I need to calculate types of following expressions:
- curry fst
- foldr const
- (foldr const) . (curry fst)
First I have:
curry :: ((a, b) -> c) -> a -> b -> c
fst :: (d, e) -> d
{} | (a,b)-> c = (d,e) -> d
{} | (a,b) = (d,e) , c = d
c -> d | (a,b) = (d,e)
c -> d , a ->d | b = e
c -> d , a ->d, b -> e | {}
then replace the result in a -> b -> c and you get d -> e -> d
Now similar to this I do:
foldr :: (f -> g -> g) -> g -> [f] -> g
const :: h -> i -> h
{} | (f -> g -> g) = h -> i -> h
{} | f = h , g = i, g = h
f -> h | g = i, g = h
f -> h, g -> i | i = h
f -> h, g -> i i->h | {}
then replace the result in g -> [f] -> g and you get i -> [h] -> i
Next, i have to do
(foldr const) . (curry fst) but i am not sure if these results are correct. I tried anyway but got stuck. So:
(foldr const) :: i -> [h] -> i and (curry fst) :: d -> e -> d
(.) :: (j -> k) -> (l -> j) -> l -> k
then i start with:
d = (j -> k) -> (l -> j) -> l -> k, e = d -> e -> d
e = (j -> k) -> (l -> j) -> l -> k -> e -> (j -> k) -> (l -> j) -> l -> k
but it feels wrong and i cannot continue... Are my first two results correct? If they are, how should i solve the last one?