haskell - How to reduce lambda calculus -


i try understand lambda calculus , read wonderful article.

on page 8, there expression: 

(λy.(x(λx.xy)))

if going substitute left outer most(λy) t,

(λy.(x(λx.xy))) t

then result be?

x(λx.xy)

i took liberty of rewriting haskell syntax

(\y -> x (\x-> x y)) (\y -> x (\x-> x y)) t  x (\x -> x t)

since y bound input \y y substituted t.

edit: noted in comment below if t contain free x important rename bound x in scope "fresh" name. name has no meaning. if say

let t = \t -> x t

then proper substitution like

x (\z -> z (\t -> x t))

where stated pigworker z chosen freshness fresh identifier replace our bound x prevent hiding free x in t.


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