# Re: [tlaplus] Temporal logic

Thanks yes, your suggestions help somewhat.

This is an awful lot of work! I have most of the proof of the invariant stack = << >> \/ \A s \in stack : s.pc \in ValidPC except for a few key assumptions like the stack # <<>> /\ pc \in ProcPC you mentioned.

I am finding certain "basic" facts about sequences to be quite difficult. It took two hours and a page of proof to verify:
seq = Cons(a, b) =>
/\ Tail(seq) = b
Which of course distils down to one or two clauses with the selection of the right sequence theorems. But now to use it them need I need stack \in Seq(S), which means I now need to prove Head(stack') \in S which I suspect will end up forcing me to prove of all the other type invariants anyway..

It's like diving down the rabbit hole. I am gaining a new appreciation for the phrase burden of proof.

On Saturday, November 15, 2014 12:11:42 AM UTC-8, Stephan Merz wrote:
Hi Chris,

despite the claim in a comment in your module, your type invariant is certainly not inductive: you'll need to describe rather precisely the interplay between the stack and the program counter. I believe you want to write something along the following lines (but I haven't checked any of this):

StackEntry ==
[ procedure : {"MergeSort"},
pc : ValidPC,
A0 : [0 .. N-1 -> 0 .. N-1],  \* is this local variable really necessary?
mid : Nat,  \* 0 .. N-1 ? (similar for the following)
k : Nat,
i : Nat,
j : Nat,
lo : Nat,
hi : Nat ]

MainPC == {"MS", "Done"}
ProcPC == {"l_1", "l_msl", ..., "l_rtn"}

TypeOK ==
/\ ...
/\ stack \in Seq(StackEntry)
/\ \/ stack = <<>> /\ pc \in MainPC
\/ /\ stack # <<>> /\ pc \in ProcPC
/\ Last(stack).pc = "Done"   \* return address from main procedure call
/\ \A i \in 1 .. Len(stack)-1 : stack[i].pc \in {"l_msh", "l_sl1"}
\* return addresses from recursive calls

By the way, you may want to consider simplifying your model by removing some of the auxiliary computations, e.g. replace the while loop at l_sl2 by something like

B := [i \in 0 .. mid-lo |-> A[i+lo]]

and similar for l_sh2. It will be pretty obvious in a second step to refine this assignment into your while loop.

Hope this helps to get you on track,

Stephan

On 13 Nov 2014, at 20:00, chri...@xxxxxxxxx wrote:

OK thanks. That confirms I am looking in the right direction then.

I realised last night I could get partly simplify the proof of pc \in ValidPC if I prove the invariant stack = << >> /\ Head(stack).pc \in Valid PC, which should be straightforward. But then I still need to be able to prove stack # << >> in states that return, which is where I think I will need more complex temporal logic. Essentially I need to be able to prove that state X came from A, B or C and in A, B, C stack' # << >> therefore state X stack # << >>.

I have attached my work, it's in a bit of a mess but <4>2 is where I am working.

Regards,
Chris.

On Wednesday, November 12, 2014 11:51:28 PM UTC-8, Stephan Merz wrote:
Hi Chris,

reasoning about ENABLED is still not supported by TLAPS, unfortunately. For the moment you'll have to assert statements like step <1>2 in your proof of z1_liveness. Moreover, temporal reasoning beyond propositional temporal logic is not yet possible, and you will typically need more complex temporal reasoning, such as combining leads-to and well-founded orderings, for realistic liveness properties. Full support for temporal reasoning, and then for ENABLED, is on the top of our to-do list.

Concerning your first question, proving type correctness for your MergeSort algorithm should not be difficult in principle. Of course, you'll need type assertions for the complete state space – for a recursive function, this includes a suitable type correctness predicate for the call stack (which contains the return address from a recursive call). I'm afraid I cannot be more specific without seeing the spec.

Best,
Stephan

On 13 Nov 2014, at 02:49, chri...@xxxxxxxxx wrote:

Hi, two questions:

Firstly I implemented a model for MergeSort, based on the BubbleSort example. It is a recursive function so I am using PlusCal procedure and calling it recursively. It model checks OK with TLC.

I am having trouble proving my TypeOK properties I assume this is because following the data flow into the function call requires some proof using temporal properties. For example, in states which return, I cannot prove pc \in ValidPC. Is this expected to be a hard problem? I can't find any examples that use recursive function calls, and very few that involve this type of temporal proof.

Secondly, I am finding it harder than expected to prove temporal properties. I have traced it down and it seems that I sometimes need to assert ENABLED without proof. I notice a comment in one of the Paxos proofs that TLAPS cannot reason about ENABLED, is this currently true?

My test example follows, note z1_liveness <1>2:

VARIABLE z1

z1_init == z1 = 1
z1_next == z1' = z1 + 1
z1_spec == z1_init /\ [][z1_next]_z1 /\ WF_z1(z1_next)

z1_inv == z1 \in Int

THEOREM z1_safety == z1_spec => []z1_inv
<1>1. z1_init => z1_inv
BY SMT DEF z1_spec, z1_init, z1_inv
<1>2. z1_inv /\ [z1_next]_z1 => z1_inv'
BY SMT DEF z1_inv, z1_next
<1> QED
BY <1>1, <1>2, PTL DEF z1_spec

THEOREM z1_liveness == z1_spec => (<> <<z1_next>>_z1)
<1> SUFFICES ASSUME z1_spec PROVE (<> <<z1_next>>_z1)
BY PTL
<1>1. []z1_inv
BY z1_safety, PTL
<1>2. [](ENABLED <<z1_next>>_z1)
(**** BY SMT DEF z1_spec, z1_init, z1_next ****) PROOF OMITTED
<1>3. []([]ENABLED <<z1_next>>_z1 => (<> <<z1_next>>_z1))
BY PTL DEF z1_spec
<1> QED
BY <1>1, <1>2, <1>3, PTL DEF z1_spec, z1_inv, z1_next, z1_init

--
You received this message because you are subscribed to the Google Groups "tlaplus" group.
To unsubscribe from this group and stop receiving emails from it, send an email to tlaplu...@xxxxxxxxxxxxxxxx.
To post to this group, send email to tla...@xxxxxxxxxxxxxxxx.

--
Stephan Merz

--
You received this message because you are subscribed to the Google Groups "tlaplus" group.
To unsubscribe from this group and stop receiving emails from it, send an email to tlaplus+u...@googlegroups.com.
To post to this group, send email to tla...@xxxxxxxxxxxxxxxx.