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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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