Abstract
An increasing amount of contemporary philosophy of mathematics posits, and theorizes in terms of special kinds of mathematical modality. The goal of this paper is to bring recent work on higher-order metaphysics to bear on the investigation of these modalities. The main focus of the paper will be views that posit mathematical contingency or indeterminacy about statements that concern the ‘width’ of the set theoretic universe, such as Cantor’s continuum hypothesis. Within a higher-order framework I show that contingency about the width of the set-theoretic universe refutes two orthodoxies concerning the structure of modal reality: the view that the broadest necessity has a logic of S5, and the ‘Leibniz biconditionals’ stating that what is possible, in the broadest sense of possible, is what is true in some possible world. Nonetheless, I suggest that the underlying picture of modal set-theory is coherent and has attractions.
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\(\blacklozenge \exists x(V_{\alpha }x \wedge Fx)\) implies by Theorem 15 that there is a mathematically possible world proposition \(w \le \exists x(V_{\alpha }x \wedge Fx)\), and since \(\Diamond \exists x(w \wedge V_{\alpha }x \wedge Fx)\) there is a world property W entailing \(\lambda x(w \wedge V_{\alpha }x \wedge Fx)\).
Models of this will interpret a with an individual that does not belong to the domain of quantification at any world. It is quite easy to generate an extensional model of Free Classicism in which \(\forall _ex(Fx \rightarrow Gx)\) (and thus \(\Box \forall _ex\Box (Fx \rightarrow Gx)\)), Fa and \(\lnot Ga\) are all true.
Boole [12], p20 project Gutenberg.
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This paper has benefited greatly from discussions with Zach Goodsell, Chris Scambler, Joel Hamkins and Jeff Russell on issues relating to this paper.
Appendices
A Appendix: Proofs of Theorems
Theorem 5 \((\textsf{C5})\)
\(\forall X({\text {Nec}}_\infty X \rightarrow \textsf{BF}_X^\sigma )\)
Proof
\(\textsf{C5}\) contains the broad Barcan formula, \(\textsf{BF}^\sigma _\Box \).
Suppose that X is infinitely closed and that \(\forall _\sigma x\, X(Fx)\). We want to show that \(X(\forall _\sigma x Fx)\). Since X is infinitely closed, it suffices to show that anything entailing every X-necessary proposition also entails \(\forall _\sigma x Fx\). Suppore r entails every X-necessary proposition. Since Fx is X-necessary for every x, \(\forall _\sigma x.\Box (r\rightarrow Fx)\). By the broad Barcan formula, \(\Box \forall _\sigma x(r\rightarrow Fx)\) and so \(\Box (r \rightarrow \forall _\sigma x Fx)\). Thus r entails \(\forall _\sigma xFx\) as required. Since X is closed under entailment, \(X(\forall _\sigma xFx)\). \(\square \)
Theorem 7 \((\textsf{C}^{\blacksquare \in })\)
Given \(\textsf{BF}^e_\Box \) (for broad necessity), being of stage \(\alpha \) (i.e. \(V_\alpha \)) is rigid for every ordinal \(\alpha \).
Proof
As we have noted (Theorem 5), \(\textsf{BF}^e_\Box \) for broad necessity implies the Barcan formula for \(\blacksquare \), \(\textsf{BF}^e_\blacksquare \). Subsequent uses of the word ‘possibly’ and ‘necessarily’ in the proof refer to \(\blacklozenge \) and \(\blacksquare \).
The proof is by transfinite induction. \(V_0\) is necessarily empty, and so vacuously rigid.
Suppose that \(\alpha \) is an ordinal, and for each \(\beta \in \alpha \), \(V_\beta \) is rigid. We want to show that \(V_\alpha \) is rigid. Suppose \(\blacklozenge \exists x(V_{\alpha }x \wedge Fx)\). We must show \(\exists x(V_\alpha x \wedge \blacklozenge Fx)\).
The Barcan formula ensures there is an x such that \(\blacklozenge (V_\alpha x \wedge Fx)\), but we have no guarantee that x is in fact \(V_\alpha \), or even if it is a set. Instead of directly showing x is a set, we’ll define another set by separation containing the elements that would have belonged to x if x had been a set, and show that this is a \(V_\alpha \) set that is possibly F.
We can now show that \(x'\) is identical x as follows. Given the mathematical necessity of Set Extensionality it suffices to show that necessarily if x is a set, x coextensive with \(x'\): \(\blacksquare ({\text {Set}}x \rightarrow \forall y.(y\in x \leftrightarrow y\in x')\). We break this up into two claims:
-
1.
