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$(\Box p \wedge \Box q) \rightarrow \Box(p\wedge q)$ is valid in K. But $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$ is not.

I'm not sure what a frame that validates $(\Diamond p \wedge \Diamond q) \rightarrow \Diamond(p\wedge q)$ would look like. I wonder if $\Diamond$ reduces to $\Box$ in such a system? I'd appreciate any help on this.

eyet
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1 Answers1

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Intuitively, $$\mbox{"$(\Diamond p\wedge\Diamond q)\rightarrow\Diamond(p\wedge q)$ is valid"}$$ says that I can never partition the worlds I see into two different (nonempty) pieces: if I could, consider making $p$ true and $q$ false on one half of the partition and $p$ false and $q$ true on the other half of the partition.

So suppose $\mathfrak{F}$ is a frame validating $(\Diamond p\wedge\Diamond q)\rightarrow\Diamond(p\wedge q)$; how many worlds can each world in $\mathfrak{F}$ possibly see?

No world in $\mathfrak{F}$ can see more than one world.

And in fact it's easy to show that this is an exact characterization.

Noah Schweber
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  • Thanks. Is the idea that if a world were to see more than one world, then it would have been possible to partition those worlds into two different nonempty pieces? Also, does $\Diamond p$ entail $\Box p$ in that case? It seems to me that yes, because if $p$ is true at a world I see then it's true at all the worlds I see. – eyet May 14 '20 at 00:55
  • @eyet Yes to both. Note however that these frames do not validate $\Box p\rightarrow\Diamond p$, so $\Diamond$ is strictly stronger than $\Box$ in this class of frames! – Noah Schweber May 14 '20 at 01:04
  • Thanks again. As for not validating $\Box p \rightarrow \Diamond p$, is that because such frames allow for worlds that see nothing, in which case if $p$ is true at w that sees nothing, then $\Box p$ would be (trivially) true at w but $\Diamond p$ would not? – eyet May 14 '20 at 01:53
  • @eyet That's exactly right. – Noah Schweber May 14 '20 at 02:59
  • Thank you very much for your help! – eyet May 14 '20 at 04:16