Abstract
Hodes (2021) “looked under the hood” of the familiar versions of the classical propositional modal logic K and its intuitionistic counterpart (see Plotkin & Sterling 1986). This paper continues that project, addressing some familiar classical strengthenings of K (D, T, K4, KB, K5, Dio (the Diodorian strengthening of K) and GL), and their intuitionistic counterparts (see Plotkin & Sterling 1986 for some of these counterparts). Section 9 associates two intuitionistic one-step proof-theoretic systems to each of the just mentioned intuitionistic logics, this by adding for each a new rule to those which generated IK in Hodes (2021). For the systems associated with the intuitionistic counterparts of D and T, these rules are “pure one-step”: their schematic formulations does not use □ or ♢. For the systems associated with the intuitionistic counterparts of K4, etc., these rules meet these conditions: neither □ nor ♢ is iterated; none use both □ and ♢. The join of the two systems associated with each of these familiar logics is the full one-step system for that intuitionistic logic. And further “blended” intuitionistic systems arise from joining these systems in various ways. Adding the 0-version of Excluded Middle to their intuitionistic counterparts yields the one-step systems corresponding to the familiar classical logics. Each proof-theoretic system defines a consequence relation in the obvious way. Section 10 examines inclusions between these consequence relations. Section 11 associates each of the above consequence relations with an appropriate class of models, and proves them sound with respect to their appropriate class. This allows proofs of some failures of inclusion between consequence relations. (Sections 10 and 11 provide an exhaustive study of a variety of intuitionistic modal logics.) Section 12 proves that the each consequence relation is complete or (for those corresponding to GL) weakly complete, that relative to its appropriate class of models. The Appendix presents three further results about some of the intuitionistic consequence relations discussed in the body of the paper.
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Notes
If \(\nvdash \boldsymbol {0}\top \) we didn’t need to have 0R0. If Y contains Dio\(_{\square },\) to show that F is an IDio\(_{\square }\)-frame we use this fact: if uiRiv for i ∈ 2, then 0R+v.
We could have kept R a little smaller for certain choices of Y, as follows: \( R=\left \{ \begin {array}{ll} R^{\ast } & \text {if Y=D or Y=Dio}_{\square }, \\ R^{\ast }\cup R_{2} & \text {if Y contains }4_{\%}\text { but not T}_{\square }, \\ R^{\ast }\cup \{\left \langle 1,0\right \rangle \} & \text {if Y contains T} _{\square }\text { but not }4_{\%}, \\ R^{\ast }\cup \{\left \langle 1,0\right \rangle \}\cup R_{2} & \text {if Y contains T}_{\square }\text { and }4_{\%}. \end {array} \right . \)(Abovereplace ‘%’ by ‘♢’, ‘\(\square \)’ or make it blank.)
We need 1Rv0 for right-completeness.
Were we to define F so as to make it a B♢-, B\(_{\square }\) -, 5♢-, 5\(_{\square }\)-, or Dio♢-frame, it isn’t clear how we could insure (*).
If the replacement for ‘Y’ contained ‘T♢’ we would need to have 0R0 or 0R1, which would block this point.
Recall that \(\vdash _{IY}\subseteq \) ⊩IGL if Y = K or 4; so those choices of Y are also covered by this case.
This case covers those Y such that \(\vdash _{IT_{\%}}\subseteq _{IY}\vdash \subseteq \) ⊩IS5 for \(\%\in \{\square ,\Diamond \}\).
Y = GL covers the Y = 4 case, since \(\vdash _{I4}\subseteq \) ⊩IGL; but since \(\vdash _{IS4}\nsubseteq \) ⊩IGL we need to list S4 separately.
References
Boolos, G. (1979). The Unprovability of Consistency. Cambridge: Cambridge University Press.
Hodes, H.T. (2021). One-step Modal Logics, Intuitionistic and Classical: Part 1. Journal of Philosophical Logic.
Plotkin, G., & Stirling, C. (1986). A Framework for Intuitionistic Modal Logics. In J.Y. Halpern (Ed.) Theoretical Aspects of Reasoning About Knowledge (pp. 399–406). Morgan Kaufmann Publishers.
Popkorn, S. (1994). First Steps in Modal Logic. Cambridge: Cambridge University Press.
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Thanks to Philip Sink, and to the referee, for reading several drafts of this paper and catching errors.
