# Re: [tlaplus] TLA Proof System - the syntax has impact on provability ?

Hello Harold,

you are certainly not doing any mistake, it just means that there was some little progress between the official release and the forthcoming version. We are planning for a new release in spring but have to fix a few bugs that we are aware of.

Best,
Stephan

> On 18 Mar 2019, at 19:26, haroldas.giedra@xxxxxxxxx wrote:
>
> I tried to prove:
>
> LEMMA lemma5 ==  C \subseteq [s : Nat]
>      /\
>      D = {[s|->x.s] : x \in C}
>      =>
>      C = {[s|->x.s] : x \in D}
> BY Z3
>
> TLA Proof System showed:
>
> status: z3: failed (timeout)
>
> Am I doing something wrong ?
>
> I can not change records to set Nat, because the original problem consists of records [s: SET1, st: SET2, d: SET3]. The example I gave is just simplified version of the problem.
>
> Maybe you know, when the new version of TLA Proof System will be released ?
>
> Thank you,
> Harold
>
> On Monday, 18 March 2019 14:31:46 UTC+2, Stephan Merz  wrote:
>> My version proves this "BY OBVIOUS", but it may be because I'm using a pre-release version. "BY Z3" may work for the official release of TLAPS, assuming that you have Z3 installed. The proof (and perhaps the spec) would become easier if C and D were just sets of natural numbers instead of one-field records.
>>
>> Stephan
>>
>>> On 18 Mar 2019, at 12:29, haroldas.giedra@xxxxxxxxx wrote:
>>>
>>>
>>> Thank you very much for your answer, Stephan.
>>>
>>> The example I gave came from the problem I want to solve. I want to prove such lemma:
>>>
>>> VARIABLE C, D
>>>
>>> LEMMA lemma5 ==
>>>     C \subseteq [s : Nat]
>>>     /\
>>>     D = {[s|->x.s] : x \in C}
>>>     =>
>>>     C = {[s|->x.s] : x \in D}
>>>
>>> Is it possible to prove such lemma in TLA Proof System ?
>>>
>>> Harold
>>>
>>> On Monday, 18 March 2019 10:07:54 UTC+2, Stephan Merz  wrote:
>>>> Hello,
>>>>
>>>> theorem proving is indeed driven by syntax and unfortunately, TLAPS's automatic backends may fail to prove lemmas that are true, even if their truth is obvious to a human. Recognizing the elements of the set expression [s:{0}] requires reasoning about the extension of the set, whereas the set { [s |-> 0] } is just a singleton whose sole element is explicitly given.
>>>>
>>>> I am happy to report that your lemma1 will (also) be proved by SMT in the upcoming release, as well as
>>>>
>>>> LEMMA lemma3 ==
>>>> ASSUME NEW C, NEW D,
>>>>        C = {[s|->0]},
>>>>        D = {[s|->x.s] : x \in C}
>>>> PROVE  C = D
>>>> OBVIOUS
>>>>
>>>> The proof of lemma2 currently seems to require a somewhat roundabout proof:
>>>>
>>>> LEMMA lemma2 ==
>>>> ASSUME NEW C, NEW D,
>>>>        C = [s : {0}],
>>>>        D = {[s|->x.s] : x \in C}
>>>> PROVE  C = {[s|->x.s] : x \in D}
>>>> <1>1. C = {[s |-> 0]}  OBVIOUS
>>>> <1>2. QED  BY ONLY <1>1, D = {[s|->x.s] : x \in C}
>>>>
>>>> where step <1>1 is proved by Isabelle and step <1>2 by either SMT or Isabelle (but both backends fail if the assumption about C is not hidden). Thanks for reporting this, we'll try to add a suitable rewrite rule for automating the handling of similar expressions.
>>>>
>>>> Stephan
>>>>
>>>>
>>>>> On 17 Mar 2019, at 20:50, haroldas.giedra@xxxxxxxxx wrote:
>>>>>
>>>>>
>>>>> Hello,
>>>>>
>>>>> I have example of two lemmas which have the same meaning but slightly differs in the syntax.
>>>>>
>>>>> LEMMA lemma1 == C = {[s|->0]}
>>>>>    /\
>>>>>    D = {[s|->x.s] : x \in C}
>>>>>    =>
>>>>>    C = {[s|->x.s] : x \in D}
>>>>> OBVIOUS
>>>>>
>>>>> LEMMA lemma2 == C = [s : {0}]
>>>>>    /\
>>>>>    D = {[s|->x.s] : x \in C}
>>>>>    =>
>>>>>    C = {[s|->x.s] : x \in D}
>>>>> OBVIOUS
>>>>>
>>>>> lemma1 has been proved by "TLA - Isabelle", but "TLA - SMT", "TLA - Zenon", "TLA - Isabelle" fail to prove lemma2.
>>>>>
>>>>> I don't understand why TLA proof system fails to prove lemma2, which is different from lemma1 only in the syntax of TLA records. Meaning:
>>>>>
>>>>> C = {[s|->0]}
>>>>> and
>>>>> C = [s : {0}]
>>>>>
>>>>> which are the same for me.
>>>>>
>>>>> Harold
>>>>>
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