\(\blacksquare ({\text {Set}}x \rightarrow \forall y(y\in x' \rightarrow y\in x)\)
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2.
\(\blacksquare ({\text {Set}}x \rightarrow \forall y(y\in x\rightarrow y\in x')\)
We establish 1 first. From the definition of membership in \(x'\), we immediately have \(\forall y\in x'\blacksquare ({\text {Set}}x \rightarrow y\in x)\). Since Sets are Rigid, it follows that \(x'\)-restricted quantification satisfies BF, so we can infer \(\blacksquare \forall y\in x'({\text {Set}}x\rightarrow y\in x)\). By applying first-order logic under the scope of \(\blacksquare \), this is equivalent to 1.
To establish 2, it suffices to show \(\forall y \blacksquare ({\text {Set}}x \rightarrow (y\in x \rightarrow y\in x'))\) by the Barcan formula. Let y be an arbitrary individual. Now either \(y\in x'\) or \(y\notin x'\). Suppose the former. Then by the rigidity of set membership y is necessarily in \(x'\) and so \(\blacksquare ({\text {Set}}x \rightarrow (y\in x \rightarrow y\in x'))\) follows. Suppose, then, that \(y\notin x'\). By the definition of \(x'\) this would mean that \(\blacklozenge ({\text {Set}} x\wedge y\notin x)\). It follows that \(\blacksquare ({\text {Set}}x \rightarrow (y\in x \rightarrow y\in x'))\), for if this were false we’d have \(\blacklozenge y\in x\) and by Sets are Rigid we could conclude \(\blacksquare ({\text {Set}}x\rightarrow y\in x)\) contradicting the previous line. (It is here that we must use the stronger version of Sets are Rigid outlined in footnote 36 if we are not assuming the convergence axiom.)
By the necessity of Extensionality, we have shown that \(\blacksquare ({\text {Set}}x \rightarrow x=x')\). \(\blacklozenge (V_\alpha x \wedge Fx)\) thus entails \(\blacklozenge Fx'\). And by construction \(x'\) is \(V_\alpha \) so \(\exists x(V_{\alpha }x \wedge \blacklozenge F x)\) as required. \(\square \)
Lemma 23
\((\textsf{C}^{\blacksquare \in })\). Sets are mathematically necessarily distinct: \(\forall _e xy({\text {Set}}x\wedge {\text {Set}}y \rightarrow x\ne y \rightarrow \blacksquare x\ne y)\)
Proof
Suppose the claim is false for contradiction. Choose x to be \(\in \)-minimal such that x possibly identical to some set it is distinct from. Choose y to be \(\in \)-minimal such that it is distinct from, but possibly identical to x.
Since x and y are distinct we may suppose, without loss of generality, that there is some set z belonging to x but not belonging to y. By Set Rigidity, \(\blacksquare z\in x\). So \(\blacklozenge z\in y\), since \(\blacklozenge x=y\). Since \(\blacklozenge \exists z'\in y.z'=z\) it follows by Set Rigidity that \(\exists z'\in y\blacklozenge z'=z\). Since x is an \(\in \)-minimal failure of the necessity of distinctness, z cannot be possibly identical to anything distinct from it. It follows that whatever member of y that is possibly identical to z is in fact identical to z, so that z is a member of y after all, a contradiction. \(\square \)
Theorem 8 \((\textsf{C}^{\blacksquare \in })\)
Suppose \(A(\overline{x})\) is a first-order formula with free variables \(\overline{x}\). If all the quantifiers in \(A(\overline{x})\) are restricted to rigid properties of sets, then A is modally absolute with respect to the parameters \(\overline{y}\).
Proof
By Set Rigidity, \(x\in y\) is modally absolute, since if y is a set and \(x \in y\) then by Set Rigidity x is necessarily in y. And if \(x\not \in y\) and y is a set, then by the necessity of distinctness of sets x could not be identical to a member of y. The necessity of identity and distinctness for sets ensures the modal absoluteness of \(x=y\). Suppose A and B are modally absolute. If for any sequence of sets \(\overline{x}\), \(A(\overline{x})\) and \(B(\overline{x}\), then the modal absoluteness of A and B ensures that \(\blacksquare A(\overline{x})\) and \(\blacksquare B(\overline{x})\) and so \(\blacksquare (A(\overline{x})\wedge B(\overline{x}))\). Similarly if \(\lnot (A(\overline{x})\wedge B(\overline{x}))\) either \(\lnot A(\overline{x})\) or \(\lnot B(\overline{x})\) and so given the modal absoluteness of A and B we have either \(\blacksquare \lnot A(\overline{x})\) or \(\blacksquare \lnot B(\overline{x})\) and in either case \(\blacksquare \lnot (A\wedge B)\) as required. The disjunction case is a dualization of the above, and the negation case is trivial.
