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*From*: Apostolis Xekoukoulotakis <apostolis.xekoukoulotakis@xxxxxxxxx>*Date*: Mon, 18 Mar 2019 16:41:15 +0200*References*: <CAFEV_-im2PbUEpqkipW=GprWeqtr6OAw7QovVMpbcjyz=M9EJw@mail.gmail.com> <29bbe489-7365-429a-b054-9fcf9c5b264d@googlegroups.com> <CAFEV_-jKLP9rB12EOkHDRH0xad-An+vxjfSmFy9HZcZDvubV9Q@mail.gmail.com> <20aba619-6db1-4090-9f4d-07b6c09caea8@googlegroups.com>

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 , thenif ⊨ ((PC /\ PA) => PB) , then ⊨ (PC => (PA -▹ PB))Consider closed sets1 . 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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**Follow-Ups**:**[tlaplus] Spurious coverage warning in TLC***From:*Jonathan Ostroff

**References**:**[tlaplus] Conjoining Specifications lemma1.2***From:*Apostolis Xekoukoulotakis

**[tlaplus] Re: Conjoining Specifications lemma1.2***From:*Leslie Lamport

**Re: [tlaplus] Re: Conjoining Specifications lemma1.2***From:*Apostolis Xekoukoulotakis

**Re: [tlaplus] Re: Conjoining Specifications lemma1.2***From:*Ron Pressler

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