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POGenerating truth tables for logic expressions in Haskell
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copied!<p>The first part is an evaluation function that has the following type signature:</p> <pre><code>evaluate :: Logic Expr -> [(Variable, Bool)] -> Bool </code></pre> <p>This takes a logic expression and a list of assignment pairs as input and returns the value of the expression according to the Boolean assignment provided. The assignment list is a distinct list of pairs where each pair contains a variable and its Boolean assignment. That is, if you pass to the function the expression A &and; B and the assignment A = 1 and B = 0, your function must return 0 (this comes from Digital Logic Design, 0 corresponds to false, and 1 corresponds to true).</p> <p>This is what I managed to do so far:</p> <pre><code>type Variable = Char data LogicExpr = V Variable | Negation LogicExpr | Conjunction LogicExpr LogicExpr | Disjunction LogicExpr LogicExpr | Implication LogicExpr LogicExpr evaluate :: LogicExpr -> [(Variable,Bool)] -> Bool evaluate (V a) ((x1,x2):xs) | a==x1 = x2 | otherwise = (evaluate(V a)xs) evaluate (Negation a) l | (evaluate a l)==True = False | otherwise = True evaluate (Conjunction a b) l = (evaluate a l)&&(evaluate b l) evaluate (Disjunction a b) l = (evaluate a l)||(evaluate b l) evaluate (Implication a b) l | (((evaluate b l)==False)&&((evaluate a l)==True)) = False | otherwise = True </code></pre> <p>The next part is to define <code>generateTruthTable</code>, which is a function that takes a logic expression as input and returns the truth table of the expression in the form of a list of lists of assignment pairs. That is, if you pass to the function the expression E = A &and; B, your function must return A = 0, B = 0, E = 0 | A = 0, B = 1, E = 0 | A = 1, B = 0, E = 0 | A = 1, B = 1, E = 1.</p> <p>I'm not exactly familiar with the syntax so I don't know how to return the list.</p>

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