Construct a frame $\mathfrak{F} = (W,R)$ that refutes $\square \bot$

Question

(Blackburn, Modal Logic, Section 1.3) Construct a frame $\mathfrak{F} = (W,R)$ that refutes $\square \bot$, to show that $\square \bot$ is not a valid formula.

I think I've solved this, but I'm not sure if my understanding of $\bot$ is correct. My solution is as follows -

Consider $W = \{a\}$, and $R$ such that $Raa$ (self-loop). If $\square \bot$ holds at $a$, i.e. $\mathfrak{F},a \Vdash \square \bot$ then $\mathfrak{F},a \Vdash \bot$ also holds (by definition of $\square$). However, $\bot$ cannot hold anywhere, so $\square\bot$ is not a valid formula (We have constructed a frame in which it is not valid).

Is my understanding of $\bot$ right? Is $\bot$ something that can never be true, i.e. is always false?

Answer 1

Yes.   $\bot$ is valued as false by its definition in propositional logic, so cannot be valued as true by any world in any frame of modal logic.

When $\mathsf W=\{a\}$ and $\operatorname R=\{\langle a,a\rangle\}$, then $a\Vdash\square\varphi$ if and only if $a\Vdash\varphi$.   Since $a\nVdash\bot$ by definition, therefore $a\nVdash\square\bot$ in such a frame and thus $\square\bot$ is not a valid formula there in.

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More.

• The meaning of $~a\Vdash\varphi~$ is that "formula $\varphi$ is valued true in world $a$".
• The meaning of $~a\Vdash\square\varphi~$ is that "formula $\varphi$ is valued true in all worlds accessible from $a$."
• A valid formula is one that is valued true in all worlds of the frame.
• $\bot$ is valued false. No worlds can value $\bot$ as true.
• Thus $\square\bot$ is only valued true in worlds which have no accessible worlds.
• Therefore $\square\bot$ is only a valid formula in frames which have no accessible worlds.

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