sat1 f) 
unsat2 f = not (sat2 f) 

-- examples of tautologies: de Morgan1,2 
-- examples of satisfiable formulas: xor, impl, ite 

-- example of contradiction (unsatisfiable formulas): contr1 
contr1 x = eval (And x (Not x)) 

谢谢！

Answer 1

你可以写NAND在其他门的条款，但更好的办法可能是直接定义它：

-- Nand 
nand x y = not' (and' x y) 

-- Nand - by definition 
nand' F _ = T 
nand' _ F = T 
nand' _ _ = F 

什么是最大的逻辑同义反复？ Modus ponens当然！

modusPonens p q = (p `and'` (p `impl` q)) `impl` q 
prove_modusPonens = taut2 modusPonens 

这里有一些简单的公式：

f0 p q = p `and'` q 
satisfy_f0 = sat2 f0 

f1 p = p `and'` (not' p) 
satisfy_f1 = sat1 f1