Abstract
Most ordinary objects - cats, humans, mountains, ships, tables, etc. - have indeterminate mereological boundaries. If the theory of mereology is meant to include ordinary objects at all, we need it to have some space for mereological indeterminacy. In this paper, we present a novel degree-theoretic semantics - Boolean semantics - and argue that it is the best degree-theoretic semantics for modeling mereological indeterminacy, for three main reasons: (a) it allows for incomparable degrees of parthood, (b) it enforces classical logic, and (c) it is compatible with all the axioms of classical mereology. Using Boolean semantics, we will also investigate the connection between vagueness in parthood and vagueness in existence/identity. We show that, contrary to what many have argued, the connection takes neither the form of entailment nor the form of exclusion.
Article PDF
Similar content being viewed by others
Avoid common mistakes on your manuscript.
Notes
Our theory can be easily generalized to first order languages with function symbols, as functions can always be treated as relations that satisfy special conditions.
See [29] for a detailed model theory on Boolean-valued models.
Here and in the following, when the context is clear, we use \(\llbracket a_{i} = a_{j} \rrbracket ^{\mathfrak {A}}\) to abbreviate \(\llbracket = \rrbracket ^{\mathfrak {A}}(a_{i}, a_{j})\), and similarly for cases of the relation symbols.
In the case when \({\mathscr{L}}\) is a first-order language, this line can simply be ignored, for obvious reasons. And similarly for 2(c), 3(f) and 3(g) below.
We omit the superscripts when the context is clear.
References
Akiba, K. (2004). Vagueness in the world. Noû,s, 38(3), 407–429.
Barnes, E., & Williams, J.R.G. (2009). Vague parts and vague identity. Pacific Philosophical Quarterly, 90(2), 176–187.
Bell, J.L. (2011). Set theory: boolean-valued models and independence proofs. Oxford University Press.
Calosi, C., & Wilson, J. (2019). Quantum metaphysical indeterminacy. Philosophical Studies, 176(10), 2599–2627.
Cook, M. (1986). Indeterminacy of identity. Analysis, 46(4), 179–186.
Donnelly, M. (2009). Mereological vagueness and existential vagueness. Synthese, 168(1), 53–79.
Eklund, M. (2008). Deconstructing ontological vagueness. Canadian Journal of Philosophy, 38(1), 117–140.
Evans, G. (1978). Can there be vague objects. Analysis, 38(4), 208.
Fine, K. (1975). Vagueness, truth and logic. Synthese:265–300.
Giraud, T. (2013). An abstract mereology for meinongian objects. Journal of Philosophical Studies, 6(25), 177–210.
Gruszczyński, R., & Pietruszczak, A (2008). Full development of Tarski’s geometry of solids. Bulletin of Symbolic Logic, 14(4), 481–540.
Hovda, P. (2009). What is classical mereology? Journal of Philosophical Logic, 38(1), 55–82.
Keefe, R. (2000). Theories of vagueness. Cambridge University Press.
Lewis, D. (1986). On the plurality of worlds. Blackwell Oxford.
Mariani, C., Michels, R., & Torrengo, G. (2021). Plural metaphysical supervaluationism. Inquiry:1–38.
McGee, V., & McLaughlin, B. (1995). Distinctions without a difference. Southern Journal Philosophy, 33(Supplement), 203–251.
McGee, V., & McLaughlin, B. (2000). The lessons of the many. Philosophical Topics, 28(1), 129–151.
McGee, V., & McLaughlin, B. (2022). Terrestrial logic. Oxford University Press.
Merricks, T. (2005). Composition and vagueness. Mind, 114(455), 615–637.
Morreau, M. (2002). What vague objects are like. Journal Philosophy, 99(7), 333–361.
Priest, G. (2014). Much ado about nothing. Australasian Journal Logic, vol 11(2).
Rasiowa, H. (1963). The mathematics of metamathematics.
