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Is there an intuitive characterization of the correspondence for the modal logical formula $\square (\alpha \rightarrow \square \alpha) \rightarrow (\square \alpha \vee \square \lnot \alpha)$?

In the presence of $D$ the schema is slightly weaker than the schema $\square (\square \alpha \rightarrow \alpha) \rightarrow (\square \alpha \vee \square \lnot \alpha)$ considered by Raymond Turner in his book Truth and Modality for Knowledge Representation (1991) and named Turner's schema by Andrea Cantini in the latter's book Logical Frameworks for Truth and Abstraction (1996).

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    $\begingroup$ If I'm reading it right, that formula says "every world which sees some world, sees a world which sees a world other than itself." Is this right? $\endgroup$ Mar 31, 2014 at 3:11
  • $\begingroup$ I don't think so. What you state seems like the first order condition $\forall x (\exists y xRy \rightarrow \exists z (xRz \wedge \exists w (zRw \wedge z \neq w)))$. Is that right? $\endgroup$ Mar 31, 2014 at 3:36
  • $\begingroup$ Oops! I am very sorry! There is a disjunction in the consequent and not a conjunction! I'll edit. $\endgroup$ Mar 31, 2014 at 3:54
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    $\begingroup$ Now it seems to be "Any world which sees only worlds not seeing any worlds other than themselves, sees at most one world." (Or, "No world sees two distinct worlds, each of which sees no other worlds than itself.") Is this right, now? (Also, what exactly do you mean by "characterization of the correspondence?" $\endgroup$ Mar 31, 2014 at 4:11
  • $\begingroup$ Is it not the same as: If a world sees two worlds it sees a world which sees another world? $\endgroup$ Mar 31, 2014 at 4:35

2 Answers 2

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The scheme you write $(*)$ can be interpreted as saying, "No world sees two distinct worlds, each of which sees no world other than itself."

Turner's scheme $(T)$, by contrast, can be interpreted as "Any world which sees only self-seeing worlds, sees at most one world."

In the presence of no additional axioms, these two schemes are incomparable: the Kripke model $\mathcal{K}_0=(\{A, B, C\}; \{(A, B), (A, C)\})$ satisfies $(T)$ but not $(*)$, and the Kripke model $\mathcal{K}_1=(\{A, B, C\}, \{(A, A), (A, B), (A, C), (B, A), (B, B), (B, C), (C, A), (C, B), (C, C)\})$ satisfies $(*)$ but not $(T)$. [Here, a Kripke model is presented as $(\{worlds\}, \{"sees"\})$.]

Note that both schemes are valid in any Kripke frame of size $<3$, so the examples given above are minimal.

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  • $\begingroup$ I think you mean "sees only" rather than "only sees" in your second paragraph. $\endgroup$ Mar 31, 2014 at 11:23
  • $\begingroup$ @JoelDavidHamkins, do those two phrases have different meanings? (Regardless, I think your phrasing is clearer, so I've changed it.) $\endgroup$ Mar 31, 2014 at 17:29
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    $\begingroup$ "I only see $X$" means I see $X$, but don't do anything else to $X$. "I see only $X$" means I see $X$, and I don't see anything else. "Only I see $X$" means I see $X$ and nobody else does. But I apologize for picking on this; it is of course completely minor. (Meanwhile, once you are attuned to this kind of "only" abuse, you will hear it everywhere!) It is like when they announce on the subway, "walk up at next station; all doors will not open." But of course, they mean "not all doors will open". Similar mistakes with "only" abound. $\endgroup$ Mar 31, 2014 at 17:40
  • $\begingroup$ Yesterday, I was on a number six train, which is local, and they announced "this train will make express stops above Grand Central", but of course they meant "this train will make only express stops above Grand Central", since the local train always makes the express stops, in addition to the local-only stops. $\endgroup$ Mar 31, 2014 at 18:09
  • $\begingroup$ Maybe to see this more straightforwardly it is easier to switch to diamonds: the equivalent formula is $$ (\Diamond\alpha\land\Diamond\neg\alpha)\to\Diamond(\alpha\land\Diamond\neg\alpha) $$ $\endgroup$ Mar 31, 2014 at 18:38
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Let $\mathrm{Z}$ be the formula in question, which can be rewritten as: $(\Diamond\alpha\land\Box(\alpha\rightarrow\Box\alpha))\rightarrow\Box\alpha$.

