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Lambda Calculus - Reductions #1
Lambda Calculus - Reductions #1
+ 7 comments If you are new to lambda calculus following video & links will help
+ 0 comments I started coding a lexical parser for this. Then read a comment that we just have to write plain text answer for the given expression. Facepalm.
+ 1 comment (Function)argument e.g (λx.(+1)x)3 = (λx.+1 x)3 = (+ 1 3) = 4
((λx.(x y))(λz.z)) = (λx.(x y))(λz.z) = ((λz.z)y) = (λz.z)y = y
+ 1 comment Hi, I still don't understand why application can be written in the statement of the function?
E.g.
λx.(x y)
Shouldn't it be written as
(λx.x)y
?
+ 0 comments The problem with how to rename a variable is difficult. Love spells UK
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