Implementation of a Monad with proof

The Probably Monad is same as the Maybe Monad, so this is how I had it defined:

data Probably x = Nope | Kinda x deriving (Show)

instance Monad Probably where
Kinda x >>= f = f x
Nope >>= f = Nope
return = Kinda

add x y = x >>= (\x -> y >>= (\y -> return (x + y)))
-- Example: add (add (Kinda 1) (Kinda 2)) (Kinda 4)

Proof:

1. Left identity:
return a >>= f ≡ f a
“return a” by definition is “Kinda a”, and so the bind operator is defined as “Kinda a >>= f ≡ f a”.
So with these two definitions, we have shown that “return a >>= f ≡ f a”.

2. Right identity:
m >>= return ≡ m
If we let m equal “Kinda n”, then we have “Kinda n >>= return ≡ Kinda n”, but by definition of the binding operator,
we have that “Kinda n >>= return = return n”, and so “return n”. So, by definition of return, we have “Kinda n”, which is m.

3. Associativity:
(m >>= f) >>= g ≡ m >>= (\x -> f x >>= g)
If we let m equal “Kinda n”, then we have:
LHS: (Kinda n >>= f) >>= g ≡ (f n) >>= g
RHS: Kinda n >>= (\x -> f x >>= g) ≡ (\x -> f x >>= g) n ≡ f n >>= g

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s