Robert,
I think the two possible behaviors are:
A) subst( lambda([x],g(x)), 'f , f(q) ) => lambda([x],g(x))(q)
B) subst( lambda([x],g(x)), 'f , f(q) ) => g(q)
(B) is the current behavior. It seems wrong in principle, because the
transformation lambda([x],g(x))(q) => g(q) is performed by evaluation, not
by simplification; that is, '(lambda([x],g(x))(q)) does *not* simplify to
g(q).
Another example of how (B) is inconsistent with other cases of subst:
subst( print , 'f , f(23) ) => print(23) <<< does not print because
it does not eval
subst( lambda([x],print(x)), 'f, f(23) ) => 23 <<< prints and returns
23 because it *does* eval
However, (B) is a handy idiom, which Barton in particular is very fond
of....
-s
On Wed, Jun 12, 2013 at 3:14 PM, Robert Dodier <robert.dodier at gmail.com>wrote:
> On Wed, Jun 12, 2013 at 12:07 PM, Barton Willis <willisb at unk.edu> wrote:
>
> > If the substitute a lambda form for a function idiom stops working
> > as I want it to, that's OK. I'll just write my own substitute :)
>
> How do you want it to work?
>
> For what it's worth, there isn't any proposal to disallow such a
> substitution -- only that simplification of the result
> should not evaluate the lambda expression. So presumably
> evaluating the result should give the same result as one gets
> from subst(lambda ...) now.
>
> best
>
> Robert Dodier
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