Sainsbury, R. (1989). What is a vague object? Analysis, 49(3), 99–103.
Smith, N.J. (2005). A plea for things that are not quite all there: or, is there a problem about vague composition and vague existence? Journal Philosophy, 102(8), 381–421.
Tarski, A. (1956). Foundations of the geometry of solids. Logic, Semantics, Metamathematics, 52, 24–29.
Van Inwagen, P. (1990). Material beings. Cornell University Press.
Weatherson, B. (2003). Many many problems. Philosophical Quarterly, 53 (213), 481–501.
Williamson, T. (2002). Vagueness. Routledge.
Wu, X. (2022). Boolean-valued models and their applications. PhD thesis, Massachusetts Institute of Technology.
Acknowledgments
For their helpful feedback on earlier drafts of this paper, I would like to thank Vann McGee, Agustín Rayo and Stephen Yablo. I would also like to thank the editor and two anonymous referees of this journal. Additionally, I am grateful to the participants of the WIP seminar at MIT, especially to Jack Spencer.
Funding
Open Access funding provided by the MIT Libraries
Author information
Authors and Affiliations
Corresponding author
Additional information
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Appendices
Appendix A: Preliminaries on Boolean Model Theory
Definition A.1
Let \({\mathscr{L}}\) be an arbitrary first-order/second-order language. For simplicity, we assume that \({\mathscr{L}}\) has no function symbols/variables, but only relation symbols/variables, individual constants/variables.Footnote 1 Let B be a complete Boolean algebra. A B-valued modelFootnote 2\(\mathfrak {A}\) for the language \({\mathscr{L}}\) consists of:
-
1.
A universe A of elements;
-
2.
The B-value of the identity symbol: a function \(\llbracket = \rrbracket ^{\mathfrak {A}}: A^{2} \rightarrow B\);
-
3.
The B-values of the relation symbols: (let P be a n-ary relation) \(\llbracket P \rrbracket ^{\mathfrak {A}}: A^{n} \rightarrow B\);
-
4.
The B-values of the constant symbols: (let c be a constant) \(\llbracket c \rrbracket ^{\mathfrak {A}} \in A\).
And it needs to satisfy:
-
1.
For the B-value of the identity symbolFootnote 3: for any a1,a2,a3 ∈A
$$ \begin{array}{@{}rcl@{}} \llbracket a_{1} = a_{1} \rrbracket^{\mathfrak{A}} = 1_{B} \end{array} $$(1)$$ \begin{array}{@{}rcl@{}} \llbracket a_{1} = a_{2} \rrbracket^{\mathfrak{A}} = \llbracket a_{2} = a_{1} \rrbracket^{\mathfrak{A}} \end{array} $$(2)$$ \begin{array}{@{}rcl@{}} \llbracket a_{1} = a_{2} \rrbracket^{\mathfrak{A}} \sqcap \llbracket a_{2} = a_{3} \rrbracket^{\mathfrak{A}} \leqslant \llbracket a_{1} = a_{3} \rrbracket^{\mathfrak{A}} \end{array} $$(3) -
2.
For the B-value of relation symbols: let P be an n-ary relation; for any 〈a1,…,an〉,〈b1,…,bn〉∈An,
$$ \begin{array}{@{}rcl@{}} \llbracket P(a_{1},\ldots, a_{n}) \rrbracket^{\mathfrak{A}} \sqcap (\bigsqcap\limits_{1 \leqslant i \leqslant n}\llbracket a_{i} = b_{i} \rrbracket^{\mathfrak{A}}) \leqslant \llbracket P(b_{1},\ldots, b_{n}) \rrbracket^{\mathfrak{A}} \end{array} $$(4)
Definition A.2
Let \(\mathfrak {A}\) be a B-valued model of \({\mathscr{L}}\). For any n ∈ω, we define \({D^{n}_{A}}\) as the following set: \({D^{n}_{A}} = \{R: A^{n} \rightarrow B | \text {for any} \langle a_{1},\ldots , a_{n} \rangle , \langle b_{1},\ldots , b_{n} \rangle \in A^{n}, R(a_{1},\ldots , a_{n}) \sqcap (\bigsqcap _{1 \leqslant i \leqslant n}\llbracket a_{i} = b_{i} \rrbracket ^{\mathfrak {A}}) \leqslant R(b_{1},\ldots , b_{n}) \}\). We call the \({D^{n}_{A}}\)’s the second-order domains of \(\mathfrak {A}\). For each n ∈ω, we call \({D^{n}_{A}}\) the n-ary second-order domain of \(\mathfrak {A}\).