Appendix: More About Intuitionistic Modal Logics
Appendix: More About Intuitionistic Modal Logics
For A.1 and A.2, let ⊩ = ⊩IY for \(\vdash _{IGL_{\Diamond }}\nsubseteq \) ⊩IY and \(\vdash _{IGL_{\square }}\nsubseteq \) ⊩IY.
1.1 A.1 Observation
⊩ has the disjunction property, i.e. for any φi∈2 ∈ Fml, if ⊩0(φ0 ∨ φ1) then ⊩0φ0 or ⊩0φ1.
Proof
Assume the if-clause. Assume that for both i ∈ 2 \(\nvdash \boldsymbol {0} \varphi _{i}\). By Completeness for IY we may fix an IY-model \({\mathscr{M}} _{i} \) and ui so that \({\mathscr{M}}_{i},u_{i}\nvDash \varphi _{i}\). Let \( F^{{\mathscr{M}}_{i}}=\left \langle W_{i},R_{i},\sqsubseteq _{i}\right \rangle \). Without loss of generality we may assume that W0 ∩ W1 = {} and 0∉W0 ∪ W1. Let W = {0}∪ W0 ∪ W1, \(R=R_{0}\cup R_{1}\cup \{\left \langle 0,0\right \rangle \},\) and
Check that \(F=\left \langle W,R,\sqsubseteq \right \rangle \) is an IY-frame. Footnote 1 For v ∈ W and π ∈ S, let
Let \({\mathscr{M}}=\left \langle F,\mathcal {V}\right \rangle \). For both i ∈ 2 and any u ∈ Wi and v, uRv iff uRiv, and also \(u\sqsubseteq v\) iff \(u\sqsubseteq _{i}v\). So \({\mathscr{M}},u_{i}\nvDash \varphi _{i}\). By the if-clause and the soundness of ⊩ with respect to IY-models, \( {\mathscr{M}},u\models (\varphi _{0}\vee \varphi _{1})\). Fix i so that \( {\mathscr{M}},u\models \varphi _{i}\). By the Persistence Lemma \({\mathscr{M}} ,u_{i}\models \varphi _{i},\) a contradiction. The then-clause follows. □
If 12.5 is true, the previous argument applies for Y = GL♢ and Y= GL\(_{\square }. \)
1.2 A.2 Observation
Consider any φ ∈ Fml. (1) If \(\nvdash \boldsymbol {0} \Diamond \top ,\) then \(\ \boldsymbol {0}\lnot \square \varphi \vdash \boldsymbol {0}\Diamond \lnot \varphi \) iff \(\vdash \boldsymbol {0}\lnot \lnot \square \varphi \). (2) Assume that the replacement for ‘Y’ is constructed using ‘T♢’ and any of the above names. Then \(\boldsymbol {0} \lnot \square \varphi \vdash \boldsymbol {0}\Diamond \lnot \varphi \) iff \( \boldsymbol {0}\lnot \square \varphi \vdash \boldsymbol {0}\lnot \varphi \). (3) Assume that the replacement for ‘Y’ is constructed using ‘D’, ‘T\(_{\square }\)’, ‘T’, ‘4♢’, ‘4\(_{\square }\)’, ‘4’ or ‘Dio\(_{\square }\)’. Then \(\boldsymbol {0}\lnot \square \varphi \vdash \boldsymbol {0 }\Diamond \lnot \varphi \) iff either \(\vdash \boldsymbol {0}\Diamond \lnot \varphi \) or \(\vdash \boldsymbol {0}\lnot \lnot \square \varphi \).