If I understand correctly "No world sees two distinct worlds, each of which sees no world other than itself" from the previous answer, then this is a necessary, but not sufficient, condition for a frame to validate $\mathrm{Z}$. For example, the frame $F=(\{A,B,C,D\},\{(A,B),(B,C),(A,D)\})$ satisfies the condition, yet $\mathrm{Z}$ is invalid on $F$ for the valuation $V(\alpha)=\{A,B,C\}$.

A necessary and sufficient condition for a frame to validate $\mathrm{Z}$ is that the following holds for every world $w$:

$C(\mathrm{Z})$: If $w$ sees $w'$ then every $w''\not=w'$ seen by $w$ can see $w'$ in a finite number of steps through worlds seen by $w$.

Proof: Let $S(w)$ be the set of worlds seen by $w$ and $w_0\in S(w)$. For any $w_i\in S(w)$ for which $w_i\not=w_0$, let $L_{w_0}(w_i)>0$ be the length of the shortest path from $w_i$ to $w_0$ through worlds from $S(w)$; we write $L_{w_0}(w_i)=\infty$ if no such path exists.

Sufficiency: Assume $C(\mathrm{Z})$ holds. We attempt to invalidate $\mathrm{Z}$ at $w$ and show that this is impossible. If $S(w)=\varnothing$ then $\mathrm{Z}$ is obviously valid at $w$. If $w$ sees some world, then to invalidate $\mathrm{Z}$ at $w$ we need $V(\Diamond\alpha,w)=1$, $V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ and $V(\Box \alpha,w)=0$, so we need $V(\alpha,w_0)=0$ for some $w_0$ in $S(w)$. For the remaining $w_i\in S(w)$ (if any), $C(\mathrm{Z})$ implies $0<L_{w_0}(w_i)<\infty$. Then:

  • For any $w_1\in S(w)$ with $L_{w_0}(w_1)=1$, if $V(\alpha,w_1)=1$ then $V(\alpha\rightarrow\Box\alpha,w_1)=0$ (because $w_1$ sees $w_0$ where $V(\alpha,w_0)=0$), hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$. So we need to take $V(\alpha,w_1)= 0$ at all such $w_1$.

  • For any $w_2\in S(w)$ with $L_{w_0}(w_2)=2$, if $V(\alpha,w_2)=1$ then $V(\alpha\rightarrow\Box\alpha,w_2)=0$ (because $w_2$ sees some $w_1\in S(w)$ with $L_{w_0}(w_1)=1$ where we took $V(\alpha,w_1)=0)$, hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$. So we need to take $V(\alpha,w_2)=0$ at all such $w_2$.

We use induction to conclude that to invalidate $\mathrm{Z}$ at $w$ we need to take $V(\alpha,w_i)=0$ at all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$, otherwise $V(\Box(\alpha\rightarrow\Box\alpha),w)=0$ and $\mathrm{Z}$ is valid at $w$. But since there are no other worlds in $S(w)$ we have $V(\Diamond\alpha,w)=0$, therefore $\mathrm{Z}$ is still valid at $w$.

Necessity: Assume $w$ sees some $w_0$ and there are also worlds $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$. Then we take:

  • $V(\alpha,w_i)=0$ for all $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$,

  • $V(\alpha,w_x)=1$ for all $w_x\in S(w)$ with $L_{w_0}(w_x)=\infty$, as well as $V(\alpha,w_y)=1$ for any $w_y$ seen by such $w_x$. This is possible because no such $w_x$ sees a $w_i\in S(w)$ with $L_{w_0}(w_i)<\infty$ (otherwise it would itself have $L_{w_0}(w_x)<\infty$).

For this valuation $V(\alpha\rightarrow\Box\alpha,w_x)=1$, hence $V(\Box(\alpha\rightarrow\Box\alpha),w)=1$ continues to hold, as well as $V(\Box\alpha,w)=0$. But this time $V(\Diamond\alpha,w)=1$ and $\mathrm{Z}$ is invalid at $w$.

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