Given a B-valued model \(\mathfrak {A}\) for \({\mathscr{L}}\), we define satisfaction in \(\mathfrak {A}\) as follows. Let Var be the set of all variables. An assignment s on \(\mathfrak {A}\) is a function with domain Var such that (1) for any individual variable vi, s(vi) ∈A, and (2) for any relation variable Xi of arity n, \(s(X_{i}) \in {D^{n}_{A}}\).Footnote 4 We define the value of a term/an atomic open formula in \(\mathfrak {A}\) under assignment s in the standard way. For complex formulas,
-
1.
\(\llbracket \neg \phi \rrbracket ^{\mathfrak {A}}[s] = - \llbracket \phi \rrbracket ^{\mathfrak {A}}[s]\).
-
2.
\(\llbracket \phi \land \psi \rrbracket ^{\mathfrak {A}}[s] = \llbracket \phi \rrbracket ^{\mathfrak {A}}[s] \sqcap \llbracket \psi \rrbracket ^{\mathfrak {A}}[s]\).
-
3.
\(\llbracket \exists v_{i} \phi \rrbracket ^{\mathfrak {A}}[s] = \bigsqcup _{a \in A} \llbracket \phi \rrbracket ^{\mathfrak {A}}[s(v_{i}/a)]\).
-
4.
\(\llbracket \exists X_{i} \phi \rrbracket ^{\mathfrak {A}}[s] = \bigsqcup _{R \in {D^{n}_{A}}} \llbracket \phi \rrbracket ^{\mathfrak {A}}[s(X_{i}/R)]\).
The values of the quantified formulas are well-defined as B is complete. We say that ϕ is a first-order formula when ϕ has no second order variables.
Theorem A.1
Let \(\mathfrak {A}\) be a B-valued model for \({\mathscr{L}}\). For any formula ϕ(v1,…,vn) in \({\mathscr{L}}\), any assignments \(s, s^{\prime }\) on \(\mathfrak {A}\),
Proof
By a straightforward induction on the complexity of ϕ(v1,…,vn). □
Appendix B: Soundness and Completeness
Definition B.1
Let T be a theory in a language \({\mathscr{L}}\). Let \(\mathfrak {A}\) be a B-valued model of \({\mathscr{L}}\). \(\mathfrak {A}\) is a model of T just in case for any ϕ ∈T, \(\llbracket \phi \rrbracket ^{\mathfrak {A}} = 1_{B}\).
Definition B.2
Let T be a theory and ϕ be a sentence in a language \({\mathscr{L}}\). ϕ is a Boolean-consequence of T, in symbols, T⊧Bϕ just in case for any Boolean valued model \(\mathfrak {A}\), if \(\mathfrak {A}\) is a model of T, then \(\mathfrak {A}\) is a model of ϕ.
In the rest of this section we assume that \({\mathscr{L}}\) is a first-order language.
Theorem B.1
Let T be a theory and ϕ be a sentence in \({\mathscr{L}}\). If T ⊩ϕ, then T⊧Bϕ.
Proof
See [22] or [29] for a detailed proof. □
Corollary B.1.1
Let ϕ be a theorem of first order logic. Then in any Boolean valued model \(\mathfrak {A}\), \(\llbracket \phi \rrbracket ^{\mathfrak {A}} = 1\).
