Re: [isabelle] lemma about finitely-branching finite sequences



Hi John,

On 16/06/14 16:48, John Wickerson wrote:
Thanks very much for your reply Andreas.

I had actually tried the idea of using "well-foundedness of ↝ restricted to the states reachable from C", but couldn't make it work. I've just tried it again, encouraged by your suggestion, but I'm still struggling.

Specifically, I'm phrasing my lemma like this (↝* is an abbreviation I defined for the reflexive transitive closure of ↝):

lemma
   assumes "wf {(C2,C1). C ↝* C1 ∧ C1 ↝ C2}"
   shows "finite (traces C)"

I use wf_induct:

proof (rule wf_induct[OF assms])

That reduces the problem to the following (I've made the assumption slightly more readable):

   fix C1
   assume 1: "∀C2. C ↝* C1 ∧ C1 ↝ C2 ⟶ finite (traces C2)"
   show "finite (traces C1)"

My specific problem is that before I can exploit fact "1", I need to show that C1 is indeed reachable from C. I simply don't have that fact anywhere!
At the moment, your induction predicate is "%C. finite (traces C)". You have to generalise it to arbitrary states reachable from C, i.e.,

lemma
  assumes "wf {(C2,C1). C ↝* C1 ∧ C1 ↝ C2}"
  and "C ↝* C1"
  shows "finite (traces C1)"
using assms
proof(induction C1 rule: wf_induct_rule)

Do you have any further tips that might help me? Many many thanks!
Do not use any of the basic proof rules for wf to establish the well-foundedness assumption. The resulting proof oblications are usually too complicated. Rather show that the relation "{(C2,C1). C ↝* C1 ∧ C1 ↝ C2}" is a subset of "measure f" for some measure function f on states (like the longest execution starting in the state as I suggested in my previous mail). Then, you get well-foundedness from the rules wf_subset and wf_measure.

Andreas


On 16 Jun 2014, at 13:57, Andreas Lochbihler <andreas.lochbihler at inf.ethz.ch> wrote:

Hi John,

On 16/06/14 14:00, John Wickerson wrote:
I have a relation "↝" representing the reductions of a small-step operational semantics. By construction, every configuration can only reduce to finitely-many next configurations.

Given an initial configuration C, I would like to prove that if every execution starting from C gets stuck after a finite number of steps, then there are only finitely-many executions of C0.

I'm having quite a bit of difficulty getting my head round how to prove this. I can do it if I assume that the entire "↝" relation is well-founded, but that's too strong an assumption,
Your assumption is that all executions starting in C eventually get stuck. You can prove that this is equivalent to the well-foundedness of ↝ restricted to the states reachable from C. Then, you can use well-founded induction to show finiteness, probably similar to what you already have.

To show well-foundedness, you can take the length of the longest execution starting in a state as measure function.

Hope this helps,
Andreas

since I need it to be possible for *some* executions to diverge, just not those that start from C.

I have a sense that my lemma, if it is indeed true, will have been proven before, perhaps in the context of graph theory, or computability theory.

I'd really appreciate any hints the Isabelle community might have for how I might prove this, or where/whether it has already been proven.

Thanks!
John

ps. In case more precision is appropriate...

I'm defining executions like this:

definition executions :: "config ⇒ (nat ⇒ config option) set"
where
  "executions C ≡ {π.  π 0 = Some C ∧ (∀i.
  case π i of None ⇒ π (i+1) = None | Some C ⇒
  if reduce C=[] then π (i+1) = None
                 else π (i+1) ∈ Some ` set (reduce C))}"


and my lemma is:

lemma
  assumes "∀π ∈ executions C. ∃i > 0. finite_seq i π"
  shows "finite (executions C)"

and I define finite_seq like so:

fun finite_seq where
  "finite_seq 0 π = (∀i. π i = None)"
| "finite_seq (Suc i) π = (π 0 ≠ None ∧ finite_seq i (λi. π (Suc i)))"







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