Now suppose \(B(y\overline{x})\) is modally absolute, and \(\lambda y.A(y\overline{x})\) is a rigid property of sets (\(\lambda y.A(y\overline{x})\) entails \({\text {Set}}\)). We will show the modal absoluteness of \(\forall _ey(A(y\overline{x})\rightarrow B(y\overline{x})\). Let \(\overline{x}\) be a sequence of sets, and suppose \(\forall _ey(A(y\overline{x})\rightarrow B(y\overline{x})\). By the modal absoluteness of B we can conclude \(\forall _ey(A(y\overline{x})\rightarrow \blacksquare B(y\overline{x}))\), and by the rigidity of A we can get \(\blacksquare \forall _ey(A(y\overline{x})\rightarrow B(y\overline{x})\). On the other hand, if \(\lnot \forall _ey(A(y\overline{x})\rightarrow B(y\overline{x})\) then for some set y, \((A(y\overline{x})\wedge \lnot B(y\overline{x})\). By the modal absoluteness of B, \(\blacksquare \lnot B(y\overline{x})\) and by the rigidity of A, \(\blacksquare A(y\overline{x})\) so \(\blacksquare \exists _ey(A(y\overline{x})\wedge \lnot B(y\overline{x}))\), as required. The existential case involves dualizing this argument. \(\square \)
Theorem 9 \((\textsf{C}^{\blacksquare \in })\)
Given the truth of the theorems of \(\textsf{C}^{\blacksquare \in }\), the following formulas are modally absolute.
-
1.
being an ordinal.
-
2.
being a limit ordinal.
-
3.
being the smallest limit ordinal, the successor of the smallest limit ordinal, the successor of the succcessor of the smallest limit ordinal...
moreover, the properties in 3. are rigid.
Proof
\(\alpha \) is an ordinal if and only if \(\alpha \) is (i) transitive \(\forall x\in \alpha \forall y\in x.y\in \alpha )\) and (ii) linearly ordered by membership (\(\forall x\in \alpha \forall y\in \alpha (x\ne y \rightarrow x\in y\vee y\in x)\). All the quantifiers in these definitions are restricted by conditions of the form \(\in z\), which is rigid by Set Rigidity, and entails sethood (by the definition of Set as \(\lambda y\exists x.y\in x\)). Thus they are all modally absolute.
\(\alpha \) is a limit ordinal if it is an ordinal and additionally \(\forall x\in \alpha \exists y\in \alpha (x\in y)\) and \(\exists x\in \alpha \). These have the same property. \(\alpha \) is the smallest limit ordinal iff it is a limit ordinal, and for every \(x\in \alpha \) x is not a limit ordinal. \(\alpha \) is the successor of the smallest limit ordinal iff every member of \(\alpha \) is either belongs to the smallest limit ordinal or is identical to it. Again, all quantifiers are restricted by membership to some set.
Finally we can show that the properties in 3 are rigid. Let \(\omega \) be the set that is actually the smallest limit ordinal. By the modal absoluteness, \(\omega \) is necessarily the smallest limit ordinal, and uniquely so, since is a theorem of ZF that if two sets are the smallest limit ordinal they are identical. Suppose it is possible that something is the smallest limit ordinal is also F. Then it is possible that \(\omega \) is F, and thus there is an actual smallest limit ordinal, \(\omega \), which is possibly F. Similar strategies apply to the other properties listed in 3. \(\square \)
Theorem 12 \((\textsf{C5}^{\blacksquare \in })\)
-
1.
\(\blacksquare CH \vee \blacksquare \lnot CH\).
-
2.
\(\forall _e x({\text {Uncountable}}x \rightarrow \blacksquare {\text {Uncountable}}x)\)
Proof
Let \(V_{\omega +2}y\) be the property ‘\(\lambda y\).for some set \(\alpha \), \(\alpha \) is the successor of the successor of the smallest limit ordinal, and y is \(V_\alpha \)’. Using the results above, it is easily seen that this property is rigid.
