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Exercise 1.8.2 in Fitting and Mendelson's "First Order Logic" asks to show that $\mathcal{M} \models \Box \phi \rightarrow \Box \Box \phi$ for all $\phi$ if and only if the accessibility relation of $\mathcal{M}$ is transitive. This is reiterated in the answer to this question but I have been unable to prove the only if part and believe I have a counterexample:

Let the universe of $\mathcal{M}$ be $\{\Gamma_i: I\in\mathbb{N}\}$ with relation $R = \{(i,i+1): i \in \mathbb{N}\}$. Clearly, $R$ is not transitive. Let $\Vdash$ be identical for each $\Gamma_i$. Then, each for each $i,j,\phi$ we have that $\Gamma_i \Vdash \phi$ if and only if $\Gamma_j \Vdash \phi$, so $\Box \phi \rightarrow \Box\Box \phi$ is valid in $\mathcal{M}$. The proof uses standard induction on height of $\phi$.

What am I missing here?

OwlBandicoot
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1 Answers1

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I realized the problem 30 seconds after posting: The definition of validity in a Frame requires truth across all valuations. Here is the proof:

Fix $\Gamma,\Delta,\Omega$ in the universe if $\mathcal{M}$ with $\Gamma R \Delta$ and $\Delta R \Omega$. Pick a valuation $\Vdash$ so that for $\Gamma R \Lambda$, $\Lambda \Vdash P$ and for $\Gamma \not\text{R} \Lambda$, $\Lambda \not\Vdash P$ for some Propositional Variable $P$. Then, $\Gamma \Vdash \Box P$, so $\Gamma \Vdash \Box\Box P$, so $\Omega \Vdash P$. By construction of $\Vdash$, $\Gamma R\Omega$.

OwlBandicoot
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