# 99 questions/Solutions/46

### From HaskellWiki

(**) Define predicates and/2, or/2, nand/2, nor/2, xor/2, impl/2 and equ/2 (for logical equivalence) which succeed or fail according to the result of their respective operations; e.g. and(A,B) will succeed, if and only if both A and B succeed.

A logical expression in two variables can then be written as in the following example: and(or(A,B),nand(A,B)).

Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables.

The first step in this problem is to define the Boolean predicates:

-- Negate a Boolean argument not' :: Bool -> Bool not' True = False not' False = True -- True if both a and b are True and',or',nor',nand',xor',impl',equ' :: Bool -> Bool -> Bool and' True True = True and' _ _ = False -- True if a or b or both are True or' False False = False or' _ _ = True -- Negation of 'or' nor' a b = not' $ or' a b -- Negation of 'and' nand' a b = not' $ and' a b -- True if either a or b is true, but not if both are true xor' True False = True xor' False True = True xor' _ _ = False -- True if a implies b, equivalent to (not a) or (b) impl' a b = (not' a) `or'` b -- True if a and b are equal equ' True True = True equ' False False = True equ' _ _ = False

The explicit implementation of logic functions above could be shortened using Haskell's builtin equivalents:

and' a b = a && b or' a b = a || b nand' a b = not (and' a b) nor' a b = not (or' a b) xor' a b = and' (or' a b) (nand' a b) impl' a b = or' (not a) b equ' a b = a == b

Some could be reduced even further through Pointfree style:

and' = (&&) or' = (||) equ' = (==)

Anyway, the only remaining difficulty is to generate the truth table. This approach accepts a Boolean function `(Bool -> Bool -> Bool)`, then calls that function with all four combinations of two Boolean values:

table :: (Bool -> Bool -> Bool) -> IO () table f = mapM_ putStrLn [show a ++ " " ++ show b ++ " " ++ show (f a b) | a <- [True, False], b <- [True, False]]

The table function in Lisp supposedly uses Lisp's symbol handling to substitute variables on the fly in the expression. I chose passing a binary function instead because parsing an expression would be more verbose in haskell than it is in Lisp. Template Haskell could also be used :)