The continuum hypothesis can be formulated in such a way that all quantifiers are restricted by the predicate \(V_{\omega +2}\). Since this predicate is rigid, CH is modally absolute: \(CH \rightarrow \blacksquare CH\) and \(\lnot CH \rightarrow \blacksquare CH\). This establishes 1.
Let x be an uncountable set, and suppose that \(\alpha \) is an ordinal such that \(x\in V_\alpha \). Then the claim that \(x\in V_\alpha \) and is uncountable is equivalent to the claim that \(x\in V_\alpha \) and there is no set of ordered pairs belonging to \(V_{\alpha +3}\) that is an injective function from the smallest limit ordinal to x. All of the quantifiers in this claim are similarly restricted to rigid properties. \(\square \)
Theorem 15 \((\textsf{C}^\blacksquare )\)
Given \(\textsf{LB}^t\), \(\blacklozenge p \leftrightarrow \exists w({\text {World}}w \wedge w\le p \wedge \blacklozenge w)\)
Proof
Mathematical Necessity states that anything entailed by the \(\blacksquare \)-necessities must be itself \(\blacksquare \)-necessary. So any \(\blacklozenge \)-possibility is such that its negation is not entailed by the \(\blacksquare \)-necessities.
Thus if \(\blacklozenge p\), \(\blacksquare \nleq \lnot p\). That is, for some r such that \(\forall q(\blacksquare q \rightarrow r\le q)\), \(r\nleq \lnot p\). This means \(\Diamond (r\wedge p)\), so by \(\textsf{LB}^t\), there is a world proposition w that entail \(r\wedge p\). We finally can see that w must be \(\blacklozenge \)-possible. For if not, then \(\blacksquare \lnot w\), and since r entails every \(\blacksquare \)-necessity, \(r\le \lnot w\). But since \(w\le r\), \(w\le \lnot w\), contradicting the assumption that w is a world.
The right-to-left direction is obvious. \(\square \)
Theorem 16 \((\textsf{C}^{\blacksquare \in })\)
\(\textsf{LB}^{t\rightarrow t}\) and \(\textsf{LB}^t\) imply that \(V_\alpha \) is rigid for every ordinal \(\alpha \).
Proof
The proof is by transfinite induction. \(V_0\) is necessarily empty, and so vacuously rigid.
Suppose that \(\alpha \) is an ordinal, and for each \(\beta \in \alpha \), \(V_\beta \) is rigid. We want to show that \(V_\alpha \) is rigid. Suppose \(\blacklozenge \exists x(V_{\alpha }x \wedge Fx)\). We must show \(\exists x(V_\alpha x \wedge \blacklozenge Fx)\).
Since \(\lambda x(V_{\alpha }x\wedge Fx)\) is broadly possibly instantiated, it follows by the Leibniz Biconditionals, \(\textsf{LB}^{e\rightarrow t}\), that there is a world property W that that entails it, and by Proposition 15 it will be a world property that is mathematically possibly instantiated.Footnote 1 We can use this world property to define the actual member of \(V_{\alpha }\) that’s possibly F explicitly:
Roughly W singles out a merely possible set. \(x'\) is the set of ys in \(V_\alpha \) that would have belonged to the merely possible object picked out by W if it had existed. We can now show that \(x'\) is identical to the merely possible W: i.e. we show \(\blacksquare \forall x(Wx \rightarrow x=x')\). Given the mathematical necessity of Set Extensionality and the mathematical possibility of W it suffices to show that necessarily whatever is W is coextensive with \(x'\): \(\blacksquare \forall x(Wx \rightarrow \forall y.(y\in x \leftrightarrow y\in x')\). We break this up into two claims:
-
1.
\(\blacksquare \forall x(Wx \rightarrow \forall y(y\in x' \rightarrow y\in x)\)
-
2.
\(\blacksquare \forall x(Wx \rightarrow \forall y(y\in x\rightarrow y\in x')\)
We establish 1 first. From the definition of membership in \(x'\), we immediately have \(\forall y\in x'\blacksquare \forall x(Wx \rightarrow y\in x)\). Since Sets are Rigid, it follows that \(x'\)-restricted quantification satisfies BF, so we can infer \(\blacksquare \forall y\in x'\forall x(Wx \rightarrow y\in x)\). By applying first-order logic under the scope of \(\blacksquare \), this is equivalent to 1.
