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Let 'L' and 'M' denote the necessity and possibility operators. In Modal Logic, the following theorems hold:

  1. L(p and q) <--> (Lp and Lq)
  2. (Lp or Lq) --> L(p or q)
  3. M(p or q) <--> (Mp or Mq)
  4. M(p and q) --> (Mp and Mq)

Do these theorems hold in infinitary modal logic, i.e.

  1. L(p1 and p2 and ...) <--> (Lp1 and Lp2 and ...)
  2. (Lp1 or Lp2 or ...) --> L(p1 or p2 or ...)
  3. M(p1 or p2 or ...) <--> (Mp1 or Mp2 or ...)
  4. M(p1 and p2 and ...) --> (Mp1 and Mp2 and ...)?
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  • researchgate.net/publication/… isn't behind a paywall(?) so it would be worth checking out. I'll skim it to see if they have anything, although it'll be a while before I have anything to post accordingly.
    – Kristian Berry
    Jun 20 at 20:34
  • Bard says: "The distribution property of modal operators is not always valid in finite modal logic, but it is always valid in infinitary modal logic. This is because in infinitary modal logic, we can always find a bisimulation between two models that satisfy the same infinitary modal formulas." IDK if that's true, though, Bard might be hallucinating again.
    – Kristian Berry
    Jun 20 at 21:26
  • @KristianBerry. Thank you for your comments. Sorry, who is Bard? Also, does Bard's quote imply that the conditional sign in (6) and (8) in my post can be replaced by a biconditional?
    – Beginner
    Jun 21 at 16:12
  • Bard is Google's chatbot. I've had the word "bisimulation" in my head for some years now re: a Hamkins response to a post of mine over on MathOF, but I am not entirely sure what it is even now. It seems like it is the kind of thing that would allow you to use a biconditional as such, though.
    – Kristian Berry
    Jun 21 at 16:28
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    @KristianBerry Just to let you know, I posted on MO mathoverflow.net/questions/449334/…
    – Beginner
    Jun 21 at 17:19

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