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There are three answers to your question, one captious and one unhelpful and one abstract: <h2>The captious answer</h2> <pre><code>instance Monoid (Ftree a) where mempty = Empty mappend = Branch </code></pre> This is an instance of the <code>Monoid</code> type class, but does not satisfy any of the required properties. <h2>The unhelpful answer</h2> What Monoid do you want? Just asking for a monoid instance without further information is like asking for a solution without giving the problem. Sometimes there is a natural monoid instance (e.g. for lists) or there is only one (e.g. for <code>()</code>, disregarding questions of definedness). I don’t think either is the case here. BTW: There would be an interesting monoid instance if your tree would have data at internal nodes that combines two trees recursively... <h2>The abstract answer</h2> Since you gave a <code>Monad (Ftree a)</code> instance, there is a generic way to get a <code>Monoid</code> instance: <pre><code>instance (Monoid a, Monad f) => Monoid (f a) where mempty = return mempty mappend f g = f >>= (\x -> (mappend x) `fmap` g) </code></pre> Lets check if this is a Monoid. I use <code><> = mappend</code>. We assume that the <code>Monad</code> laws hold (I did not check that for your definition). At this point, recall the <a href="http://www.haskell.org/haskellwiki/Monad_Laws" rel="nofollow">Monad laws written in do-notation</a>. Our <code>mappend</code>, written in do-Notation, is: <pre><code>mappend f g = do x <- f y <- g return (f <> g) </code></pre> So we can verify the monoid laws now: Left identity <pre><code>mappend mempty g ≡ -- Definition of mappend do x <- mempty y <- g return (x <> y) ≡ -- Definition of mempty do x <- return mempty y <- g return (x <> y) ≡ -- Monad law do y <- g return (mempty <> y) ≡ -- Underlying monoid laws do y <- g return y ≡ -- Monad law g </code></pre> Right identity <pre><code>mappend f mempty ≡ -- Definition of mappend do x <- f y <- mempty return (x <> y) ≡ -- Monad law do x <- f return (x <> mempty) ≡ -- Underlying monoid laws do x <- f return x ≡ -- Monad law f </code></pre> And finally the important associativity law <pre><code>mappend f (mappend g h) ≡ -- Definition of mappend do x <- f y <- do x' <- g y' <- h return (x' <> y') return (x <> y) ≡ -- Monad law do x <- f x' <- g y' <- h y <- return (x' <> y') return (x <> y) ≡ -- Monad law do x <- f x' <- g y' <- h return (x <> (x' <> y')) ≡ -- Underlying monoid law do x <- f x' <- g y' <- h return ((x <> x') <> y') ≡ -- Monad law do x <- f x' <- g z <- return (x <> x') y' <- h return (z <> y') ≡ -- Monad law do z <- do x <- f x' <- g return (x <> x') y' <- h return (z <> y') ≡ -- Definition of mappend mappend (mappend f g) h </code></pre> So for every (proper) Monad (and even for every applicative functor, as Jake McArthur pointed out on #haskell), there is a Monoid instance. It may or may not be the one that you are looking for.
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