# 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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