Let me try another attempt for explaining the issue. To make things simpler, let's leave aside the issue of refinement mapping, that is, consider just the identity refinement mapping. We want to prove

(0) InitL /\ [][NextL]_varsL => InitH /\ [][NextH]_varsH

Obviously, this can be reduced to proving

(1) InitL => InitH

(2) NextL \/ varsL' = varsL => NextH \/ varsH' = varsH

However, the obligation (2) is very likely going to be unprovable because really, we only have to prove that implication for all reachable states (i.e., reachable in runs of the low-level spec) rather than for completely arbitrary states, and that's where the invariant comes in.

Let's look at a stupid example: our high-level specification has one variable initialized to zero and that keeps growing while remaining an even number:

Even == {n \in Int : n % 2 = 0}

InitH == x = 0

NextH == x' > x /\ x' \in Even

Our low-level specification has two variables x and y that evolve as follows:

InitL == x = 0 /\ y = 0

NextL == x' = x+y /\ y' = y+2

Condition (2) now requires us to show

(x' = x+y /\ y'=y+2) \/ (x'=x /\ y'=y)

=> (x' > x /\ x' \in Even) \/ x' = x

but this implication is not true and therefore cannot be proved: we have no information about the "types" of x and y, so they could be strings for example. Even if they are integers, we could have x=42 and y=7, x' = 49 and y'=9, and the right-hand side will be false. And indeed, such a state cannot be reached because the low-level spec ensures that x and y are always even. Therefore we define

Inv == x \in Even /\ y \in Even

and relax our proof obligations to be

(1') InitL => InitH /\ Inv

(2') Inv /\ (NextL \/ varsL' = varsL) => Inv' /\ (NextH \/ varsH' = varsH)

which still implies our high-level goal (0). I leave proving (1') and (2') for our example as an exercise to you.

Indeed, Inv can be an arbitrary state predicate and has to be "invented" by the system designer / verifier. But it must be an invariant of the low-level specification, and FALSE is unlikely to be one.

You may also want to read section 6.8 of the Hyperbook.

Hope this helps,

Stephan

I have to admit I'm a bit confused by the explanation in the text too and don't quite see what Inv is representing. For example, can I set Inv to anything that allows me to prove the formula, what if I set it to False?

Thanks

On Sunday, March 10, 2019 at 12:52:43 PM UTC-7, Leslie Lamport wrote:

(1) If we remove the Inv from the formula Inv /\ Next => ..., it would assert

that a step starting in any state that satisfies Next satisfies "..." --

for example a state in which memQ is a sequence of imaginary numbers.

I have no idea if that assertion is true for such a starting state.

However, it suffices to prove the assertion for steps starting in a

reachable state. Conjoining the invariant Inv allows you to prove

the assertion only for reachable states. You have to choose Inv so

it asserts what is true about reachable states that makes the

implication true. To do this, you have to understand why the theorem

you're trying to prove is true.

(2) That mapping isn't derived; you have to invent it. The sentence

beginning "Intuitively" that starts on line 9 of page 63 tells you

what condition that substitution must satisfy. To be able to choose

the necessary mapping, you need to understand why the theorem you're

trying to prove is true .

Leslie

On Wednesday, March 6, 2019 at 10:53:35 PM UTC-8, Oliver Yang wrote:

Hi All,

In Section 5.8 of book "Specifying Systems", the "Proving Impl" is introduced. I have a rough understanding of refinement mapping, which

essentially maps states of Spec A to the states of Spec B. However, I have a hard time understanding "step simulation".

1) What's the purpose of introducing the invariant Inv in Formula 5.3? What are we trying to achieve here?

2) How do we derive the mapping: omem = vmem, octl = ..., obuf = buf? It looks like we jumped to the conclusion without showing any proof?

Thanks,

Oliver

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