Use s-combinator for questions related to creating a function which does generic partial function application
When the s-combinator S is applied to a set of combinators x y z, it yields a combinator that applies x to y after first substituting z into each of them. For example:
Can an analog of the S combinator be expressed in Haskell using only standard functions (without defining it by equation) and without using lambda (anonymous function)? I expect it to by of type (a -> b -> c) -> (a -> b) -> a -> c.
For example, an…
I'm starting to learn lambda calculus and I need to implement I, S, K combinators in Erlang.
Of course, S, K, I stands for:
S = λxyz.xz(yz) K = λxy.x I = λx.x
I have no problem understanding I=SKK transformation on paper (like presented here: To…
Suppose there are some data types to express lambda and combinatorial terms:
data Lam α = Var α -- v
| Abs α (Lam α) -- λv . e1
| App (Lam α) (Lam α) -- e1 e2
deriving (Eq,…
I am new to lambda calculus and struggling to prove the following.
SKK and II are beta equivalent.
where
S = lambda xyz.xz(yz)
K = lambda xy.x
I = lambda x.x
I tried to beta reduce SKK by opening it up, but got nowhere, it becomes to messy. Dont…
I'm having trouble converting the lambda for flip into the SKI combinators (I hope that makes sense). Here is my conversion:
/fxy.fyx
/f./x./y.fyx
/f./x.S (/y.fy) (/y.x)
/f./x.S f (/y.x)
/f./x.S f (K x)
/f.S (/x.S f) (/x.K x)
/f.S (/x.S f) K
/f.S (S…
I am attempting to implement a simply typed lambda calculus type checker. When running sanity tests I tried typing (S K K) and my type checker throws this error:
TypeMismatch {firstType = t -> t, secondType = t -> t -> t}
The offending term is…
I have the equation Y = FY (fixed point equation). How to get of it the equation for F through other combinator (in particular S- combinator with first fixed parameter)?
Context
I started teaching myself lambda calculus last night and I am trying to determine if what I understand so far is correct.
Understanding
SKK is equivalent to the Identity combinator, I.
Where L stands for lambda:
S = LxLyLz((xz)(yz))
K =…
Hello I am having trouble proving these combinators S K = K I
The steps with the brackets [] are just telling you the step i am doing. For example [λxy.x / x] in λyz.x z(y z) means I am about to substitute (λxy.x) for every x in the expression…
I was tasked with showing that
S(KK)I = K
Now since S takes three arguments, I was simply stuck at the beginning not knowing how to tackle this. I see two arguments, namely (KK) and I but there is no third one I can "spot". What happens in this…
