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Re: [PVS-Help] Unrecognized equality
Here's another equality that should be straighforward, but PVS doesn't
seem to understand:
[-1] t!1(e!1) o fold(t!1, U!1) = fold(t!1, U!1) o t!1(e!1)
[-2] (fold(s!1, U!1) o fold(t!1, U!1)) o t!1(e!1) =
fold(s!1, U!1) o (fold(t!1, U!1) o t!1(e!1))
[-3] fold(s!1, U!1) o fold(t!1, U!1) =
fold(LAMBDA (k: D): s!1(k) o t!1(k), U!1)
|-------
[1] U!1(e!1)
{2} s!1(e!1) o fold(s!1, U!1) o (fold(t!1, U!1) o t!1(e!1)) =
s!1(e!1) o t!1(e!1) o (fold(s!1, U!1) o fold(t!1, U!1))
Rule? (replace -2 (2) :dir RL)
No change on: (REPLACE -2 (2) :DIR RL)
fold_distributive.2 :
[-1] t!1(e!1) o fold(t!1, U!1) = fold(t!1, U!1) o t!1(e!1)
[-2] (fold(s!1, U!1) o fold(t!1, U!1)) o t!1(e!1) =
fold(s!1, U!1) o (fold(t!1, U!1) o t!1(e!1))
[-3] fold(s!1, U!1) o fold(t!1, U!1) =
fold(LAMBDA (k: D): s!1(k) o t!1(k), U!1)
|-------
[1] U!1(e!1)
{2} s!1(e!1) o fold(s!1, U!1) o (fold(t!1, U!1) o t!1(e!1)) =
s!1(e!1) o t!1(e!1) o (fold(s!1, U!1) o fold(t!1, U!1))
What I want is for [2] to become:
{2} s!1(e!1) o (fold(s!1, U!1) o fold(t!1, U!1)) o t!1(e!1) =
s!1(e!1) o t!1(e!1) o (fold(s!1, U!1) o fold(t!1, U!1))
What could be going on? Unlike my last mysterious equality, doing a
check of 'show-expanded-sequent' doesn't yield any red flags...
jerome
On Tue, Jun 24, 2008 at 03:25:55PM -0700, Jerome wrote:
>Does anyone know why the following doesn't produce Q.E.D in PVS?
>
> {-1} strict_subset?(K!1, J!1) IMPLIES
> fold(s!1, J!1) = fold(s!1, K!1) o fold(s!1, difference(J!1, K!1))
> [-2] strict_subset?(K!1, J!1)
> |-------
> [1] fold(s!1, J!1) = fold(s!1, K!1) o fold(s!1, difference(J!1, K!1))
>
> Rerunning step: (ASSERT)
> Simplifying, rewriting, and recording with decision procedures,
> this simplifies to:
> composable :
>
> {-1} fold(s!1, J!1) = fold(s!1, K!1) o fold(s!1, difference(J!1, K!1))
> [-2] strict_subset?(K!1, J!1)
> |-------
> [1] fold(s!1, J!1) = fold(s!1, K!1) o fold(s!1, difference(J!1, K!1))
>
>After I make ASSERT call I would think the proof was finished... am I
>missing something? Thanks
>
>jerome
>