`stable-under-simplificationp`

flag
Major Section: MISCELLANEOUS

A problem with the example in using-computed-hints-6 is that exactly one simplification occurs between each (effective) firing of the hint. Much more commonly we wish to fire a hint once a subgoal has become stable under simplification.

A classic example of this is when we are dealing with an interpreter for some state machine. We typically do not want the ``step'' function to open up on the symbolic representation of a state until that state has been maximally simplified. We will illustrate with a simple state machine.

Let us start by defining the step function, `stp`

, and the
corresponding `run`

function that applies it a given number of times.

(defun stp (s) (+ 1 s)) (defun run (s n) (if (zp n) s (run (stp s) (- n 1))))The step function here is trivial: a state is just a number and the step function increments it. In this example we will not be interested in the theorems we prove but in how we prove them. The formula we will focus on is

(thm (equal (run s 7) xxx))This is not a theorem, of course. But we want to test our advice on non-theorems because we do not want the advice to work only for proofs that succeed. (In the past, we gave advice about using computed hints and that advice caused the theorem prover to run forever when given formulas that it couldn't prove -- but most of the time the system is presented with formulas it cannot prove!)

Furthermore, without some kind of additional rules, the `(run s 7)`

expression in the conjecture above will not expand at all, because ACL2's
heuristics do not approve.

In fact, we do not want to take chances that `run`

will be
expanded -- we want to control its expansion completely.
Therefore, disable `run`

.

(in-theory (disable run))

Now, what do we want? (That is always a good question to ask!) We want
`(run s 7)`

to expand ``slowly.'' In particular, we want it to expand
once, to `(run (stp s) 6)`

. Then we want the `stp`

to be expanded and
fully simplified before the `run`

expression is expanded again. That is,
we want to force the expansion of `run`

whenever the goal is stable under
simplification. This is sometimes called ``staged simplification.''

We can achieve staged simplification for any given function symbol by defining the functions shown below and then using a simple computed hint:

(thm (equal (run s 7) xxx) :hints ((stage run)))

By inspecting how `stage`

is defined you can see how to extend it,
but we explain as we go. To experiment, you can just paste the
definitions (and defmacro) below into your ACL2 shell and then
try the `thm`

command.

First, define this pair of mutually recursive functions.
`Find-first-call`

finds the first call of the function symbol `fn`

in a given term.

(mutual-recursion (defun find-first-call (fn term) ; Find the first call of fn in term. (cond ((variablep term) nil) ((fquotep term) nil) ((equal (ffn-symb term) fn) term) (t (find-first-call-lst fn (fargs term))))) (defun find-first-call-lst (fn lst) ; Find the first call of fn in a list of terms. (cond ((endp lst) nil) (t (or (find-first-call fn (car lst)) (find-first-call-lst fn (cdr lst)))))))

We will arrange for the computed hint to generate an `:EXPAND`

hint for the first call of `fn`

, whenever the goal becomes
stable under simplification. If no call is found, the
hint will do nothing. To make sure the hint will not loop
indefinitely (for example, by forcing `fn`

to expand only to
have the rewriter ``fold'' it back up again), we will provide
the hint with a bound that stops it after some number of
iterations. Here is the basic function that creates the
`expand`

hint and replaces itself to count down.

(defun stage1 (fn max clause flg) ; If the clause is stable under simplification and there is a call of ; fn in it, expand it. But don't do it more than max times. (let ((temp (and flg (find-first-call-lst fn clause)))) (if temp (if (zp max) (cw "~%~%HINT PROBLEM: The maximum repetition count of ~ your STAGE hint been reached without eliminating ~ all of the calls of ~x0. You could supply a larger ~ count with the optional second argument to STAGE ~ (which defaults to 100). But think about what is ~ happening! Is each stage permanently eliminating a ~ call of ~x0?~%~%" fn) `(:computed-hint-replacement ((stage1 ',fn ,(- max 1) clause stable-under-simplificationp)) :expand (,temp))) nil)))

Suppose that when `stage1`

is called, `fn`

is the function we want to
expand, `max`

is the maximum number of iterations of this expansion,
`clause`

is the current goal clause, and `flg`

is the value of the
`stable-under-simplificationp`

flag. Then if `clause`

is stable and we
can find a call of `fn`

in it, we ask whether `max`

is exhausted. If
so, we print an ``error message'' to the comment window with `cw`

and
return `nil`

(the value of `cw`

). That `nil`

means the hint does
nothing. But if `max`

is not yet exhausted, we return a new hint. As
you can see above, the hint replaces itself with another `stage1`

hint
with the same `fn`

and a decremented `max`

to be applied to the new
`clause`

and the then-current value of `stable-under-simplificationp`

.
The hint also contains an `:expand`

directive for the call of `fn`

found.

Thus, if the computed hint was:

(stage1 'run 5 clause stable-under-simplificationp)

and `(run s 7)`

occurs in the clause, then it will
generate

(:computed-hint-replacement ((stage1 'run 4 clause stable-under-simplificationp)) :expand ((run s 7)))which will in turn replace the old

`stage1`

hint with
the new one and will apply `:expand ((run s 7))`

to
the current goal.We can make this more convenient by defining the macro:

(defmacro stage (fn &optional (max '100)) `(stage1 ',fn ,max clause stable-under-simplificationp))Note that the macro allows us to either provide the maximum bound or let it default to 100.

Henceforth, we can type

(thm (equal (run s 7) xxx) :hints ((stage run)))to stage the opening of

`run`

up to 100 times, or we can write
(thm (equal (run s 7) xxx) :hints ((stage run 5)))to stage it only 5 times. In the latter example, the system with print a ``error message'' after the fifth expansion.

Note that if we executed

(set-default-hints '((stage run)))then we could attack all theorems (involving

`run`

) with
staged simplification (up to bound 100), without typing an
explicit hint.
(thm (equal (run s 7) xxx))

Using techniques similar to those above we have implemented ``priority phased
simplification'' and provided it as a book. See community book
`books/misc/priorities.lisp`

. This is an idea suggested by Pete Manolios,
by which priorities may be assigned to rules and then the simplifier
simplifies each subgoal maximally under the rules of a given priority before
enabling the rules of the next priority level. The book above documents both
how we implement it with computed hints and how to use it.

Here is another example of using the `stable-under-simplificationp`

flag to
delay certain actions. It defines a default hint, see DEFAULT-HINTS,
which will enable non-linear-arithmetic on precisely those goals
which are stable-under-simplificationp. It also uses the `HISTORY`

and
`PSPV`

variables to determine when toggling non-linear-arithmetic is
appropriate. These variables are documented only in the source code. If
you start using these variables extensively, please contact the developers
of ACL2 or Robert Krug (`rkrug@cs.utexas.edu`

) and let us know how we can
help.

(defun nonlinearp-default-hint (stable-under-simplificationp hist pspv) (cond (stable-under-simplificationp (if (not (access rewrite-constant (access prove-spec-var pspv :rewrite-constant) :nonlinearp)) '(:computed-hint-replacement t :nonlinearp t) nil)) ((access rewrite-constant (access prove-spec-var pspv :rewrite-constant) :nonlinearp) (if (not (equal (caar hist) 'SETTLED-DOWN-CLAUSE)) '(:computed-hint-replacement t :nonlinearp nil) nil)) (t nil)))