# Re: [tlaplus] Reusing TypeOK theorem in TLAPS

Thanks!

Karolis

On Wed, Jan 22, 2020 at 2:43 PM Stephan Merz <stephan.merz@xxxxxxxxx> wrote:
>
> Hello,
>
> in short, in order to use a temporal-level theorem inside the proof of another one, your proof should be written as follows:
>
> THEOREM ThmA == Spec => []A
>
> THEOREM ThmB == Spec => []B
> <1>1. Init (* /\ A *) => B
> <1>2. B /\ [Next]_vars (* /\ A /\ A' *) => B'
> <1>. QED  BY <1>1, <1>2, ThmA, PTL DEF Spec
>
> The commented parts indicate where you may make use of the invariant proved in ThmA.
>
> As a rule of thumb, do not decompose temporal-level implications into ASSUME ... PROVE steps, but make sure that all hypotheses used for proving a fact that will be passed to the PTL proof method are constant level or [] formulas. The reason is that PTL will try to apply the necessitation rule F |- []F to promote any fact to a boxed formula, but it can do so only if there are no state- or action-level hypotheses in the context where that fact was established. For example, the steps <1>1 and <1>2 of your proof of SpecIInvB are established in the context of the assumption Spec, which is not a [] formula (due to the initial condition), and that's indicated by the "non-[]" warnings that you see in the proof obligation. As a result, the preservation of the invariant (step <1>2) cannot be "boxed" as it would need to be in order to apply temporal induction.
>
> Regards,
> Stephan
>
>
> > On 21 Jan 2020, at 15:45, Karolis Petrauskas <k.petrauskas@xxxxxxxxx> wrote:
> >
> > Hello,
> >
> > I'm trying to reuse the "Spec => []TypeOK" theorem in other theorems
> > regarding inductive invariants, and have stuck.
> > Bellow is a small example to reproduce my problem.
> >
> > ------------------------------ MODULE TestPTL ------------------------------
> > EXTENDS Naturals, TLAPS
> > VARIABLES x
> > Init == x = 0
> > Next == x' = x + 1
> > Spec == Init /\ [][Next]_x
> > --------------------------------
> > TypeOK == x \in Nat
> > THEOREM SpecTypeOK == Spec => []TypeOK
> >  <1>1. Init => TypeOK BY DEF Init, TypeOK
> >  <1>2. TypeOK /\ [Next]_x => TypeOK' BY DEF TypeOK, Next
> >  <1>q. QED BY <1>1, <1>2, PTL DEF Spec
> > --------------------------------
> > IInv == x >= 0
> > THEOREM SpecIInvA == TypeOK /\ Spec => []IInv
> >  <1> SUFFICES ASSUME TypeOK PROVE Spec => []IInv OBVIOUS
> >  <1>1. Init => IInv BY DEF Init, IInv
> >  <1>2. IInv /\ [Next]_x => IInv' BY DEF IInv, Next, TypeOK
> >  <1>q. QED BY <1>1, <1>2, PTL DEF Spec
> >
> > THEOREM SpecIInvB == Spec => []IInv
> >  <1>a. SUFFICES ASSUME Spec PROVE Spec => []IInv OBVIOUS
> >  <1>b. TypeOK BY <1>a, SpecTypeOK, PTL
> >  <1>1. Init => IInv BY DEF Init, IInv
> >  <1>2. IInv /\ [Next]_x => IInv' BY <1>b DEF IInv, Next, TypeOK
> >  <1>q. QED BY <1>1, <1>2, PTL DEF Spec
> > =============================================================================
> >
> > Theorems SpecTypeOK and SpecIInvA are proved successfully, and that's OK.
> > In SpecIInvA I want to avoid having TypeOK as an assumption, therefore the
> > theorem SpecIInvB is formulated identically, just omitting the TypeOK
> > before the implication.
> > TypeOK is needed in <1>2, therefore I proved it in <1>b using SpecTypeOK.
> > The problem here is that in this case the last step (<1>q) is not
> > proved anymore, and I don't understand why.
> > It looks the same, as in SpecIInvA.
> > The state at this step is:
> >
> > ASSUME NEW VARIABLE x,
> >       Init => IInv (* non-[] *),
> >       IInv /\ [Next]_x => IInv' (* non-[] *)
> > PROVE  Init /\ [][Next]_x => []IInv
> >
> > Karolis Petrauskas
> >
> > --
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