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I'm new to lambda calculus and was wondering if transforming the lambda expression

$v\lambda v.v$

into

$v(\lambda v.)v$

produces the same expression. Could someone help out?

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2  
Your second version is not a valid expression in the $\lambda$-calculus. $\lambda$-abstraction is over a term: given a term $v$ its $\lambda$-abstraction is ($\alpha$-equivalent to) $\lambda x . v$. With parentheses, your initial expression $v \lambda v . v$ should be $v (\lambda v . v)$. –  Benedict Eastaugh Mar 11 '13 at 15:38

1 Answer 1

up vote 2 down vote accepted

I'm reposting Benedict Eastbaugh's comment CW:

Your second version is not a valid expression in the $λ$-calculus. $λ$-abstraction is over a term: given a term $v$ its $λ$-abstraction is ($α$-equivalent to) $λx.v$. With parentheses, your initial expression $vλv.v$ should be $v(λv.v)$.

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