Is `x >> pure y` equivalent to `liftM (const y) x`

Question

The two expressions

y >> pure x
liftM (const x) y

have the same type signature in Haskell. I was curious whether they were equivalent, but I could neither produce a proof of the fact nor a counter example against it.

If we rewrite the two expressions so that we can eliminate the x and y then the question becomes whether the two following functions are equivalent

flip (>>) . pure
liftM . const

Note that both these functions have type Monad m => a -> m b -> m a.

I used the laws that Haskell gives for monad, applicatives, and functors to transform both statements into various equivalent forms, but I was not able to produce a sequence of equivalences between the two.

For instance I found that y >> pure x can be rewritten as follows

y >>= const (pure x)
y *> pure x
(id <$ y) <*> pure x
fmap (const id) y <*> pure x

and liftM (const x) y can be rewritten as follows

fmap (const x) y
pure (const x) <*> y

None of these spring out to me as necessarily equivalent, but I cannot think of any cases where they would not be equivalent.

Answer 1

The other answer gets there eventually, but it takes a long-winded route. All that is actually needed are the definitions of liftM, const, and a single monad law: m1 >> m2 and m1 >>= \_ -> m2 must be semantically identical. (Indeed, this is the default implementation of (>>), and it is rare to override it.) Then:

liftM (const x) y
= { definition of liftM* }
y >>= \z -> pure (const x z)
= { definition of const }
y >>= \z -> pure x
= { monad law }
y >> pure x

* Okay, okay, so the actual definition of liftM uses return instead of pure. Whatever.