Thank you Stephan. I really need to think it thru, I just can't get it reason out yet in my head. I was wondering if the first formula is weaker than []<><<A>>_v, is it stronger than SF_v(A)?Hi Andrew, you are right....my derivation got short cut too much between = and =>. I am not sure if another derivation attempt of mine below makes sense or not.[](<>[]P => <>Q)
[](<>[]P => <>Q) => ([]<>[]P => []<>Q)
<>[]P => []<>Q
~<>[]P \/ []<>Q
[]~[]P \/ []<>Q
[](~[]P \/ <>Q) <= []~[]P \/ []<>Q
[]([]P => <>Q) <=> WF
= WFThanks and best regards,Chris (Zitro)--On Thursday, November 13, 2025 at 12:15:12 PM UTC-8 Stephan Merz wrote:I meant []~ENABLED. As soon as the action is enabled at some point, your condition requires <<A>>_v to eventually occur.StephanOn 13 Nov 2025, at 19:18, Chris Ortiz <zitro...@xxxxxxxxx> wrote:Hi Stephan,Thank you. Do you mean never enabled (which I think is []~ENABLED) , or it should be not always enabled (which is ~[]ENABLED)?Best regards,Chris (Zitro)--On Thursday, November 13, 2025 at 9:44:15 AM UTC-8 Chris Ortiz wrote:Thank you Andrew and Stephan. I was thinking the same way but I don't trust my intuition as I don't know how to formally prove it.For the second formula.[](<>[]ENABLED <<A>>_v => <><<A>>_v)[]<>[]ENABLED <<A>>_v => []<><<A>>_v<>[]ENABLED<<A>>_v => []<><<A>>_v~<>[]ENABLED<<A>>_v \/ []<><<A>>_v[]~[]ENABLED<<A>>_v \/ []<><<A>>_v[](~[]ENABLED<<A>>_v \/ <><<A>>_v)[]([]ENABLED<<A>>_v => <><<A>>_v) which is WF_v(A)So it means the second formula is just a tautology of WF_v(A), or equivalent.For the first formula[](ENABLED <<A>>_v => <><<A>>_v)[]ENABLED<<A>>_v => []<><<A>>_v~[]ENABLED<<A>>_v \/ []<><<A>>_v <--- This is where I mind got short-circuitedAs per Andrew, if this is unconditional fairness then this should be equivalent to []<><<A>>_v, for which I don't know how to derive.Thanks and best regards,Chris (Zitro)On Thursday, November 13, 2025 at 8:59:57 AM UTC-8 Andrew Helwer wrote:Thanks, Stephan! Another interesting question that occurs to me is whether the first formula is equivalent to unconditional fairness, since presumably a step must be enabled before it can be taken. Thus the ENABLED A precondition seems redundant.AndrewOn Wed, Nov 12, 2025 at 11:34 PM Stephan Merz <stepha...@xxxxxxxxx> wrote:As Andrew suspects, the second formula is equivalent to weak fairness:THEOREMASSUME NEW ACTION A, NEW STATE vPROVE WF_v(A) <=> [](<>[]ENABLED <<A>>_v => <><<A>>_v)BY PTLStephanOn 13 Nov 2025, at 01:51, Andrew Helwer <andrew...@xxxxxxxxx> wrote:[](ENABLED <<A>>_v => <><<A>>_v) is equivalent to []((ENABLED A) ~> <><<A>>_v) which means if an A step is enabled (like at all), eventually it will be taken; moreover, it will also be taken after every time an A step is enabled in the future. This is like an even stronger fairness assumption, but looking through the texts I have it does not seem to be a standard one. The usual next strongest fairness assumption in temporal logic is unconditional fairness, which is just []<><<A>>_v. There are a lot of fairness definitions out there though so I'm sure someone has covered & named it.For the second one, since P => <>P, it seems to be implied by weak fairness. So it is at most as strong as weak fairness. Is it any weaker than weak fairness? I can't think of how it could be, but then we should be able to rewrite it into weak fairness. And there I am somewhat at a loss.AndrewOn Wed, Nov 12, 2025 at 4:09 PM Chris Ortiz <zitro...@xxxxxxxxx> wrote:Hello,WF_v(A) == []([]ENABLED <<A>>_v => <><<A>>_v) andSF_v(A) == []([]<>ENABLED <<A>>_v => <><<A>>_v)Is there[]( ENABLED <<A>>_v => <><<A>>_v) and[](<>[]ENABLED <<A>>_v => <><<A>>_v ?And what could they mean?Thanks and best regards,Chris (Zitro)--
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