Theorem B.2
Let T be a theory in \({\mathscr{L}}\). T is consistent if and only if for some complete Boolean Algebra B, T has a B-valued model \(\mathfrak {A}\).
Proof
See [22] or [29] for a detailed proof. □
Corollary B.2.1
Let B be any complete Boolean algebra. A theory T has a B-valued model just in case every finite subset of T has a B-valued model.
Theorem B.3
Let T be a theory and ϕ be a sentence in a first order language \({\mathscr{L}}\). If T⊧Bϕ, then T ⊩ϕ.
Corollary B.3.1
Let T be a theory and ϕ be a sentence in a first order language \({\mathscr{L}}\). T⊧Bϕ if and only if T ⊩ϕ.
Appendix C: Equivalence Between Systems
In this section we prove the two promised theorems in Section Five (Theorem 5.1 and Theorem 5.2).
Theorem C.1
CM is equivalent to Tarski’s system, which is the theory closed under the following two axioms:
Proof
We first show that CM entails Tarski’s system. (Transitivity) is already in CM. For (UniqueFusionExistence), let X1 be such that ∃v1X1(v1). By (Fusion), ∃v2(FU(v2,X1)). Let \(v_{3} \leqslant v_{2}\). If E(v3), then we are done. Suppose ¬E(v3). By (NoZero), \(\forall v_{4} \forall v_{5} (v_{4} \precsim v_{5})\). Hence trivially \(v_{3} \precsim v_{3}\) and \(v_{3} \precsim v_{1}\).
For the other direction, we can just use the standard argument that these axioms are all theorems of Tarski’s system. See, for example, [12]. □
Theorem C.2
The (second-order) theory of complete Boolean algebra (CBA) is equivalent to MCM plus Anti-symmetry plus the following axiom:
Proof
For the direction that the latter system entails CBA, it suffices to show that (Reflexivity), (SupremumExistence), (Complementation) and (Distribution) are all theorems of the latter system. See [12] for proofs. For the other direction, the only axiom worth mentioning is (Fusion). We will show that if Sup(v1,X1), then FU(v1,X1). That v1 is a upper bound of X1 is obviously the case. We only need to show that \(\forall v_{2} (v_{2} \precsim v_{1} \land E(v_{2}) \rightarrow \exists v_{3} (X_{1}(v_{3}) \land v_{3} \circ v_{2}))\). Suppose the antecedent. Assume for reductio that \(\forall v_{4} (X(v_{4}) \rightarrow v_{2} \sqcap v_{4} = 0)\). Then by infinite distribution, v2 ⊓v1 = 0. Since E(v2), v2 ≠ 0. Hence . Contradiction. □
Appendix D: The SE Models
In this section we prove the following result:
Theorem D.1
In any SE model, Transitivity, Supplementation, Fusion, Atomicity and NoZero all have value 1.
Theorem D.2 (Transitivity)
\(\mathfrak {S}^{B}_{S} \models \forall v_{1} \forall v_{2} \forall v_{3} (v_{1} \precsim v_{2} \land v_{2} \precsim v_{3} \rightarrow v_{1} \precsim v_{3})\).
Lemma D.2.1
For any f ∈M, \(\llbracket E(f) \rrbracket = \bigsqcup _{a \in S} f(a) = 1\).Footnote 5
Proof
\(\llbracket E(f) \rrbracket =\llbracket \exists v_{2} (\neg f \precsim v_{2}) \rrbracket = \bigsqcup _{g \in M} \bigsqcup _{a \in S} f(a) \sqcap - g(a)\). We want to show that \(\bigsqcup _{g \in M} \bigsqcup _{a \in S} f(a) \sqcap - g(a) = \bigsqcup _{a \in S} f(a)\). For any a ∈S, let ga be the function from S to B that takes a to 1 and every b ≠ a to 0. Obviously ga ∈M. Pick some a ∈S, then it is easy to see for any b ≠ a ∈S, \(f(a) \leqslant \bigsqcup _{c \in S} f(c) \sqcap - g^{b}(c)\). Hence \(f(a) \leqslant \llbracket E(f) \rrbracket \). For the other direction, pick some g ∈M. Obviously \(\bigsqcup _{a \in S} f(a) \sqcap - g(a) \leqslant \bigsqcup _{a \in S} f(a)\). □
Lemma D.2.2
For any f1,f2 ∈M, \(\llbracket f_{1} \circ f_{2} \rrbracket = \bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a)\).
