# [tlaplus] Re: Conjoining Specifications lemma1.2

• From: Leslie Lamport <tlaplus.ll@xxxxxxxxx>
• Date: Sat, 16 Mar 2019 14:46:41 -0700 (PDT)
• References: <CAFEV_-im2PbUEpqkipW=GprWeqtr6OAw7QovVMpbcjyz=M9EJw@mail.gmail.com>

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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