Why S4 logic should have kripke models with reflexive and transitive relations?

Question

From here and here you can see that in S4 modal logic we should have Kripke models (S,R,V) which have a relationship R on their set of states S so that R is reflexive and transitive on S! I know that the set of axioms we obliged on our logic forces us to have our models this way but actually, I don't know how and why!

Answer 1

Welcome to MSE!

It helps to think about what the bonus axioms are saying.

• Axiom $T$ says $\square \varphi \to \varphi$
• Axiom $4$ says $\square \varphi \to \square \square \varphi$

What do these conditions say on frames?

---

Let's start with $T$. We know that for any valuation we like, we need to guarantee that $\square p \to p$ holds at every world. That is, no matter what the valuation, if $w$ only sees $p$-worlds, then $w$ must be a $p$-world too.

First, it should be clear that reflexive frames have this property. Since $w$ sees itself, if every world $w$ can see is a $p$-world, then $w$ itself must be a $p$ world.

But now say you look at a frame that isn't reflexive. Then fix a world $w$ that doesn't see itself. Now look at the valuation where $w$ thinks $\lnot p$ and every other world thinks $p$. Then since $w$ doesn't see itself, $w$ must think that $\square p$ is true! So $w$ thinks $\square p$ and $\lnot p$, and axiom $T$ fails!

So the frames validating $T$ are exactly the reflexive ones.

---

Let's move on to $4$. This says that if everywhere I can get in 1 step is a $p$-world, then everywhere I can get in $2$ steps is a $p$-world too.

Again, it should be clear that transitive frames have this property. After all, the definition of transitivity is that anywhere I can get in multiple steps is somewhere I can already get in 1 step.

As a (fun?) exercise, can you show that any nontransitive frame doesnt have this property? You'll want to argue like we did before with reflexivity. Fix a nontransitive frame. Then you can find worlds $x, y, z$ with $xRy$, $yRz$, but $x \not R z$. Can you cook up a valuation which makes axiom $4$ fail on this frame?

---

I hope this helps ^_^

Answer 2

HallaSurvivor's answer is absolutely right. Let me add a bit more, however, about the general context surrounding it.

The relevant connection between modal axiom systems and Kripke frame properties is very particular:

Definition:

• A Kripke frame $(W,R)$ validates a modal sentence $\sigma$ iff for every valuation $\nu$ the sentence $\sigma$ is true at every world according to $(W,R,\nu)$.

• A class $\mathfrak{F}$ of Kripke frames validates a modal sentence $\sigma$ iff every frame in $\mathfrak{F}$ validates $\sigma$.

The precise theorem connecting $S4$ and the class of transitive and reflexive frames, then, is the following multi-part result:

Theorem:

• A frame $(W,R)$ validates every theorem of $S4$ iff that frame is reflexive and transitive.
• If $\sigma$ is not a theorem of $S4$, then there is a reflexive and transitive frame not validating $\sigma$

It is in this sense that "$S4\approx$ reflexive + transitive."

In their answer above, HallaSurvivor proved the first bulletpoint (actually they proved more than that - they analyzed the role of reflexivity on its own and transitivity on its own). And this was the point relevant to your question.

The second bulletpoint is a bit more subtle. Basically, it says that $S4$ completely captures the nature of reflexive and transitive frames: there's nothing "missing" from $S4$ in this regard. So we really have a very tight connection between a modal theory ($S4$) and a frame property (reflexive-and-transitive)!

(In more technical language, the first bulletpoint says that $S4$ is sound with respect to the reflexive-and-transitive frame semantics (right-to-left direction) and moreover this is the maximal frame semantics with respect to which $S4$ is sound (left-to-right direction), while the second bulletpoint says that $S4$ is complete with respect to the reflexive-and-transitive frame semantics.)

---

It's separately worth noting that we can rephrase things in terms of operators:

• Frames-to-theories: Given a class of frames $\mathfrak{F}$, let $Th(\mathfrak{F})$ be the set of modal sentences validated by $\mathfrak{F}$.

• Theories-to-frames: Given a set of modal sentences $T$, let $Fr(T)$ be the set of frames validating every sentence in $T$.

We always have $Th(Fr(T))\supseteq T$ and $Fr(Th(\mathfrak{F}))\supseteq \mathfrak{F}$. However, in general these may be strict inclusions. The two-part theorem above can be reformulated as:

Let $\overline{S4}$ be the set of theorems of $S4$ and let $\mathfrak{X}$ be the class of reflexive and transitive Kripke frames. Then $$Th(Fr(\overline{S4}))=\overline{S4}\quad\mbox{and}\quad Fr(Th(\mathfrak{X}))=\mathfrak{X}.$$

Think of this as saying that the connection between (the theorems of) $S4$ and the class of reflexive and transitive Kripke frames is as good as we could hope - there's no "loss of information" in replacing one with the other, in a precise sense.

(The term "Galois connection" is relevant here, and it's a good exercise to try and figure out in what sense the pair of operators $(Th,\mathfrak{F})$ forms a Galois connection.)