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Disappearing Diamonds: Fitch-Like Results in Bimodal Logic

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Abstract

Augment the propositional language with two modal operators: □ and ■. Define \(\blacklozenge \) to be the dual of ■, i.e. \(\blacklozenge \equiv \neg \blacksquare \neg \). Whenever (X) is of the form φψ, let (X\(^{\blacklozenge } \) ) be \(\varphi \rightarrow \blacklozenge \psi \). (X\(^{\blacklozenge } \) ) can be thought of as the modally qualified counterpart of (X)—for instance, under the metaphysical interpretation of \(\blacklozenge \), where (X) says φ implies ψ, (X\(^{\blacklozenge } \) ) says φ implies possiblyψ. This paper shows that for various interesting instances of (X), fairly weak assumptions suffice for (X\(^{\blacklozenge } \)) to imply (X)—so, the modally qualified principle is as strong as its unqualified counterpart. These results have surprising and interesting implications for issues spanning many areas of philosophy.

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Acknowledgments

I owe thanks to Jeff Russell, Gabriel Uzquiano, Tim Williamson, and an anonymous reviewer for their helpful comments.

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Correspondence to Weng Kin San.

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Appendix

Appendix

Theorem 6 shows that given KTK, (4\(^{\blacklozenge }_{\Box }\)) and (5\(^{\blacklozenge }_{\Box }\)) together give rise to an S5 logic for □. We will show that though jointly sufficient, (4\(^{\blacklozenge }_{\Box }\)) and (5\(^{\blacklozenge }_{\Box }\)) are not individually sufficient to give rise to an S5 logic for □.

First, given KTK, (4\(^{\blacklozenge }_{\Box }\)) isn’t strong enough by itself to generate an S5 logic for □:

Theorem 20

The smallest normal extension ofKTKcontaining\({(4^{\blacklozenge }_{\Box })}\)doesnot extendS5K.

Proof

As before, nothing depends on □ and ■ being distinct operators. So, it suffices to show that the smallest normal monomodal extension of S4 containing □p →♢□□p does not extend S5. And that is obvious, since □p →♢□□p is already a theorem of S4. □

Now, we show, by a semantic argument, that (5\(^{\blacklozenge }_{\Box }\)) also isn’t strong enough by itself to generate an S5 logic for □. First, some preliminaries: A Kripke frame is a structure \(\mathfrak {F}=<W, R_{\Box }, R_{\blacksquare }>\), where the domain W is a non-empty set, whose elements we shall refer to as ‘worlds’, and R⊆ (W × W) and R⊆ (W × W) are binary relations on W. A Kripke model\(\mathfrak {M}=<\mathfrak {F}, V>\) is a frame with a valuation function V which maps each propositional letter to a set of worlds. If \(\mathfrak {M}=<\mathfrak {F}, V>\), we say that \(\mathfrak {M}\) is based on \(\mathfrak {F}\). A pointed Kripke model\(<\mathfrak {M},w>\) is a model \(\mathfrak {M}\) together with a world w in the domain of \(\mathfrak {M}\) (by the domain of \(\mathfrak {M}\), we mean the domain of the frame on which \(\mathfrak {M}\) is based). Satisfaction in a pointed model is defined:

$$\begin{array}{lll} \mathfrak{M},w\Vdash p & \text{iff} & w\in V(p), \text{for every}~p\in At; \\ \mathfrak{M},w\Vdash \neg\varphi & \text{iff} & \text{not}~\mathfrak{M},w\Vdash \varphi; \\ \mathfrak{M},w\Vdash (\varphi\wedge\psi) & \text{iff} & \mathfrak{M},w\Vdash\varphi~\text{and}~\mathfrak{M},w\Vdash \psi; \\ \mathfrak{M},w\Vdash \Box\varphi & \text{iff} & \text{for every}~v\in W~\text{such that}~<w,v>\in R_{\Box}, \mathfrak{M},v\Vdash\varphi.\\ \mathfrak{M},w\Vdash \blacksquare\varphi & \text{iff} & \text{for every}~v\in W~\text{such that}~<w,v>\in R_{\blacksquare}, \mathfrak{M},v\Vdash\varphi.\\ \end{array} $$

\(\mathfrak {M}\Vdash \varphi \) iff \(\mathfrak {M},w\Vdash \varphi \) for all w in the domain of \(\mathfrak {M}\). And \(\mathfrak {F}\Vdash \varphi \) iff \(\mathfrak {M}\Vdash \varphi \) for every model \(\mathfrak {M}\) based on \(\mathfrak {F}\).

It is easy to show that any frame satisfying the condition in the antecedent of the lemma below validates (5\(^{\blacklozenge }_{\Box }\)):

Lemma 21

If \(\mathfrak {F}\vDash \forall wv (wR_{\Box } v \rightarrow \exists u (wR_{\blacksquare } u\wedge \forall t(uR_{\Box } t\rightarrow tR_{\Box } v)))\) , then \(\mathfrak {F}\Vdash \Diamond p\rightarrow \blacklozenge \Box \Diamond p\) .

Thus:

Theorem 22

LetLbe the smallest normal extension ofKTKcontaining\({(5^{\blacklozenge }_{\Box })}\).Then,Ldoesn’t extendS4K(and thus also doesn’t extendS5K).

Proof Sketch

Consider the model \(\mathfrak {M}\) below, where the arrows represent R (let R be the universal accessibility relation):

figure a

R is reflexive, so \(\mathfrak {M}\) is a KTK-model. Furthermore, checking that the condition in the antecedent of the previous lemma is satisfied is a tedious but straightforward exercise. Thus, the model is an L-model. However, \(\mathfrak {M},w\Vdash \Box p\wedge \neg \Box \Box p\), so \(\mathfrak {M}\) is a countermodel to (4). Thus, L doesn’t extend S4K and thus also doesn’t extend S5K. □

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San, W.K. Disappearing Diamonds: Fitch-Like Results in Bimodal Logic. J Philos Logic 48, 1003–1016 (2019). https://doi.org/10.1007/s10992-019-09504-0

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