Is "If ⊨□φ then ⊨φ" true in modal logic?

Question

(Under propositional modal logic, system K)
There's seems to be an obvious counterexample: A model with one world w s.t. $\phi$ is not true in $\phi$ and $\lnot R(w,w)$.

But consider this argument:
Assuming $\vDash\square\phi$, take an arbitrary model $M$ and construct a new model $M^*$ s.t. for any $w$ in $M$, add a world x s.t. $R(z,x)$. Since $M^*,x \vDash \square\phi$, $M^*,w \vDash\phi$. And since $\langle M,w\rangle$ and $\langle M^*,w\rangle$ is bisimilar, $M,w \vDash\phi$. Because $M$ and $w$ was arbitrary, $\vDash\phi$.
I don't think this argument is right because we don't know if $\langle M,w\rangle$ and $\langle M^*,w\rangle$ is bisimilar, since we don't know if adding an $x$ would change how $M^*, w$ satisfies sentential variables.

Intuitively, this claim holds when $\phi$ is a modal tautology. But I'm not sure if that's always the case given that $\vDash\square\phi$. I suspect it isn't.

Update:
The counterexample does not work because: since $\phi$ is any wff, we cannot assume that $\phi$ is not true at some world, for $\phi$ could be a tautology (i.e. $P\lor\lnot P$).

The argument that "If ⊨□φ then ⊨φ" is true does work:
Originally, I thought that it begs the question as saying there is a bisimulation between the old $w$ and the new $w$ means that any atomic letter in either $w$ receives the same truth assignment, which implies the conclusion. My professor got me to see that this is not the case.

Consider this:
Before we assume $\vDash\square\phi$, we first extend any model $M=\langle W, R, V\rangle$ (where $w\in W$) to $M^*=\langle W^*, R^*, V^*\rangle$ like so:

• $W^* = W\cup \{x\}$
• $R^* = R\cup\{\langle x, w\rangle\}$
• $V^*(u,p) = V(u,p)$ if $u\neq x$, F otherwise (so, satisfaction of any formulas in $w$ stays the same)

It is trivial that there is a bisimulation between $\langle M,w\rangle$ and $\langle M^*, w\rangle$.
Now we assume $\vDash\square\phi$. So $M^*,x\vDash\square\phi$, hence $M^*, w\vDash\phi$. Since $M$ and $w$ was arbitrary, we have that $\vDash\phi$.

It isn't necessary to do the model extension and bisimulation first, we can assume $\vDash\square\phi$ first. But this way makes it clearer that the argument doesn't beg the question.

Answer 1

We notice that

$$\Box\psi\rightarrow\psi$$

is not a theorem of $\mathbf{K}$, hence

$$\not\vdash_{\mathbf{K}}\Box\psi\rightarrow\psi$$

However,

$$\vdash\Box\psi\implies\vdash\psi$$

equivalently,

$$\Vdash\Box\psi\implies\Vdash\psi$$

is a metatheorem of $\mathbf{K}$. We may intuitively see this noticing that $\Box\psi$ says that $\psi$ holds in all worlds, therefore, the question of reflexivity has already been resolved. Indeed, the proof system presented in James Garson's Modal Logic for Philosophers includes it as an inference rule.

Johan van Benthem gives an interesting proof of it in his Modal Logic for Open Minds. I quote it here for completeness:

Proof. Suppose that $\psi$ is not provable. Then by completeness, there is a counter-model $\mathcal{M}$ with a world $w$ where $\neg\psi$ holds. Now here is a semantic trick that is used a lot in modal logic. Take any new world $v$, add it to $\mathcal{M}$ and put just one extra $R$-link, from $v$ to $w$:

The atomic valuation at $v$ does not matter. In the new model $\mathcal{M}^{+}$, $\Box\psi$ clearly false at $v$ — hence it is not universally valid, and so by soundness, is not derivable in $\mathbf{K}$.

Answer 2

That the rule is valid is trivial from the perspective of the sequent calculus for K obtained by extending g3cp with the distributive rule

$\Gamma\Rightarrow\varphi$ / $\Gamma',\Box\Gamma\Rightarrow\Box\varphi,\Delta'$

where $\Box\Gamma$ is obtained from $\Gamma$ by replacing every formula $\psi$ in $\Gamma$ with $\Box\psi$ and $\Gamma'$ and $\Delta'$ are arbitrary formulas (for admissibility of thinning).

Since this sequent calculus is cut-free, a derivation of $\Rightarrow\Box\varphi$ can only be obtained from a derivation of $\Rightarrow\varphi$, so if the former is derivable, then the latter is derivable.