Proof
By definition, \(\llbracket f_{1} \circ f_{2} \rrbracket = \llbracket \exists v_{3} (E(v_{3}) \land v_{3} \precsim f_{1} \land v_{3} \precsim f_{2}) \rrbracket \). Since every g ∈M is such that ⟦E(g)⟧ = 1, \(\llbracket f_{1} \circ f_{2} \rrbracket = \bigsqcup _{g \in M} \llbracket g \precsim f_{1} \rrbracket \sqcap \llbracket g \precsim f_{2} \rrbracket = \bigsqcup _{g \in M} \bigsqcap _{a \in S} g(a) \Rightarrow (f_{1}(a) \sqcap f_{2}(a))\). We will show that this is equal to \(\bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a) = p\).
For the \(\leqslant \) direction: Fix g ∈M. Since \(\bigsqcup _{a \in S} g(a) = 1\), \(\bigsqcap _{a \in S} g(a) \Rightarrow (f_{1}(a) \sqcap f_{2}(a)) = \bigsqcap _{a \in S} -g(a) \sqcup (f_{1}(a) \sqcap f_{2}(a)) \leqslant \bigsqcap _{a \in S} -g(a) \sqcup p = 0 \sqcup p = p\).
For the \(\geqslant \) direction: Fix a ∈S. Then it is easy to see that \(f_{1}(a) = \llbracket g^{a} \precsim f_{1} \rrbracket \), and similarly \(f_{2}(a) = \llbracket g^{a} \precsim f_{2} \rrbracket \). Hence \(f_{1}(a) \sqcap f_{1}(a) \leqslant \llbracket f_{1} \circ f_{2} \rrbracket \). □
Theorem D.3 (Supplementation)
.
Proof
Let f1,f2 ∈M. Since every g ∈M is such that ⟦E(g)⟧ = 1, we just need to show that \(-\llbracket f_{2} \precsim f_{1} \llbracket \leqslant \bigsqcup _{g \in M} \llbracket g \precsim f_{2} \rrbracket \sqcap -\llbracket g \circ f_{1} \rrbracket \). \(-\llbracket f_{2} \precsim f_{1} \llbracket = \bigsqcup _{a \in S} f_{2}(a) \sqcap -f_{1}(a)\). Fix some a ∈S. \(\llbracket g^{a} \precsim f_{1} \rrbracket = f_{2}(a)\). By the previous lemma, \(-\llbracket g^{a} \circ f_{1} \rrbracket = -(\bigsqcup _{b \in S} g^{a}(b) \sqcap f_{1}(b)) = -f_{1}(b)\). □
Theorem D.4 (Fusion)
\(\mathfrak {S}^{B}_{S} \models \forall X_{1} (\exists v_{1} X_{1}(v_{1}) \rightarrow \exists v_{2} (FU(v_{2}, X_{1}))\).
Proof
We will show that for any \(R \in {D^{1}_{M}}\), \(\llbracket \exists v_{1} R(v_{1}) \rightarrow \exists v_{2} (FU(v_{2}, R)) \rrbracket = 1\). That is, \(q = \bigsqcup _{t \in M} R(t) \leqslant \llbracket \exists v_{1} (\forall v_{2} (R(v_{2}) \rightarrow v_{2} \precsim v_{1}) \land \forall v_{3} (v_{3} \precsim v_{1} \land E(v_{3}) \rightarrow \exists v_{4} (R(v_{4}) \land v_{3} \circ v_{4}))) \rrbracket = \bigsqcup _{f \in M} ((\bigsqcap _{g \in M} R(g) \Rightarrow \llbracket g \precsim f \rrbracket ) \sqcap (\bigsqcap _{h \in M} (\llbracket h \precsim f \rrbracket \Rightarrow (\bigsqcup _{s \in M} R(s) \sqcap \llbracket h \circ s \rrbracket ))))\).
