I am new to Haskell. I am trying to understand how composition of types work.
(.) :: (b -> c) -> (a -> b) -> a -> c
fmap :: Functor f => (x -> y) -> f x -> f y
fmap . fmap :: (Functor f1, Functor f2) => (x -> y) -> f1 (f2 x) -> f1 (f2 y)
How I understood the above type information is as below.
1. (x -> y) -> f1 x -> f1 y -- First fmap
-> (x' -> y') -> f2 x' -> f2 y' -- Second fmap
2. Compare the (1) with (.) type signature then we get
In (b -> c)
b = (x -> y)
c = f1 x -> f1 y
In (a -> b)
a = (x' -> y')
b = f2 x' -> f2 y'
3. Now the result is a -> c but before that b in (b -> c) should be b in (a -> b)
(x -> y) === f2 x' -> f2 y'
That means x = f2 x' & y = f2 y'
4. Result is a -> c
(x' -> y') -> f1 x -> f1 y
Substituting (3) results here
(x' -> y') -> f1 (f2 x') -> f1 (f2 y')
This is Alpha equivalent to (x -> y) -> f1 (f2 x) -> f1 (f2 y)
To test my understanding I tried with various data types and I am success in few. Below is the type signature I can't figure out.
f = undefined :: (x -> y -> w -> z -> a) -> g x -> g y
-- Type of f . f in prelude
f . f :: (x -> (w1 -> z1 -> a1) -> w2 -> z2 -> a2) -> g (y -> x) -> g y
My approach to the above one
1. (x -> y -> w -> z -> a) -> g x -> g y
-> (x' -> y' -> w' -> z' -> a') -> g' x' -> g' y'
2. Comparing with (.) then
In (b -> c)
b = (x -> y -> w -> z -> a)
c = g x -> g y
In (a -> b)
a = (x' -> y' -> w' -> z' -> a')
b = g' x' -> g' y'
3. Comparing b from (b -> c) & (a -> b)
(x -> y -> w -> z -> a) === g' x' -> g' y'
That means x = g' x' & y -> w -> z -> a = g' y'
4. Result is a -> c
(x' -> y' -> w' -> z' -> a') -> g x -> g y
Now there is no direct y I can substitute in g y
My approach of seeing the composition is not working out for me.
Type derivation is a purely mechanical process.
(.) :: (b -> c ) -> (a -> b ) ->
(a -> c)
fmap₂ :: Functor f => (t3 -> t4) -> (f t3 -> f t4)
fmap₁ :: Functor g => ((t1 -> t2) -> (g t1 -> g t2))
-------------------------------------------------------------------------------------------
a ~ (t1 -> t2) b ~ (g t1 -> g t2)
b ~ (t3 -> t4) c ~ (f t3 -> f t4) b ~ (t3 -> t4 )
-------------------------------------------------------------------------------------------
fmap₂ . fmap₁ :: (Functor f, Functor g)
=> a -> c
~ (t1 -> t2) -> (f t3 -> f t4 )
~ (t1 -> t2) -> (f (g t1) -> f (g t2))
This is much nicer with (.)'s cousin, (>>>) fmap₁ fmap₂ = fmap₂ . fmap₁:
(>>>) :: (a -> b ) ->
( b -> c ) ->
(a -> c )
fmap₁ :: Functor g => (t1 -> t2) -> (g t1 -> g t2)
fmap₂ :: Functor f => (t3 -> t4 ) -> (f t3 -> f t4)
------------------------------------------------------------------------------
a ~ (t1 -> t2) b ~ (g t1 -> g t2)
b ~ (t3 -> t4 ) c ~ (f t3 -> f t4)
------------------------------------------------------------------------------
fmap₁ >>> fmap₂ :: (Functor f, Functor g)
=> a -> c
~ (t1 -> t2) -> (f t3 -> f t4 )
~ (t1 -> t2) -> (f (g t1) -> f (g t2))
To your specific problem:
f = undefined :: (x -> y -> w -> z -> a) -> g x -> g y
f₂ . f₁ = f₁ >>> f₂ ::
(x -> y -> w -> z -> a) -> ((->) (g x) (g y ))
((->) x2 ((->) y2 (w2 -> z2 -> a2))) -> (g2 x2 -> g2 y2)
---------------------------------------------------------------------------------------
x2 ~ g x g ~ ((->) y2) y ~ (w2 -> z2 -> a2)
---------------------------------------------------------------------------------------
(x -> y -> w -> z -> a) -> ( g2 x2 -> g2 y2) ~
(x -> y -> w -> z -> a) -> ( g2 (g x) -> g2 y2) ~
(x -> (w2 -> z2 -> a2) -> w -> z -> a) -> ( g2 (y2 -> x) -> g2 y2)
which is what you got from GHCi, up to the renaming of variables.