Prove if $\Box \Box p \to \Box p$ is valid on Frame $F=\langle W,R \rangle$ then $R$ is dense

Question

Prove if $\Box \Box p \to \Box p$ is valid on Frame $F=\langle W,R \rangle$ then $R$ is dense.

Where dense means $\forall w,v \in W: wRv \implies \exists u \in W : wRu\ \&\ uRv$.

I know we are supposed to come up with a valuation such that when $wRv$ and $M,w \models \Box \Box p \to \Box p$ we must have the denese-ness property but I'm not sure where to even begin since $\Box\Box p$ requires length 2 path.

Answer 1

We have to be careful about what things we're quantifying over at what time when dealing with modal logic. Sequitur answered a question of mine eight months ago that gives the standard names for the different parts of a modal model.

I interpret $\square\square p \to \square p$ holds on the frame to mean:

For all models $M$ with frame $\langle W, R \rangle$ and all worlds $w$, it holds for all propositions $p$ that $M, w \models \square \square p \to \square p$

In particular, a frame $\langle W, R \rangle$ does not include a valuation, which means we get to pick our valuations in very obnoxious ways to rule out features of the accessibility relation that we don't like.

First, let's note that $\square \square p \to \square p$ enforces the same conditions as $\lozenge p \to \lozenge\lozenge p$.

Proof:

• $\square\square p \to \square p \;\;\;\;$ is our hypothesis.
• $\square\square \lnot \lnot p \to \square \lnot \lnot p \;\;\;\;$ holds by double negation introduction.
• $\lnot \lozenge \lozenge \lnot p \to \lnot \lozenge \lnot p \;\;\;\;$ holds since $\square$ and $\lozenge$ are dual.
• $\lozenge \lnot p \to \lozenge \lozenge \lnot p \;\;\;\;$ holds by contrapositive elimination.

And note that $\lozenge p \to \lozenge \lozenge p$ enforces the same frame condition since $p$ is universally quantified, so we can add or remove $\lnot$ as long as we do it uniformly.

With that out of the way, let's pick two not necessarily distinct worlds, $u$ and $v$, such that $uRv$.

World $u$ is our starting point.

Pick a variable, let's say $a$ and declare $a$ to be true at world $v$ and false everywhere else.

$M, u \models \lozenge a$ is true, since we can get from $u$ to $v$, the only world where $a$ holds.

Therefore, $M, u \models \lozenge \lozenge a$ must also be true, i.e. there must exist an intermediate world $w$ such that $uRw$ and $wRv$. This is the only way to satisfy $\lozenge \lozenge a$, since $a$ is false everywhere except at world $v$.

Therefore $R$ is dense.