op(box(x)) --> box, but is(op(box(x)) = box) --> false.
Subject: op(box(x)) --> box, but is(op(box(x)) = box) --> false.
From: Stavros Macrakis
Date: Sun, 6 Nov 2011 13:07:46 -0500
The most obvious (and I think simplest) way to do this is:
boxop: op(box(x))$
is(op(...)=boxop);
But I agree that this is a bug.
-s
On Sun, Nov 6, 2011 at 05:39, Barton Willis <willisb at unk.edu> wrote:
> My boxp function returns false for all CRE inputs; Richard's
> is(op(box(x))=?mbox) should work correctly for
> a CRE input. You could do
>
> :lisp(setf (get 'mbox 'op) '$box)
>
> Then is(op(box(x))='box) --> true and is(op(rat(box(x)))='box) --> true.
>
> --Barton
>
> -----maxima-bounces at math.utexas.edu wrote: -----
>
> try is(op(box(x))=?mbox)
>
> RJF
>
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