Proof
For (1), assume that \(\nvdash \boldsymbol {0}\Diamond \top ;\) so IY-frames can have dead-ends. Right to left is trivial. Assume the left-side. Assume that \(\nvdash \boldsymbol {0}\lnot \lnot \square \varphi ,\) By Completeness for IY we may fix an IY-model \({\mathscr{M}}\) and u so that \({\mathscr{M}} ,u\nvDash \lnot \lnot \square \varphi \). Let \(F^{{\mathscr{M}}}=\left \langle W,R,\sqsubseteq \right \rangle \) and \(\mathcal {V}=\mathcal {V}^{{\mathscr{M}}}\). So we may fix a \(u^{\prime }\sqsupseteq u\) so that \({\mathscr{M}},u^{\prime }\models \lnot \square \varphi \). Without loss of generality assume that 0∉W. Let \(W^{\prime }=\{0\}\cup W,\) \(\sqsubseteq ^{\prime }=\) \( \sqsubseteq \cup \{\left \langle 0,v\right \rangle \mid u^{\prime }\sqsubseteq v\},\) and \(F^{\prime }=\left \langle W^{\prime },R,\sqsubseteq ^{\prime }\right \rangle \). For \(v\in W^{\prime }\) and π ∈ S, let
Let \({\mathscr{M}}^{\prime }=\left \langle F^{\prime },\mathcal {V}^{\prime }\right \rangle \). By an easy induction, (*) for every v ∈ W and every ψ ∈ Fml, \({\mathscr{M}}^{\prime },v\models \psi \) iff \({\mathscr{M}} ,v\models \psi \). If \({\mathscr{M}}^{\prime },0\models \square \varphi \) then \( {\mathscr{M}}^{\prime },u^{\prime }\models \square \varphi ,\) and then by (*) \( {\mathscr{M}},u^{\prime }\models \square \varphi ,\) a contradiction. So \( {\mathscr{M}}^{\prime },0\nvDash \square \varphi \). Assume that \(0\sqsubseteq ^{\prime }v\) and \({\mathscr{M}}^{\prime },v\models \square \varphi ;\) so v≠ 0, and so \(v\sqsupseteq u^{\prime };\) so by (*) \({\mathscr{M}},v\models \square \varphi ,\) a contradiction. So \({\mathscr{M}}^{\prime },0\models \lnot \square \varphi \). Since 0 is a dead-end in \(F^{\prime },\) \({\mathscr{M}} ^{\prime },0\nvDash \Diamond \lnot \varphi ,\) contrary to the left-side. (1) follows.
For (2) going right to left, use 10.1(4). Assume the left-side. Assume that \( \boldsymbol {0}\lnot \square \varphi \nvdash \boldsymbol {0}\lnot \varphi \). By Completeness for IY we may fix an IY-model \({\mathscr{M}}\) and u so that \( {\mathscr{M}},u\models \lnot \square \varphi \) but \({\mathscr{M}},u\nvDash \lnot \varphi \). For \(W,R,\sqsubseteq \) and \(\mathcal {V}\) as above, let \(R^{\prime }=R\cup \{\left \langle 0,u\right \rangle \}\) and let \(W^{\prime },\sqsubseteq ^{\prime }\) and \(\mathcal {V}^{\prime }\) be as above; let Let \({\mathscr{M}} ^{\prime }=\left \langle F^{\prime },\mathcal {V}^{\prime }\right \rangle \). (*) carries over from the previous paragraph. As above, \({\mathscr{M}}^{\prime },0\models \lnot \square \varphi \). For any v, if \(0R^{\prime }v\) then v = u; so \({\mathscr{M}},u\nvDash \Diamond \lnot \varphi ,\) contrary to the left-side. (2) follows.
For (3), assume that ‘Y’ is replaced appropriately. Right to left is trivial. Assume the left side. By (1) we lose no generality by assuming that ⊩0♢⊤. Assume that \(\nvdash \boldsymbol {0} \lnot \lnot \square \varphi \) and \(\nvdash \boldsymbol {0}\Diamond \lnot \varphi \). By Completeness for IY we may fix IY-models \({\mathscr{M}}_{i\in 2}\) and ui∈2 so that \({\mathscr{M}}_{0},u_{0}\nvDash \Diamond \lnot \varphi \) and \({\mathscr{M}}_{1},u_{1}\nvDash \lnot \lnot \square \varphi \). Let \(F^{{\mathscr{M}}_{i}}=\left \langle W_{i},R_{i},\sqsubseteq _{i}\right \rangle \) and \(\mathcal {V}_{i}=\mathcal {V}^{{\mathscr{M}}_{i}}\) for