Finding a condition on models that validate theorems

Question

I want to find a condition on standard models for which $\Diamond A \rightarrow \Box A$ is a theorem. I can see that the condition

$\forall w_1,w_2,w_3\in W$ if $w_1Rw_2$ and $w_1Rw_3$ then $w_2=w_3$

would be a sufficient condition but it seems to me it wouldn't be a necessary condition. Does anyone know a necessary and sufficient condition?

Answer 1

Mauro Allegranza's answer shows that your condition is not only sufficient but also necessary, but the actual proof is left as an exercise. Here's a solution to that exercise. I'll prove the contrapositive; I assume your condition is not satisfied, so some $w_1$ is $R$-related to two distinct elements $w_2$ and $w_3$, and I'll show that that $\Diamond A\to\Box A$ isn't valid in such a model. Interpret $A$ as being true at $w_2$ and false at $w_3$. (It doesn't matter how $A$ is interpreted anywhere else.) Then $\Diamond A$ is true at $w_1$ (because $A$ is true at $w_2$) but $\Box A$ is false at $w_1$ (because $A$ is false at $w_3$), so the implication $\Diamond A\to\Box A$ is false at $w_1$.

Answer 2

It seems to me that the condition you have stated is an iff condition.

See :

• Brian Chellas, Modal Logic : An Introduction (1980), page 85-on for the $G^{k,l,m,n}$ schema :

$$\Diamond^k \Box^l A \to \Box^m \Diamond^l A.$$

Thus, your formula is the case with $1,0,1,0$.

Then, see EXERCISE 3.38 [page 90] :

Prove that $1,0,1,0$-incestuality is the same as :

partial functionality : if $\alpha R \beta$ and $\alpha R \gamma$, then $\beta=\gamma$

where [page 88] we say that :

$R$ is $k, l, m, n$-incestual if and only if for every $\alpha, \beta, \gamma \in \mathcal M$,

if $\alpha R^k \beta$ and $\alpha R^m \gamma$, then for some $\delta \in \mathcal M$, $\beta R^l \delta$ and $\gamma R^ n \delta$.

Theorem3.8. The schema $G^{k,l,m,n}$ is valid in the class of $k, l, m, n$-incestual standard models.