# Tag Info

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Try this.... It should help build up the intuition step by step http://pages.cs.wisc.edu/~horwitz/CS704-NOTES/2.LAMBDA-CALCULUS-PART2.html Alternatively since it's not really clear what you mean, maybe an example of what kind of abstraction you want for something simpler than sum?

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The thing to notice here is that your function sum is defined on lists, which are inductively defined. Theoretically, the inductive definition of a list defines for every type T a term match_list :: T -> (a -> [a] -> T) -> ([a] -> T) satisfying the property match_list s t [] = s match_list s t (x::xs) = (t x xs) in addition, defining a ...

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One way to not mess up with variable names in $\lambda$-calculus is to use De Brujin indexes. With this, $\alpha$-equivallence becomes a really obvious syntactic equality, for instance: \begin{align*} \lambda x.x &\rightarrow \lambda.0 \\ \lambda z.z &\rightarrow \lambda.0 \\ \lambda x .\lambda y. x &\rightarrow \lambda.\lambda.1 \\ \end{align*} ...

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If I inline (u => (f)(n => u(u)(n))) into (u => u(u)) I get: (u => f(n => u(u)(n))) (u => f(n => u(u)(n))) Which is exactly the Z-Combinator. From wikipedia: let Z = f => (u => f(v => u(u)(v))) (u => f(v => u(u)(v))) My derivation: let fix = f => (u => f(n => u(u)(n))) (u => f(n => u(u)(n))...

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The other answers are good, I just wanted to make it explicit that currying for dependent types is $$\textstyle \prod (x : A) . \prod (y : B(x)) . C(x, y) \ \cong \ \prod (p : \sum (x : A) . B(x)) . C(\pi_1(p), \pi_2(p))$$ which is more suggestive in Agda-style notation:  (x : A) \to (y : B(x)) \to C(x,y) \ \cong\ (p : (x : A) \times B(x)) \to C(\pi_1(...

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Here is one example: (λx.(λx.x(λy.x)x)x)x And by applying alpha equivalnce I can rewrite it to: (λw.(λz.w(λy.z)z)w)x Is my understanding correct ? There is a small error in your rewriting: at λz.w you have changed the meaning of the expression, because now the bound variable refers to the outer lambda abstraction, instead of the inner lambda abstraction ...

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