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# Lambda Calculus - Reductions #2

# Lambda Calculus - Reductions #2

peterhry + 0 comments I was enjoying learning Haskell until now.

rablanken13 + 2 comments ((λx.((λy.(x y))x))(λz.w)) ((λy.((λz.w) y))(λz.w)) ((λz.w) (λz.w)) (λz.w) w

At the end I am left with

`(λz.w)`

. While I know that it is true that this maps any z to w, why is it legal to write`w`

instead of`(λz.w)`

?`w`

is a reference wheras`(λz.w)`

is a function.- C
coral + 1 comment Kaan_Aydinay + 0 comments *[(λz.w)/z]w = w*The Right Sort of Tokens

mekis_peter + 0 comments Up to ((λz.w) (λz.w)), everything is fine. Now (λz.w) is the argument to be substituted for 0 occurrences of z in w; so the result of substitution will be w itself.

- AS
sethia_arun + 0 comments another learning use w or W has difference in test case

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