We define fR ∈M as follows: pick some particular a ∈S, let \(f^{R}(a) = (\bigsqcup _{g \in M} R(g) \sqcap g(a)) \sqcup -q\). For any b ≠ a ∈S, let \(f^{R}(b) = \bigsqcup _{g \in M} R(g) \sqcap g(b)\).
We first show that fR is indeed in M, i.e. \(\bigsqcup _{c \in S} f^{R}(c) = 1\):
Now we show that \(\bigsqcap _{g \in M} R(g) \Rightarrow \llbracket g \precsim f^{R} \rrbracket = \bigsqcap _{g \in M} R(g) \Rightarrow (\bigsqcap _{c \in S} g(c) \Rightarrow f^{R}(c)) = 1\). Pick any g ∈M. R(g) ⇒(⊓c∈Sg(c) ⇒fR(c)) = −R(g)⊔((⊓c≠a−g(c)⊔fR(c))⊓(−g(a)⊔fR(a))) = (⊓c≠a−R(g)⊔−g(c)⊔fR(c))⊓(−R(g)⊔−g(a)⊔fR(a)). \(\bigsqcap _{c \neq a} -R(g) \sqcup -g(c) \sqcup f^{R}(c) = \bigsqcap _{c \neq a} -R(g) \sqcup -g(c) \sqcup (\bigsqcup _{h \in M} R(h) \sqcap h(c)) \geqslant \bigsqcap _{c \neq a} -R(g) \sqcup -g(c) \sqcup (-R(g) \sqcap g(c)) = 1\). \(-R(g) \sqcup -g(a) \sqcup f^{R}(a) = -R(g) \sqcup -g(a) \sqcup (\bigsqcup _{h \in M} R(h) \sqcap h(c)) \sqcup -q = 1\).
We next show that \(q \leqslant \bigsqcap _{h \in M} (\llbracket h \precsim f^{R} \rrbracket \Rightarrow (\bigsqcup _{s \in M} R(s) \sqcap \llbracket h \circ s \rrbracket ))\). Fix any h ∈M. We want to show that \(q \leqslant (\bigsqcup _{c \in S} h(c) \sqcap -f^{R}(c)) \sqcup (\bigsqcup _{d\in S}\bigsqcup _{s \in M} R(s) \sqcap s(d) \sqcap h(d)) = p\). Now it is easy to see that p = p1 ⊔p2, where \(p_{1} = (\bigsqcup _{c \neq a} h(c) \sqcap -f^{R}(c)) \sqcup (\bigsqcup _{d \neq a}\bigsqcup _{s \in M} R(s) \sqcap s(d) \sqcap h(d))\) and \(p_{2} = (h(a) \sqcap -f^{R}(a)) \sqcup (\bigsqcup _{s \in M} R(s) \sqcap s(a) \sqcap h(a))\). But \(p_{1} = (\bigsqcup _{c \neq a} h(c) \sqcap -f^{R}(c)) \sqcup (\bigsqcup _{d \neq a} f^{R}(d) \sqcap h(d)) = \bigsqcup _{c \neq a} (h(c) \sqcap -f^{R}(c)) \sqcup (f^{R}(c) \sqcap h(c)) = \bigsqcup _{c \neq a} h(c) \geqslant -h(a)\), as \(\bigsqcup _{b \in S} h(b) = 1\).
