Can someone explain the meaning of ((.)$(.)) (==) 1 (1+) 0
haskell, pointfree, syntax
Solution
The `a1` is "just another type variable". It could mean anything, including `a`, but doesn't necessarily mean anything. Most likely it is different from `a`.
`(==)` is the "forced prefix" form of `==` the regular equality operator form the `Eq` type class. Normally you'd write `a == b`, but that's just syntax sugar for `(==) a b`, the prefix application of `==`.
`1 (1+) 0` doesn't mean anything in particular in this context, each of the three subexpressions is an independent argument to "the owl", which ultimately takes four arguments.
We can walk through the reduction.
((.)$(.)) (==) 1 (1+) 0
=== [ apply ]
((.)(.)) (==) 1 (1+) 0
=== [ implicit association ]
((.)(.)(==)) 1 (1+) 0
=== [ apply the definition: (f.g) x = f (g x) ]
((.) (1 ==)) (1+) 0
=== [ implicit association ]
((.) (1 ==) (1+)) 0
=== [ apply the definition: (f.g) x = f (g x) ]
1 == (1+0)
=== [addition]
1 == 1
=== [equality]
True
As this page mentions, the owl is equivalent to a function `f`
f a b c d = a b (c d)
which is to say it applies its first argument, a function of two arguments, to its second argument and the result of applying its third argument to its fourth. For the example given `((.)$(.)) (==) 1 (1+) 0` that means you first apply `(+1)` to `0`, then combine the `1` and the `(1+0)` using `(==)` which is what happened in our reduction.
More broadly, you might think of it as a function which modifies a binary operation `a` to take a slight variation on its second argument.
Problem
On haskell.org I came across this point free style function, dubbed "the owl". ``` ((.)$(.)) ``` Its type signature is `(a -> b -> c) -> a -> (a1 -> b) -> a1 -> c`. It's equivalent to `f a b c d = a b (c d)` and apparently, `((.)$(.)) (==) 1 (1+) 0` returns `True`. So my questions are: - What does the `a1` in the type signature mean? Is it related to `a`? - Is `(==)` some kind of function equality operator? Because `0 (==) 0` throws an error in GHCi. - What does `1 (1+) 0` mean in this context? I don't see how this is even a valid expression. - Why does the expression return `True`?