Thanks for the reply, Ed.

> I'd assume that `x` didn't occur in either `u` or `v`

This is exactly the issue I'm wondering about. Since rewrite rules admit lambdas and only first-order matching, I'm wondering whether they're interpreted as you did (and I'd tend to), namely that `x` doesn't occur freely in `u`or `v`, in which case lambdas don't seem useful in rules (and yet were implemented for some reason) or they're interpreted as allowing `x` in `u` and `v`, and substitution captures. I'm still puzzled.

With a wee bit of higher-order matching, one might make `u` and `v` functions and instead write:

> foo (\ x -> fmap (u x) (v x)) = bar u v

In that case I'd expect `u` and `v` to be synthesized rather than literally matched. For instance, `foo (\ (a,b) -> fmap (+ a) [b,b,b])` would match with `u = \ (a,b) -> (+ a)` and `v = \ (a,b) -> [b,b,b]`.

-- Conal

On Sat, Dec 2, 2017 at 12:20 PM, Edward Kmett <ekmett@gmail.com> wrote:

I don't knw of a formal writeup anywhere.

But does that actually mean what you are trying to write?

With the effective placement of "forall" quantifiers on the outside for u and v I'd assume that x didn't occur in either u or v. Effectively you have some scope like forall u v. exists x. ...

Under that view, the warnings are accurate, and the rewrite is pretty purely syntactic.

I don't see how we could write using our current vocabulary that which you want.

On Sun, Dec 3, 2017 at 4:50 AM, Conal Elliott <conal@conal.net> wrote:
Is there a written explanation and/or examples of rewrite rules involving a LHS lambda? Since rule matching is first-order, I'm wondering how terms with lambda are matched on the LHS and substituted into on the RHS. For instance, I want to restructure a lambda term as follows:

> foo (\ x -> fmap u v) = bar (\ x -> u) (\ x -> v)

My intent is that the terms `u` and `v` may contain `x` and that whatever variable name is actually used in a term being rewritten is preserved so as to re-capture its occurrences on the RHS.

When I write this sort of rule, I get warnings about `x` being defined but not used.

Thanks, -- Conal

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