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A normal modal propositional logic $\Delta$ has the disjunction property if and only if

For any formulas $A_1,\dotsc,A_n$, if $\Box A_1 \vee \dotsb\vee \Box A_n \in \Delta$ then $A_k\in \Delta$ for some $k$ with $1\leq k\leq n$.

Let us say that $\Delta$ has the extended disjunction property if and only if

For any formulas $A_1,\dotsc,A_n$ and non-modal formula $A_0$, if $A_0 \vee \Box A_1 \vee \dotsb \vee \Box A_n \in \Delta$ then $A_k\in \Delta$ for some $k$ with $0\leq k\leq n$.

Does there exist a normal modal propositional logic with the disjunction property but not the extended disjunction property?

I came across the second property in some unpublished notes of Kit Fine's from the '70s, but I do not know of much discussion of it elsewhere.

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  • $\begingroup$ Perhaps I misunderstand your question, but doesn't it work to consider the modal propositional theory with four propositions, $P$, $\neg P$, $\top$, and $\perp$, with $\Box P = P$ and $\Box \neg P = \perp$? Looks to me like this satisfies the disjunction property, but not the extended disjunction property, since $\neg P \vee \Box P = \top$ and yet $P \neq \top$. Sorry if I've misunderstood your question and this isn't useful. $\endgroup$
    – user509184
    Aug 25, 2023 at 2:41
  • $\begingroup$ @user509184 That's not a normal modal logic. $\endgroup$ Aug 25, 2023 at 5:49
  • $\begingroup$ Sometimes one finds oddities of typesetting software by looking at things here. Looking at $ A_0 \vee \Box A_1 \vee ... \vee \Box A_n,$ coded as A_0 \vee \Box A_1 \vee ... \vee \Box A_n, one instantly sees that the last $\vee$ does not have the conventional amount of horizontal space to its right. I had a guess about the reason for that, and I'm not yet sure my guess was wrong, but when I changed the code to A_0 \vee \Box A_1 \vee \cdots \vee \Box A_n, that fixed the issue, so that what appeared was $A_0 \vee \Box A_1 \vee \cdots \vee \Box A_n.$ MathJax usually$\,\ldots\qquad$ $\endgroup$ Aug 25, 2023 at 19:16
  • $\begingroup$ MathJax usually treats ... as if it were \dots, unlike LaTeX which renders ... differently from \dots. I suspect there's some obvious reason for this difference in horizontal spacing after \vee, but I don't know what it is. $\qquad$ $\endgroup$ Aug 25, 2023 at 19:19
  • $\begingroup$ (In the comment from user509184 above, one sees $\Box \neg P = \perp,$ coded as \Box \neg P = \perp, whereas the code \Box \neg P = \bot yields instead $\Box \neg P = \bot,$ which is the correct amount of horizontal space after the "equals" sign. In a case like that, the reason is clear and it's what I expect. I'm guessing I'll figure the other one out. $\endgroup$ Aug 25, 2023 at 19:22

1 Answer 1

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The extended disjunction property is equivalent to plain disjunction property for all normal modal logics $\Delta$.

Assume that $\Delta$ contains $$A_0\lor\bigvee_{i=1}^n\Box A_i,$$ where $A_0$ is box-free.

Let $P$ denote the set of propositional variables occurring in the formula. For each $a\colon P\to\{0,1\}$, let $\sigma_a$ denote the substitution such that $\sigma_a(p)=p^{a(p)}$ for all $p\in P$, where $p^1=p$, $p^0=\neg p$.

Since $\Delta$ is closed under substitution and classical reasoning, $\Delta$ contains $$\bigwedge_a\Bigl(\sigma_a(A_0)\lor\bigvee_{i=1}^n\Box\sigma_a(A_i)\Bigr).\tag{$*$}\label{star}$$

If $A_0$ is a classical tautology, then $A_0\in\Delta$ and we are done. Otherwise, \eqref{star} implies $$\bigvee_a\bigvee_{i=1}^n\Box\sigma_a(A_i)\tag{$**$}\label{starstar}$$ by classical propositional reasoning: for any assignment $e$, there is $a$ such that $e(\sigma_a(A_0))=0$, thus the rest of the disjuction holds under $\sigma_a$.

Applying the disjuction property to \eqref{starstar}, $\sigma_a(A_i)\in\Delta$ for some $i=1,\dotsc,n$ and $a\colon P\to\{0,1\}$. Applying closure under substitution once more, $\Delta$ contains $\sigma_a(\sigma_a(A_i))$, which is equivalent to $A_i$ over K.

The argument above applies not just to normal modal logics, but also to quasinormal logics, and even more generally, to all logics $\Delta$ that are closed under modus ponens and box-free substitutions, and that contain the tautologies of E. Here, E is the smallest logic that includes classical logic, and is closed under modus ponens and under the rule $A\leftrightarrow B\mathrel/\Box(A\leftrightarrow B)$.

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  • $\begingroup$ I had a déjà vu feeling when writing down the argument. I think I used it before somewhere on this site. $\endgroup$ Aug 25, 2023 at 7:13
  • $\begingroup$ Oh, of course: here. $\endgroup$ Aug 25, 2023 at 7:19
  • $\begingroup$ Re, "déjà écrit"? 😄 $\endgroup$
    – LSpice
    Aug 25, 2023 at 23:07
  • $\begingroup$ Thanks! I asked that previous question, but I didn't notice the connection to your answer. My bad! $\endgroup$ Aug 26, 2023 at 22:06
  • $\begingroup$ There isn’t really much of a connection that I can see. It’s just that a similar trick works for both. $\endgroup$ Aug 27, 2023 at 6:47

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