On the other hand, let \(\bigsqcup _{s \in M} R(s) \sqcap s(a) = p_{3}\). Then p2 = (h(a) ⊓−fR(a)) ⊔(p3 ⊓h(a)) = (h(a) ⊓−p3 ⊓q) ⊔(p3 ⊓h(a)) = (h(a) ⊓q) ⊔(h(a) ⊓p3). Hence \(p = p_{1} \sqcup p_{2} \geqslant -h(a) \sqcup (h(a) \sqcap q) \sqcup (h(a) \sqcap p_{3}) \geqslant q\). □
To prove Atomicity we need some more lemmas.
Lemma D.4.1
Let f ∈M. \(\mathfrak {S}^{B}_{S} \models \forall v (E(v) \rightarrow \neg v \precnsim f)\) just in case {f(a)|a ∈S} is an antichain in B.
Proof
Right to left direction. Let f ∈M be such that {f(a)|a ∈S} is an antichain. Fix some random g ∈M. We will show that \(\llbracket E(g) \rightarrow \neg g \precnsim f \rrbracket = 1\). That is,
Fix some random a ∈S. It is easy to see that
Since {f(a)|a ∈S} is an antichain, \(f(a) \leqslant (\bigsqcap _{c \in S\setminus \{a\}} -f(c))\). Hence, \(g(a) \sqcap f(a) \leqslant \bigsqcap _{c \in S} g(c) \sqcup -f(c)\). Therefore,
Left to right direction. Let f ∈M be such that for some a,b ∈S, f(a) ⊓f(b) > 0. Define g ∈M as follows: for any c ∈S,
It is easy to see that ⟦E(g)⟧ =⟧E(f)⟧. And hence g is indeed in M. We will show that \(\llbracket E(g) \rightarrow \neg g \precnsim f \rrbracket < 1\). That is,
Observe that ⊓a∈S −g(a) = 0, as g ∈M. Also \(\bigsqcup _{b \in S} g(b) \sqcap -f(b) = 0\). And ⊓c∈Sg(c) ⊔−f(c) = g(a) ⊔−f(a) = (f(a) ⊓−f(b)) ⊔−f(a) = −f(a) ⊔−f(b) < 1, as f(a) ⊓f(b) > 0. Hence the whole thing is less than 1. □
Lemma D.4.2
Let f ∈M. \(\mathfrak {S}^{S}_{A} \models At(f)\) just in case {f(a)|a ∈S} is a maximal antichain in B.
Proof
Recall that \(At(f) = E(f) \land \forall v (E(v) \rightarrow \neg v \precnsim f)\). The result follows from the previous lemma as for any f ∈M, ⟦E(f)⟧ = 1. □
Theorem D.5 (Atomicity)
\(\mathfrak {S}^{B}_{S} \models \forall v_{1} (E(v_{1}) \rightarrow \exists v_{2} (At(v_{2}) \land v_{2} \precsim v_{1}))\).
Proof
Fix some random f ∈M. Since ⟦E(f)⟧ = 1, we need to show that \(\llbracket \exists v_{2} (At(v_{2}) \land v_{2} \precsim f) \rrbracket = 1\). Let C = {a ∈S|f(a) ≠ 0}. Enumerate C by α = |C|: C = {a1,…,aβ,...|β < α}. Define g ∈M as follows: for any c ∈S,
Hence \(\llbracket E(g) \rrbracket = \bigsqcup _{a \in C} g(a) = \bigsqcup _{a \in C} f(a) = \llbracket E(f) \rrbracket = 1\). Also, \(\llbracket g \precsim f \rrbracket = 1\). Since {g(a)|a ∈S} is an antichain, by Lemma D.4.1, \(\llbracket \forall v (E(v) \rightarrow \neg v \precnsim g) \rrbracket = 1\). Hence ⟦At(g)⟧ = ⟦E(f)⟧ = 1. □
Theorem D.6 (NoZero)
Proof
This can be proven simply by showing that ⟦¬∃v3¬(E(v3))⟧ = 1, as for any f ∈M, ⟦E(f)⟧ = 1. □
Corollary D.6.1
\(\mathfrak {S}^{B}_{S}\) is a model of ACM−.