both i ∈ 2. Without loss of generality let W0 ∩ W1 = {} and 0, 1∉W0 ∪ W1. As above we may fix a \(u_{1}^{\prime }\sqsupseteq _{1}u_{1}\) so that \({\mathscr{M}}_{1},u_{1}^{\prime }\models \lnot \square \varphi ;\) fix v1 so that \(u_{1}^{\prime }R_{1}^{+}v_{1}\) and \({\mathscr{M}}_{1},v_{1}\nvDash \varphi \). Since ⊩0♢⊤, we may fix a v so that u0R0v. Since \({\mathscr{M}} _{0},v\nvDash \lnot \varphi ,\) we may fix a \(v_{0}\sqsupseteq _{0}v\) so that \({\mathscr{M}}_{0},v_{0}\models \varphi \). Let W = {0, 1}∪ W0 ∪ W1,
\(R=R^{\ast }\cup R_{2}\cup \{\left \langle 1,0\right \rangle \},\)Footnote 2\( F=\left \langle F,R,\sqsubseteq \right \rangle ,\) \(\mathcal {V}=\mathcal {V} _{0}\cup \mathcal {V}_{1},\) and \({\mathscr{M}}=\left \langle F,\mathcal {V} \right \rangle \). Check that F is an IY-frame.Footnote 3 By easy inductions, (*) for both i ∈ 2 for every v ∈ Wi and every ψ ∈ Fml, \({\mathscr{M}} ,v\models \psi \) iff \({\mathscr{M}}_{i},v\models \psi \).Footnote 4 If \({\mathscr{M}},1\models \square \varphi \) then \({\mathscr{M}},v_{1}\models \varphi ,\) and by (*) \({\mathscr{M}} _{1},v_{1}\models \varphi ,\) a contradiction. So \({\mathscr{M}},1\nvDash \square \varphi \). If \({\mathscr{M}},0\models \square \varphi \) then by the Persistence Lemma \({\mathscr{M}},1\models \square \varphi ,\) a contradiction. So \({\mathscr{M}},0\nvDash \square \varphi ;\) so \({\mathscr{M}},0\models \lnot \square \varphi \). For any v, if 0Rv then v = v0.Footnote 5 So since \({\mathscr{M}} ,v_{0}\models \varphi ,\) \({\mathscr{M}},0\nvDash \Diamond \lnot \varphi \). This contradicts the left-side. (3) follows. □
1.3 A.3 Observation
A formula is non-modal iff it contains no occurrences of \(\square \) or ♢. Let ⊩I = non-modal intuitionistic consequence.
For any choice of Y as in §11, ⊩IY is conservative over ⊩I. In other words, for any set Γ of non-modal formulas and a non-modal formula φ, 0Γ ⊩IY0φ iff Γ ⊩Iφ.
Proof
Consider Γ and φ as described. Right to left is trivial. Assume that \({\Gamma } \nvdash _{I}\varphi \). By the completeness of (non-modal) intuitionist logic with respect to intuitionistic Kripke models, we may fix an intutionistic Kripke-model \({\mathscr{M}}=\left \langle W,\sqsubseteq ,\mathcal {V}\right \rangle \) with signature \(\mathcal {S}\) and a u ∈ W such that \({\mathscr{M}},u\models {\Gamma } \) but \({\mathscr{M}},u\nvDash \varphi \).
Consider Y = B, 5, GL or Dio.Footnote 6 Set R0 = {} and \({\mathscr{M}}_{0}=\left \langle W,R_{0},\sqsubseteq ,\mathcal {V}\right \rangle \). \( {\mathscr{M}}_{0}\) is an IY-model, and \({\mathscr{M}}_{0},u\models {\Gamma } \) but \( {\mathscr{M}}_{0},u\nvDash \varphi ;\) so \({\mathscr{M}}_{0},u\Vdash 0{\Gamma } \) but \({\mathscr{M}}_{0},u\nVdash \boldsymbol {0}\varphi \). So \(\boldsymbol {0} {\Gamma } \nvdash _{IY}\boldsymbol {0}\varphi \).
Consider Y = S5.Footnote 7 Set R1 = id|W and \({\mathscr{M}}_{1}=\left \langle W,R_{1},\sqsubseteq ,\mathcal {V }\right \rangle \). \({\mathscr{M}}_{1}\) is an IS5-model, and as above \({\mathscr{M}}_{1},u\Vdash 0{\Gamma } \) but \({\mathscr{M}}_{1},u\nVdash \boldsymbol {0}\varphi . \) So \(\boldsymbol {0}{\Gamma } \nvdash _{IY}\boldsymbol {0}\varphi \). □
Thus the modal apparatus in IY has not surreptitiously strengthened the “background” (i.e. non-modal) logic from intutionistic logic to classical logic or an intermediate logic.
