在haskell中建立重言式和可满足公式

Question

我遇到了一些困难，试图理解如何建立重言式和可满足公式。我正在处理的问题，要求我模拟一个NAND门和一个NOR门使用这些方法。

问题：

通过扩展文件proplog.hs中的代码

A-模仿nand，也不表示Not，或和和.当至少一个输入为假时，nand为真。

当两个输入都为假时，也不正确。

B-通过使用nand，nor，xor，impl，T，F生成

( a) 2个重言式

( b)可满足公式

( c)不可满足的公式

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

您可以用其他门来编写nand，但是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？当然了！

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