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Re: [tlaplus] Re: Conjoining Specifications lemma1.2



You are right, in page 7 , (259) in "Refinement mappings" it makes a distinction between a set and a property.

It also introduces the operator "Γ", which means closure under stuttering. I will use this in the proofs that need it.

I need to experiment to see if it works, most certainly though it will.

On Mon, Mar 18, 2019 at 3:21 PM Ron Pressler <ron.pressler@xxxxxxxxx> wrote:
It's not a matter of the definition of a safety property so much as the definition of a safety property plus additional constraints on the specification.

A safety property is a closed set (or class) of behaviors, but in "Refinement Mappings" only stuttering-invariant specifications are considered and "Conjoining Specifications" uses TLA, where all formulas are invariant under stuttering anyway. 

A state machine is a form of specification with a set of initial states and a next-state relation. While a state machine (without fairness conditions) is a safety property, it, too, can be either invariant under stuttering or not. Because in TLA all formulas are invariant under stuttering, in particular all state machines specified *in TLA* would be.

R


On Saturday, March 16, 2019 at 10:42:50 PM UTC, Apostolis Xekoukoulotakis wrote:
According to your paper "The existence of refinement mappings." , the safety property is a closed set , similarly to this paper.
The definition you mention now, coincides with the property generated by a state machine , which is a safety property.

I will first try to prove properties with the more general definition, and If I can't, go with the second definition.

Ok let me see if it works.


On Sat, Mar 16, 2019 at 11:46 PM Leslie Lamport <tlapl...@xxxxxxxxx> wrote:
What you missed is the fact that if a safety property is satified by a
behavior b, then it is satisfied by the behavior that has the same
first N states as b and then halts--where halting means taking nothing
but stuttering steps.  In your example, since PA and PC are safety
properties, they are both satisfied by the behavior (2, 2, 2, ...)  --
a behavior not satisfied by PB. Hence |= (PC /\ PA => PB) is false,
so the implication is true.  And I believe the lemma is true.

Leslie


On Friday, March 15, 2019 at 9:56:22 PM UTC-7, Apostolis Xekoukoulotakis wrote:
I am unable to prove this direction :

If PA , PB , PC are safety properties , then
if ⊨ ((PC /\ PA) => PB)  , then ⊨ (PC => (PA -▹ PB))

Consider closed sets
1 . PA : all sequences with prefix (2 , 4 , ....)
2 . PC : all sequences with prefix (2 , 5 , ...)
3 . PB : all sequences with prefix (3 , ....)

All sets are closed, aka they are safety properties, and the condition is true.
However , for sequence a = (2 , 5 , ...) , the result of the lemma is not true.
since a belongs at PC, and the finite prefix (2) belongs at PA, but the prefix (2) does not belong at PB, as it should.

Am I missing something? Maybe I made an error in the definitions .



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