1.4 A.4 Observation
We can push this idea further. Consider Y = S4, GL or Dio,Footnote 8 a set Δ of formulas such that for each δ ∈Δ \(\boldsymbol {0}\delta \nvdash _{IY}\boldsymbol {0}\bot ,\) and a set Θ of formulas such that \(\boldsymbol {0}{\Theta } \nvdash _{IY} \boldsymbol {0}\bot \).
For any Γ and φ as in A.3, if \(\boldsymbol {0} ({\Gamma } \cup \Diamond {\Delta } \cup \square {\Theta } )\vdash _{IY}\boldsymbol {0} \varphi \) then Γ ⊩Iφ.
Proof
Given Γ and φ as described, assume that \({\Gamma } \nvdash _{I}\varphi \). Let \({\mathscr{M}},\) u and \({\mathscr{M}}_{i}\) for i ∈ 2 be as in A.3; set \(u^{{\mathscr{M}}}=u\) and \({\mathscr{M}}^{\prime }={\mathscr{M}} _{i}\). Without loss of generality, assume that \(u^{{\mathscr{M}}}\) is the unique initial element of \(\left \langle W^{{\mathscr{M}}},\sqsubseteq ^{ {\mathscr{M}}}\right \rangle \). By our model-existence theorems, for each δ ∈Δ fix an IY-model \({\mathscr{M}}_{\delta }\) and a \(u_{\delta }\in W^{{\mathscr{M}}_{\delta }}\) such that \({\mathscr{M}}_{\delta },u_{\delta }\models \delta \). Also fix an IY-model \({\mathscr{M}}_{\Gamma }\) and a \( u_{\Gamma }\in W^{{\mathscr{M}}_{\Gamma }}\) such that \({\mathscr{M}}_{\Gamma },u_{\Gamma }\models {\Gamma } \). Without loss of generality, we can make sure that \(W^{{\mathscr{M}}},\) \(W^{{\mathscr{M}}_{\Gamma }}\) and the \(W^{{\mathscr{M}} _{\delta }}\) for δ ∈Δ are all disjoint from one another. Let
Assume that Y = T or Dio. Let
Note: for each \(v\in W^{{\mathscr{M}}}\) and w ∈ WΓ, \(vR^{\prime }\left \langle v,w\right \rangle \). We will construct an IY-model \({\mathscr{M}} ^{\ast }=\left \langle W^{\ast },R^{\ast },\sqsubseteq ^{\ast },\mathcal {V} ^{\ast }\right \rangle \) by “sewing” each \({\mathscr{M}}_{\delta \in {\Delta } }\) and copies of \({\mathscr{M}}_{\Gamma }\) onto \({\mathscr{M}}^{\prime },\) with \( R^{\prime }\) as the “seam”. For each \(v\in W^{{\mathscr{M}}}\) form \({\mathscr{M}} _{\Gamma ,v}\) from \({\mathscr{M}}_{\Gamma }\) by replacing each \(w\in W^{ {\mathscr{M}}_{\Gamma }}\) by \(\left \langle v,w\right \rangle \). Let
For v,w ∈ W∗ let \(v\sqsubseteq ^{\ast }w\) iff either (i) \( v\sqsubseteq ^{{\mathscr{M}}}w,\) or (ii) for some δ ∈Δ \( v\sqsubseteq ^{{\mathscr{M}}_{\delta }}w,\) or (iii) for some \(x\in W^{{\mathscr{M}}}\) and \(y,y^{\prime }\in W^{{\mathscr{M}}_{\Gamma }},\) \(v=\left \langle x,y\right \rangle ,\) \(w=\left \langle x,y^{\prime }\right \rangle \) and \( y\sqsubseteq ^{{\mathscr{M}}_{\Gamma }}y^{\prime }\). Claim 1: the frame \( \left \langle W^{\ast },R^{\ast },\sqsubseteq ^{\ast }\right \rangle \) is left-and right-complete. Given v,w, assume that vR∗w. The only interesting case: \(v\in W^{{\mathscr{M}}}\) and \(w\in \bigcup \nolimits _{\delta \in {\Delta } }W_{\delta }\cup \left (W^{{\mathscr{M}}}\times W^{{\mathscr{M}} _{\Gamma }}\right ) \). If \(v^{\prime }\sqsupseteq v\) then \(v^{\prime }R^{\prime }w\sqsupseteq w;\) this suffices for right-completeness. If \( w^{\prime }\sqsupseteq w\) then \(v\sqsubseteq vR^{\prime }w^{\prime };\) this suffices for left-completeness.