To establish 2, we first show \(\forall \beta \in \alpha \forall y (V_\beta y\rightarrow \blacksquare \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\). Let \(\beta \in \alpha \) and let y be an arbitrary set of rank \(\beta \). Now either \(y\in x'\) or \(y\notin x'\). Suppose the former. Then by the rigidity of set membership y is necessarily in \(x'\) and so \(\blacksquare \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\) follows. Suppose then that \(y\notin x'\). By the condition for belonging to \(x'\), this means that W doesn’t entail the property of containing y. Since W is a world property, it must entail the property of not belonging to y, and thus must also mathematically necessitate it: \(\blacksquare \forall x(Wx \rightarrow y\notin x)\). So this means \(\blacksquare \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\), by applying some straightforward logic under the \(\blacksquare \) (namely that \(y\notin x\) entails \(y\in x\rightarrow y\in x'\)).
This completes the argument that \(\forall \beta \in \alpha \forall y(V_\beta y\rightarrow \blacksquare \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\). By the inductive hypothesis, \(V_\beta \) is rigid, and so we can infer \(\forall \beta \in \alpha \blacksquare \forall y(V_\beta y\rightarrow \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\). Since \(\alpha \) is a set and sets are rigid, we can also infer \(\blacksquare (\forall \beta \in \alpha \forall y(V_\beta y\rightarrow \forall x(Wx \rightarrow (y\in x \rightarrow y\in x'))\). Thus \(\blacksquare \forall x(Wx \rightarrow \forall y(y\in x \rightarrow \exists \beta \in \alpha .V_\beta y\rightarrow y\in x')))\) applying first-order logic under \(\blacksquare \). Recall that necessarily whatever the W set is, it’s \(V_{\alpha }\): thus, necessarily, whatever the W set is, if y belongs to it, y is in \(V_\beta \) for some \(\beta \in \alpha \) (by the definition of \(V_{\alpha }\) ). That is we have (a) \( \blacksquare \forall x(Wx \rightarrow V_{\alpha }x)\), (b) \(\blacksquare \forall x(V_{\alpha }x \wedge y\in x \rightarrow \exists \beta \in \alpha .V_\beta y)\) (by definition of the V relation and the mathematical necessity of ZF). So putting this together \(\blacksquare \forall x(Wx \rightarrow \forall y(y\in x \rightarrow y\in x')))\) as required.
Since W mathematically necessitates being identical to \(x'\) (\(\blacksquare \forall x(Wx \rightarrow x=x')\), and W is mathematically possible, it follows that \(\blacklozenge Wx'\). Finally, since W entails F it follows that \(\blacklozenge Fx'\). By construction \(V_{\alpha }x'\) so \(\exists x(V_{\alpha }x \wedge \blacklozenge F x)\) as required. \(\square \)
Theorem 18 \((\textsf{C}^{\blacksquare \in }\textsf{LB}^\sigma )\)
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1.
\(\blacksquare CH \vee \blacksquare \lnot CH\).
-
2.
\(\forall _e x({\text {Uncountable}}x\rightarrow \blacksquare {\text {Uncountable}}x)\)
Theorem 19 \((\textsf{C})\)
De Re Modal Choice and \(\textsf{LB}^t\) entail \(\textsf{LB}^{\overline{\sigma }\rightarrow t}\).
Proof
We show \(\textsf{LB}^{e\rightarrow t}\), since that is the instance required for Theorem 16, however the proof generalizes trivially.
Suppose that \(\Diamond \exists xFx\). By \(\textsf{LB}^t\), there is a world proposition w such that \(w\le _t \exists xFx\). Let R be a relation which is necessarily a well-order, and consider the property of being the R minimal F while w is true: \(W:=\lambda x(w \wedge {\text {Min}}RFx)\) where \({\text {Min}} = \lambda RFx(Fx \wedge \forall y(Fy \rightarrow Rxy\vee x=y))\). Clearly W entails F. Let G be another property. Since there is at most one minimal F of a well-order, we know that \(\Box ({\text {WO}}R \rightarrow \forall x({\text {Min}}RFx \rightarrow Gx) \vee \forall x({\text {Min}}RFx\rightarrow \lnot Gx))\), and since \(\Box {\text {WO}}R\), \(\Box (\forall x({\text {Min}}RFx \rightarrow Gx) \vee \forall x({\text {Min}}RFx \rightarrow \lnot Gx))\). Since w settles every question it either entails every R-minimal F is G, or that it’s not, \(\Box ( w\rightarrow \forall x({\text {Min}}RFx\rightarrow Gx)) \vee \Box (w\rightarrow \forall x({\text {Min}}RFx\rightarrow \lnot Gx))\). Rearranging a little and appealing to the definition of W this is \(\Box \forall x(Wx\rightarrow Gx) \vee \Box \forall x(Wx\rightarrow \lnot Gx)\) \(\square \)
B Appendix: Free Logic
In this appendix we provide the necessary background for the results discussed in Section 7.
