Every normal modal logic Σ contains ¬♢⊥.

Question

The following proposition comes from this PDF: http://builds.openlogicproject.org/content/normal-modal-logic/normal-modal-logic.pdf (p.31)

It says: Every normal modal logic Σ contains ¬♢⊥.

The problem beneath it says to prove this. I don't really know how I would e able to prove this. Can someone help me?

The definition of a normal modal logic can be found under definition 3.5 it says:

A modal logic Σ is normal if it contains

□(p → q) → (□p → □q), (K)

♢p ↔ ¬□¬p (dual)

and is closed under necessitation, i.e., if φ ∈ Σ, then □φ ∈ Σ. Observe that while tautological implication is “fine-grained” enough to preserve truth at a world, the rule nec only preserves truth in a model (and hence also validity in a frame or in a class of frames).

Thank you!

Answer 1

Clue: you want to rewrite what you want to prove with a box. How?

Clue: what is the logical status of $\neg\bot$?

If the answer to the second doesn’t hit you immediately then perhaps it is propositional logic you first need to get clearer about!

Answer 2

This answer contains a hint and my guess as to the authors' intent behind the problem. This answer intentionally does not contain a complete proof/argument.

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On page 31, as part of definition 3.5, the authors of Boxes and Diamonds observe that a modal logic satisfying axiom $K$, the duality of $\lozenge$ and $\square$, and the rule of necessitation is normal and the relational semantics first described in section 1.4 are adequate for producing a sound and complete semantics of such a modal logic. The text immediately below the definition that you cited suggests to me that the authors intend readers to use a semantic proof for this problem.

We can use this result and argue using the relational semantics.

I claim $W, R, w \Vdash \lnot \lozenge \bot$ is always true.

We can prove this by looking at the truth conditions of $\lnot$, $\lozenge$, and $\bot$.

As an example, here are the truth conditions of $\lor$ and $\square$.

$W, R, w \Vdash \varphi \lor \psi$ holds if and only if $W, R, w \Vdash \varphi$ holds or $W, R, w \Vdash \psi$ holds. This makes sense: intuitively, we can assess the truth of an expression headed by $\lor$ without travelling to other worlds besides $w$.

$W, R, w \Vdash \square \varphi$ holds if and only if $W, R, u \Vdash \varphi$ holds at every possible world $u$ that's $R$-accessible from $w$. This holds vacuously if there are no accessible worlds. Assessing the truth of a statement headed by $\square$ forces us to look at worlds adjacent to $w$, which also makes intuitive sense.