# Validity of $F \supset \Box F$

I started to study propositional modal logic and Kripke semantics. I learned that for any Kripke interpration $\mathcal{M}$, we have that, if $\mathcal{M} \models A$ then $\mathcal{M} \models \Box A$. From which also follows that for all frames $\mathcal{F}$, if $\mathcal{F} \models A$, then $\mathcal{F} \models \Box A$, which again makes sense.

But then I read that $F \supset \Box F$ is not valid in general. I didn't understand this part. I mean, if the validy of $F$ implies the validity of $\Box F$, then the formula $F \supset \Box F$ should be valid.

If someone can give me a concrete counter-example to $F \supset \Box F$, I would be glad. I also want to hear whether there is any frame in which $F \supset \Box F$ is valid.

• $\varphi \to \square \varphi$ is valid with respect to the class of models having a single world. – Mauro ALLEGRANZA Nov 20 '15 at 19:17

The property :

if $\vDash \varphi$, then $\vDash \square \varphi$

is the ground for the Rule of Necessitation.

It says that $\square \varphi$ is a theorem of a normal modal logic whenever $\varphi$ is a theorem of the logic.

This does not contradict the fact that $\varphi \to \square \varphi$ is not valid.

A "tricky" but simple counterexample can be manufactured considering a propositional letter $p$.

Clearly, $\nvDash p$ and thus (vacuously) : if $\vDash p$, then $\vDash \square p$ holds.

But :

$\nvDash p \to \square p$.

• Thanks for the explanation. But I have a questions. Regarding the counter-example. What exactly is the letter $p$ there? And, why you say that $\nvDash p$? And if I understood correctly, you say that if $p$ holds then $\Box p$ holds too, but $\nvDash p \to \square p$ will never hold. Then, again I need to learn more about what is $p$. On the other hand, in your comment to my question, you have written that, $\varphi \to \square \varphi$ is valid with respect to the class of models having a single world. What exactly does this single world models mean? – modpro Nov 21 '15 at 19:49
• @modpro - $p$ is only a Propositional letter; thus, it can be TRUE or FALSE, and thus $\square p$ is FALSE. The formula $p \to \square p$ is not valid, becuase if $p$ is any "contingent" truth, it is not necessary, and thus the conditional is FALSE. – Mauro ALLEGRANZA Nov 21 '15 at 20:09
• @modpro - a counterexample is "if the grass is green, then necessarily the grass is green" which is clearly not valid : we can imagine a "possible world" where the grass is blue. That is the reason why I said that the principle $\varphi \to \square \varphi$ is valid only in models with only one world. – Mauro ALLEGRANZA Nov 21 '15 at 20:12

For a countermodel consider any 2-element K-model, where the first element accesses the second and where p is true in the first, but not in the second.

$F \rightarrow \Box F$ is valid in every S4-model satisfying the hereditary condition: If $p$ is true in $w$, then $p$ is true in $v$ for all $v$ with $wRv$.

• Can you give a concrete counter-example, I'm having hard time to imagine it. On the other hand, what is $S4$ model and hereditary condition? – modpro Nov 21 '15 at 19:50
• Let M =(W, R, V) be a K-model such that W = {w, v}, wRv, V(p, w) = 1, V(p, v) = 0 (for some atom p). So, V(□p, w) = 0 and so, V(p → □p, w) = 0. S4-models are those models with an accessibility relation that is both reflexive and transitive. In the context of the usual model theory for intuitionistic logic (which can be translated into S4) the hereditary condition means that, if some evidence , e, supports some information, i, every piece of evidence at least as strong as e supports i as well. So, looked upon in this way, the condition says that growth of evidence preserves information. – sequitur Nov 22 '15 at 15:21