Free logic replaces the law of universal instantiation with its universal closure, \(\forall _\sigma y(\forall _\sigma x\,Fx \rightarrow Fy)\). We must then also add the principle that universal quantification distributes over conditionals. We of course, may apply the analogous substitutions at other types.
Free Instantiation
\(\forall _\sigma y(\forall _\sigma x\,Fx \rightarrow Fy)\) provided y is not free in F.
Quantifier Normality
\(\forall _\sigma x(A\rightarrow B) \rightarrow (\forall _\sigma xA \rightarrow \forall _\sigma xB)\)
The remaining principles of H—Gen, and the laws governing the truth-functional connectives and \(\lambda \)—remain the same. Let FH, ‘free higher-order logic’, be the result of making these substitutions to H, ad Free Classicism, FC, the result of closing FH under the rule of equivalence.
Because the logic of the quantifiers in Free Classicism is weaker than Classicism, notions we defined using the quantification over all necessities—entailment, broad necessity, world, etc—may behave in undesirable ways. For example, a natural quantificational definition of property entailment in Free Classicism, \(\Box \forall _ex\Box (Fx \rightarrow Gx)\), is consistent with pathological situations where F entails G, a is F but a is not G.Footnote 2 However, in Classicism many of the notions that we defined in terms of the classical quantifiers can be given equivalent definitions in terms of identity, and because the logic of identity in Free Classicism is classical, we can recover the desired behaviour by using the identity-theoretic definitions instead. For instance, there is a long tradition in logic, tracing back to George Boole, of defining entailment in terms of identity. For properties F and G, F entails G when the property conjunction of F with G (i.e. \(\lambda x(Fx\wedge Gx)\)) just is F.Footnote 3 The pathological situation mentioned above cannot arise, for if F entails G and then \(F = (\lambda x.Fx\wedge Gx)\). So by Leibniz’s law \(Fa\rightarrow (\lambda x.Fx\wedge Gx)a\), and thus \(Fa\rightarrow Ga\) by \(\beta \) and propositional logic.
Proposition 24 . In Classicism, the following identities are derivable:
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1.
\(\Box = \lambda p.p=_t \top \)
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2.
\(\le _{\overline{\sigma }} = \lambda RS (R\wedge _{\overline{\sigma }} S =_{\overline{\sigma }\rightarrow t} R)\)
-
3.
\({\text {SWorld}} = \lambda w((w\ne _t \bot ) \wedge (\lambda p(w\le p \vee w\le \lnot p) =_{t\rightarrow t} \lambda p.\top ))\)
Proofs of 1 and 2 may be found in Bacon [6] p149. The first conjunct of the RHS of 3, \(w\ne _t \bot \), is equivalent to \(\Diamond w\) by 1, and the second conjunct is equivalent to \(\Box \forall _tp(\top \rightarrow (w\le p\vee w\le \lnot p))\) by 2, and thus to \(\Box \forall _tp(w\le p\vee w\le \lnot p)\).
Perhaps it is possible to augment Free Classicism with further principles that would rule out these pathological situations, but we will avoid the need for any further assumptions by adopting the identity theoretic definitions of these three notions listed in Proposition 24 as our official ones when working in Free Classicism.
We can now define a possibilist quantifier along the lines of Fine’s definition discussed in Section 3:
where F has type \(\sigma \rightarrow t\) and x has type \(\sigma \). \(\Pi _e F\) means that, when w is the true strong world proposition, the vacuous property of being such that w entails F. We have replaced Fine’s \(\Box \forall _\sigma x\Box (w \rightarrow Fx)\)—the potentially ill-behaved notion of entailment mentioned above—with the corresponding identity theoretic entailment for reasons detailed above.
We are now in a position to formulate the orthodox possible worlds metaphysics within Free Classicism. We can do this by adding to FC the Strong Leibniz Conditionals and the B schema and closing under the rule of equivalence as well as the background logical rules, remembering, of course, that \({\text {SWorld}}\), \(\le \), \(\Diamond \), etc are now given in identity theoretic terms.
