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



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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