Prove that if $\Box p \rightarrow \Box\Box p$ is valid on a frame $F=\langle W,R\rangle$ then $R$ is transitive

Question

Prove that if $\Box p \rightarrow \Box\Box p$ is valid on a frame $F=\langle W,R\rangle$, then $R$ is transitive.

Suppose $F\vDash\Box p\rightarrow \Box\Box p$, where $F=\langle W,R\rangle$.

Let $u,v,w\in W$ be arbitrary worlds such that $Ruv$ and $Rvw$.

We need to show $Ruw$.

Define $V$ such that $z\in V(p)$ if and only if $Ruz$.

Let $M= \langle W,R,V\rangle$.

By definition of $V$, $M,z\Vdash p$ for all $z$ such that $Ruz$.

So, $M,u\Vdash\Box p$.

But since $F\vDash\Box p\rightarrow \Box\Box p$ and $M$ is based on $F$, $M,u\Vdash\Box p\rightarrow \Box\Box p$.

And since $M,u\Vdash\Box p$ and $M,u\Vdash\Box p\rightarrow \Box\Box p$ we can infer $M,u\Vdash\Box\Box p$.

But $M,u\Vdash\Box\Box p$ iff $M,v\Vdash\Box p$ iff $M,w\Vdash p$--all according to definitions for $M,w\Vdash$

But then, according to the definition of $V$, $Ruw$. And we're done.

This is basically the proof from the OLP's Boxes and Diamonds (I expanded a few steps). And here's my confusion:

We make a substantial assumption about $V$ in order for the proof to work. And while I get why such assumptions won't affect $M$ being based on $F$ in the strict sense. I don't get how we can infer that all models based on $F$ will similarly require that $R$ be transitive.

Thanks for any and all help.

Answer 1

What does "$\varphi$ is valid on $F = \langle W,R\rangle$" mean? It means that for all models $M$ based on $F$, $\varphi$ is valid on $M$. In particular, for any valuation $V$, $\varphi$ is valid on $M = \langle W,R,V\rangle$. So when we use the hypothesis that $\square p\rightarrow \square\square p$ is valid on the frame $F$, we're able to pick any valuation we like!

I recommend interrogating your confusions a bit more - often clarifying the question you're asking will help to resolve it. You wrote

And while I get why such assumptions won't affect $M$ being based on $F$ in the strict sense. I don't get how we can infer that all models based on $F$ will similarly require that $R$ be transitive.

I don't know what you mean by "in the strict sense" here, and I suspect you don't either. There is only one sense of $M$ being based on $F$, which is the definition. As to why "all models based on $F$ will similarly require that $R$ be transitive" - all models based on $F$ have exactly the same relation $R$ on exactly the same set of worlds $W$. This data is part of the frame itself. The only thing a particular model adds is a valuation $V$, which has no impact on whether the relation $R$ is transitive.

Also, a correction: You wrote

$M,u\Vdash\Box\Box p$ iff $M,v\Vdash\Box p$ iff $M,w\Vdash p$.

Each of these "iff"s should be "implies". For example, what's true is that $M,u\Vdash \square\square p$ iff $M,z\Vdash \square p$ for all $z$ such that $Ruz$. Now since we have $Ruv$, in particular $M,u \Vdash \square\square p$ implies $M,v\Vdash \square p$. But $M,v\Vdash \square p$ does not imply $M,u\Vdash \square\square p$ (unless we know that $v$ is the only world accessible from $u$).