For each v ∈ W∗ and \(\gamma \in \mathcal {S},\) let
Check that \({\mathscr{M}}^{\ast }\) is an IY-model.
Assume that Y = S4 or GL. It will be convenient to have a transitive frame, which requires “sewing a wider seam”. Let
Note: for each \(v\in W^{{\mathscr{M}}}\) and \(w\in W^{{\mathscr{M}}_{\Gamma ,v}},\)\(vR^{\prime }\left \langle v,w\right \rangle \). Define \(\sqsubseteq ^{\ast }\) as in the previous case. Claim 2: \(\left \langle W^{\ast },R^{\ast },\sqsubseteq ^{\ast }\right \rangle \) is left-and right-complete, and also transitive. Left- and right-completeness follow much as claim 1 did. Given x,y,z, assume that xR∗yR∗z. If \(xR^{{\mathscr{M}}^{\prime }}y\) then Y = S4 and x = y (since i = 1), yielding xR∗z. In the other interesting case, \(xR^{\prime \prime }y\) and \(y\left (\bigcup \nolimits _{\delta \in {\Delta } }R^{{\mathscr{M}}_{\delta }}\cup \bigcup \nolimits _{v\in W^{{\mathscr{M}}}}R^{{\mathscr{M}}_{\Gamma ,v}}\right ) z; \) then \(xR^{\prime }z;\) so xR∗z. Define \(\mathcal {V}^{\ast }\) as in the previous case. Check that \({\mathscr{M}}^{\ast }=\left \langle W^{\ast },R^{\ast },\sqsubseteq ^{\ast },\mathcal {V}^{\ast }\right \rangle \) is an IY-model.
Claim under both cases: for any formula σ and u ∈ W∗: (i) if \(u\in W^{{\mathscr{M}}},\) \({\mathscr{M}}^{\ast },u\models \sigma \) iff \( {\mathscr{M}}^{\prime },u\models \sigma ;\) (ii) if for δ ∈Δ \( u\in W^{{\mathscr{M}}_{\delta }},\) \({\mathscr{M}}^{\ast },u\models \sigma \) iff \( {\mathscr{M}}_{\delta },u\models \sigma ;\) (iii) if for \(x\in W^{{\mathscr{M}}}\) and \(y\in W^{{\mathscr{M}}_{\Gamma }}\) \(u=\left \langle x,y\right \rangle ,\) these are equivalent: \({\mathscr{M}}^{\ast },u\models \sigma ;\) \({\mathscr{M}} _{\Gamma ,x},u\models \sigma ;\) \({\mathscr{M}}_{\Gamma },y\models \sigma \). Proof: induction on the construction of σ.
Thus \({\mathscr{M}}^{\ast },u^{{\mathscr{M}}}\models ({\Gamma } \cup \Diamond {\Delta } \cup \square {\Theta } ),\) but \({\mathscr{M}}^{\ast },u^{{\mathscr{M}} }\nvDash \varphi \). So \(\boldsymbol {0}({\Gamma } \cup \Diamond {\Delta } \cup \square {\Theta } )\nvdash _{IY}\boldsymbol {0}\varphi . \)□
1.5 A.5 Observation
The observation in A.4 does not extend to Y = B\(_{\square }\).
Example. Consider any \(\pi \in \mathcal {S}\). Let Γ = Θ = {}, \(\varphi =(\pi \vee \lnot \pi ),\) \({\Delta }=\{\square \varphi \}\). Recall that \( \boldsymbol {0}\Diamond \square \varphi \vdash _{IB_{\square }}\boldsymbol {0} \varphi \). So for any IB\(_{\square }\)-model \({\mathscr{M}}\) and \(u\in W^{{\mathscr{M}}},\) if \({\mathscr{M}},u\models \Diamond \square \varphi \) then \( {\mathscr{M}},u\models \varphi \). But \(\nvdash _{I}\varphi \).
Question: Does the observation in A.4 extend to Y = B♢?
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Hodes, H.T. One-Step Modal Logics, Intuitionistic and Classical, Part 2. J Philos Logic 50, 873–910 (2021). https://doi.org/10.1007/s10992-021-09607-7
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DOI: https://doi.org/10.1007/s10992-021-09607-7