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- From: Adam Chlipala <adamc AT csail.mit.edu>
- To: coq-club AT inria.fr
- Subject: Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?
- Date: Fri, 28 Jun 2013 07:33:36 -0400
On 06/27/2013 11:58 PM, t x wrote:
I'm reading Chapter 2 of CPDT.
tinstrDenote feels a bit too magical to me, especially the part about:
1) statically know the # of elements on the stack
2) statically know the type of the elements on the stack
This seems like it solves all problems in formal verification and I'm trying to understand why tinstrDenote does NOT solve all problems.
Here's a concise answer which I hope helps: The two properties you mention are decidable. Most program properties are undecidable. It also helps that the two properties above are _easy_ to decide. :)
It seems important that all programs in chapter 2 always execute a _constant_ number of steps.
It sounds like this is your intuition behind why the properties above are decidable. I agree with the factual statement, and I agree that it suggests a pattern for deciding program behaviors.
- [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, t x, 06/28/2013
- Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, Adam Chlipala, 06/28/2013
- Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, t x, 06/28/2013
- Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, Adam Chlipala, 06/28/2013
- Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, t x, 06/28/2013
- Re: [Coq-Club] (soft question) tinstrDenote, dependence on O(1) steps?, Adam Chlipala, 06/28/2013
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