- \({\textsf{SLB}}^t\):
-
\(\Diamond A \leftrightarrow \exists _t w({\text {SWorld}}w \wedge w\le A)\)
- \(\textsf{B}\):
-
\(A\rightarrow \Box \Diamond A\)
We will call the result FC5(SLB). Note that because necessitated quantificational claims are weak in Free Classicism, merely adding the necessitations of the universal closures of these principles to Free Classicism would fail to deliver identities that one could obtain from the result of closing under the rule of equivalence. We could acheive the same effect as closing under the rule of equivalence by adding a pair of identities to Free Classicism. The claim that to be possible is to be true at some possible world, and the claim that to be true entails to be necessarily possible.
- \(\textsf{SLB}^t_\lambda \):
-
\(\Diamond =_{t\rightarrow t} \lambda p(\exists _t w({\text {SWorld}}w \wedge w\le p)\)
- \(\textsf{B}_\lambda \):
-
\(\lambda p.p \le _{t\rightarrow t} \lambda p.\Box \Diamond p\)
FC5 and FC(SLB) stand for the result adding, in the same way, only one of these principles.
Lemma 25 . \(\mathsf {FC5(SLB)}\) contains \(A\rightarrow \exists w({\text {SWorld}}w \wedge w \wedge w\le _t A)\).
Proof
First we show \({\text {SWorld}}w \rightarrow \Box {\text {SWorld}}w \wedge \Box (\exists _t p.w=p)\). Since \({\text {SWorld}}w\) is the conjunction of a distinctness claim and an identity claim, the necessity of the first conjunct follows from the necessity of distinctness and the necessity of identity both of which are well-known theorems of S5 with the classical axioms of identity.Footnote 4 Using SLB, and the fact that w is necessarily possible, \(\Box \exists _t v({\text {SWorld}}v \wedge v\le w)\). It’s also necessary that for any strong world \(v\le w\), \(w\le v\). For w is necessarily a strong world, and so must entail v or \(\lnot v\) for any strong world \(v\le w\), and it couldn’t entail \(\lnot v\) since otherwise \(v\le \lnot v\) by the transitivity of entailment, contradicting the fact that v is possible. So necessarily, any strong world entailing w is identical to w, thus \(\Box \exists _t v({\text {SWorld}}v \wedge v=_tw)\).
Now we argue that every strong world, w, entails (i) w, (ii) that w is an existent strong world, and (iii) \(A\rightarrow (w\le A)\). (i) is trivial, (ii) is established above. For (iii), \(\lambda p.w \le _{t\rightarrow t} \lambda p.(w \wedge (p \rightarrow (\lambda p\top )p)\) since \(p\rightarrow \top \) is a tautology. And since \(\lambda p.\top =_{t\rightarrow t} \lambda p(w\le p \vee w\le \lnot p)\) (since w is a strong world) we have \(\lambda p.w \le \lambda p(w \wedge (p \rightarrow (w\le p \vee w\le \lnot p)))\). We also have \(\lambda p.w \le \lambda p(w \wedge p \rightarrow w\nleq \lnot p)\) since \(w \wedge p \rightarrow w\nleq \lnot p\) is a theorem of Free Classicism. Since operator entailment is closed under propositional logic, \(\lambda p.w\le \lambda p(p \rightarrow w\le p)\). Apply both these operators to A and using \(\beta \) we get w and \(A\rightarrow w\le A\), and since the former operator entails the latter, \(w \le (A\rightarrow w \le A)\).
Putting (i),(ii) and (iii) together, we have that for every strong world, w, \(w \le (w \wedge {\text {SWorld}}w \exists _t p(p=w) \wedge (A\rightarrow w \le A))\). Using the fact that entailment is closed under free logic we get \(w \le (A \rightarrow \exists _t w(w \wedge {\text {SWorld}}w \wedge w \le A)))\). Since every strong world entails \(A \rightarrow \exists _t w(w \wedge {\text {SWorld}}w \wedge w \le A))\) we can infer \(\Box (A \rightarrow \exists _t w(w \wedge {\text {SWorld}}w \wedge w \le A)))\) by SLB. \(\square \)
Theorem 26 . \(\mathsf {FC5(SLB)}\) interprets \(\mathsf {C5(SLB)}\)
Proof
We map each term M of \(\mathcal {L}\) to \(M^*\), the result of substituting each free quantifier \(\forall _\sigma \) with \(\Pi _\sigma \). We wish to show that whenever A is a theorem of Classicism, \(A^*\) is a theorem of Free Classicism\(^+\).
