在haskell中建立重言式和可满足公式

Question

我在试图理解如何构建同义词和可满足公式时遇到了一些困难。我正在研究一个问题，这个问题需要我用这样的方法来模拟NAND门和NOR门。

问题：

通过扩展文件proplog.hs 中的代码

A-模仿nand，也不使用Not、Or和and。当至少一个输入为假时，nand为真

当它的两个输入都是假时也不是真的

B-通过使用nand，nor，xor，impl，T，F构建

a） 2个同义词

b） 一个可满足的公式

c） 不令人满意的公式

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

您可以根据其他门来编写nand，但更好的方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

什么是最伟大的逻辑重言式？莫杜斯·波恩斯当然！

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

以下是一些简单的公式：

f0 p q = p `and'` q 
satisfy_f0 = sat2 f0
f1 p = p `and'` (not' p)
satisfy_f1 = sat1 f1