Haskell monadic proof

Question

For the mathematical experts (I am not, nor a Haskell expert):

m >> k = m >>= \_ -> k

This 'monad' typechecks and compiles. Is this to mathematically proof that m >> k (omitting a return value) and m >>= \\_ -> k (a monad with return, but a lambda without) are the same, or can I actually put values in? Just being curious. Not a blocking issue.

Answer 1

This is definition of function >> in infix form.

It is equivalent to more usual (>>) mk = ... .

Brackets here is to explain Haskell that we use operator in prefix form.

Answer 2

I a assume you are reading the following bit in the Prelude http://hackage.haskell.org/package/base-4.12.0.0/docs/src/GHC.Base.html#%3E%3E

This is not a proof. This is more akin to the definition of a method in an OOP interface.

There in the definition of the class Monad (class in Haskell is more like an interface); it's defined that an operator >> should be defined. And a default definition is supplied. The default definition is m >>= \\_ -> k .