Each tautology, instance of B, and instance of \(\beta \eta \) are mapped to tautologies instances of B or instances of \(\beta \eta \). Uses of modus ponens and the rule of equivalence are similarly mapped to themselves. It remains to show that UI\(^*\) and SLB\(^*\) are theorems of \(\mathsf {FC5(SLB)}\), and, for Gen, that if \((A\rightarrow B)^*\) is a theorem of Free Classicism\(^+\), so is \((A\rightarrow \forall xB)^*\).
Let’s begin with UI. We will show generally that \(\Pi _\sigma F \rightarrow Fa\). Suppose \(\Pi _\sigma F\), so that there is some truth, p, such that \(\lambda x(p \wedge Fx) =_{\sigma \rightarrow t} \lambda x.p\). Want to show Fa. \((\lambda x.p)a =_t p\) by \(\beta \), and since p is true, we can conclude \((\lambda x.p)a\). By the above identity, \(\lambda x(p \wedge Fx)a\), so \(p \wedge Fa\), and finally, Fa as required.
For the right-to-left direction of SLB\(^*\) we show the dualized contrapositive version. We will suppose that \(\Pi _t w({\text {SWorld}}w \rightarrow w\le A)\) and show \(\Box A\). Expanding the definition of \(\Pi \), the true strong world, v, is such that \(\lambda w.v \le \lambda w.({\text {SWorld}}w \rightarrow w\le A)\). Applying \(\forall _t\) to both sides we see that the claim that everything is such that v (i.e. \(\forall _t p.v)\)) entails that every strong world is entails A (i.e \(\forall _tw({\text {SWorld}}w \rightarrow w\le A)\)). Since v is true, everything is such that v, and so every strong world entails A. By SLB, \(\Box A\). For the converse of SLB\(^*\) suppose \(\Sigma _t w({\text {SWorld}}w \wedge w\le A)\)—i.e. \(\lambda w.v \nleq \lambda w(Sw \rightarrow w \nleq A)\) where v is a true strong world. We want to show \(\Diamond A\). It suffices to show \(\exists _t u({\text {SWorld}}u \wedge u\le A)\). Suppose for contradiction that \(\forall _t u({\text {SWorld}}u \rightarrow u\nleq A)\). By lemma there is a strong world v that is true and entails \(\forall _t u({\text {SWorld}}u \rightarrow u\nleq A)\), delivering also the corresponding entailment between vacuous operators: \(\lambda w.v \le \lambda w \forall u({\text {SWorld}}u \rightarrow u \nleq A)\). Since being a strong world entails existence, we have \(\lambda w.v \le \lambda w({\text {SWorld}}w \rightarrow \exists _tr.r=w)\). Since the right-hand-sides of entailments are closed under free logical consequences, we have \(\lambda w.v \le \lambda w({\text {SWorld}}w \wedge \exists _tr.r=w \rightarrow w\nleq A)\) and so \(\lambda w.v \le \lambda w({\text {SWorld}}w \rightarrow w\nleq A)\). This contradicts our assumption.
For Gen it suffices to show that whenever we have a proof of \(A\rightarrow B\) where x is not free in B there is also a proof of \(A\rightarrow \Pi _\sigma xB\). Since we can prove \(A\rightarrow B\), we can prove \((\lambda x(A\rightarrow B))y \leftrightarrow (\lambda x.\top )y\) using \(\beta \) and so by the rule of equivalence we then have \(\lambda x(A\rightarrow B) = \lambda x.\top \).
Now we will show that \(A\rightarrow \exists w(w \wedge {\text {SWorld}}w \wedge \lambda x.w \le \lambda x.B\). Suppose A, and let w be the true strong world entailing A (appealing to Lemma 25). So \(w\wedge \lnot A =_t \bot \). Clearly \(\lambda x.w\le \lambda x(A\rightarrow B)\) since \(\lambda x.w \le \lambda x.\top \).
\(\lambda x(w\wedge (A \rightarrow B)) =_{\sigma \rightarrow t} \lambda x.w\). The left-hand-side is \(\lambda x.((w\wedge \lnot A) \vee (w\wedge B))\) using Boolean equivalences that can be obtained from the Rule of Equivalence. Since x isn’t free in A and \(w\wedge \lnot A = \bot \) we can infer the the left-hand-side is \(\lambda x.(w\wedge B)\) by Leibniz’s law and Boolean equivalences. So \(\lambda x(w\wedge B) = \lambda x.w\) as required. \(\square \)
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Bacon, A. Mathematical Modality: An Investigation in Higher-order Logic. J Philos Logic 53, 131–179 (2024). https://doi.org/10.1007/s10992-023-09728-1
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DOI: https://doi.org/10.1007/s10992-023-09728-1