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copied!<p>The trick is to make a type class for which you will define an instance for functions, and an instance for the return type. The fact that it's a <code>Bool</code> is not a problem at all.</p> <p>We're trying to write a function which takes a variadic argument and returns a <code>Bool</code>, so we'll define a type class with such a function.</p> <pre><code>class Stmt a where tautology :: a -> Bool </code></pre> <p>Next, we define an instance for the return type of the variadic function. In this case, that's <code>Bool</code>.</p> <pre><code>-- A Bool is a tautology if it's True. instance Stmt Bool where tautology = id </code></pre> <p>The key part is the next instance for functions that take a <code>Bool</code> argument, and whose return type is some type from our class. That way, this instance will be applied multiple times if a function takes multiple arguments.</p> <pre><code>-- A function is a tautology if it always returns a tautology. instance Stmt b => Stmt (Bool -> b) where tautology f = tautology (f True) && tautology (f False) </code></pre> <p>Writing it this way requires <code>FlexibleInstances</code> because of the <code>Bool</code> in the second instance head. To do the same with pure Haskell 98, we'll need to use a suitably-constrained type variable instead. We can for example use <code>Bounded</code> and <code>Enum</code> (there are instances for both for <code>Bool</code>), or you can make your own class that will let you construct the appropriate inputs.</p> <pre><code>instance (Enum a, Bounded a, Stmt b) => Stmt (a -> b) where tautology f = all (tautology . f) [minBound .. maxBound] </code></pre> <p>And we're done. Let's try it out:</p> <pre><code>> tautology $ \x y -> (not x && not y) == not (x && y) False > tautology $ \x y -> (not x && not y) == not (x || y) True </code></pre>

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