Appendix E: The VE Models
In this section we prove the following result:
Theorem E.1
In any VE model, Transitivity, Supplementation, Fusion and Atomicity all have value 1, but NoZero has value 0.
Transitivity is proven in the same way as before.
Lemma E.1.1
For any f ∈N, \(\llbracket E(f) \rrbracket = \bigsqcup _{a \in S}f(a)\).
Proof
The same proof as in that of Lemma D.2.1. □
Lemma E.1.2
For any f1,f2 ∈N, \(\llbracket f_{1} \circ f_{2} \rrbracket = \bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a)\).
Proof
For this proof and many followings, we need to consider two cases. Case one is when \(\bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a) = 0\). Then for any a ∈S, f1(a) ⊓f2(a) = 0. Then \(\llbracket f_{1} \circ f_{2} \rrbracket = \bigsqcup _{g \in N} \bigsqcup _{a \in S} g(a) \sqcap \bigsqcap _{b \in S} g(b) \Rightarrow (f_{1}(b) \sqcap f_{2}(b)) = \bigsqcup _{g \in N} \bigsqcup _{a \in S} g(a) \sqcap \bigsqcap _{b \in S} - g(b) = 0\).
Case two is when \(\bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a) > 0\). Then define f ∈N such that for any a ∈S, f(a) = f1(a) ⊓f2(a). It is easy to see that \(\llbracket f \precsim f_{1} \rrbracket = \llbracket f \precsim f_{2} \rrbracket = 1\).
\(\llbracket f_{1} \circ f_{2} \rrbracket = \llbracket \exists v (E(v) \land v \precsim f_{1} \land v \precsim f_{2}) \rrbracket = \bigsqcup _{g \in S^{B}} \llbracket E(g) \land g \precsim f_{1} \land g \precsim f_{2} \rrbracket \).
Fix some random g ∈SB, \(\llbracket E(g) \land g \precsim f_{1} \land g \precsim f_{2} \rrbracket = \bigsqcup _{a \in S} g(a) \sqcap \bigsqcap _{b \in S} (g(b) \Rightarrow (f_{1}(b) \sqcap f_{2}(b))) \leqslant \bigsqcup _{a \in S} g(a) \sqcap (g(a) \Rightarrow (f_{1}(a) \sqcap f_{2}(a))) \leqslant \bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a) = \llbracket E(f) \rrbracket = \llbracket E(f) \land f \precsim f_{1} \land f \precsim f_{2} \rrbracket \). Hence \(\bigsqcup _{g \in S^{B}} \llbracket E(g) \land g \precsim f_{1} \land g \precsim f_{2} \rrbracket = \llbracket E(f) \land f \precsim f_{1} \land f \precsim f_{2} \rrbracket = \llbracket E(f) \rrbracket = \bigsqcup _{a \in S} f_{1}(a) \sqcap f_{2}(a)\). □
Theorem E.2 (Supplementation)
.
Proof
Let f1,f2 ∈N. We want to show that .
Again, there are two cases. If , then we are done. If then define f ∈N such that for any a ∈S, f(a) = −f1(a) ⊓f2(a). We can easily show that \(\llbracket f \precsim f_{2} \rrbracket = 1\). Also, \(\llbracket E(f) \rrbracket = \bigsqcup _{a \in S} -f_{1}(a) \sqcap f_{2}(a)\), by Lemma E.1.1, and \(\llbracket \neg f_{1} \circ f \rrbracket = -(\bigsqcup _{a \in S} f_{1}(a) \sqcap -(f_{1}(a) \sqcap f_{2}(a))) = 1\), by Lemma E.1.2. Hence . □
Theorem E.3 (Fusion)
\(\mathfrak {S}^{B}_{V} \models \forall X_{1} (\exists v_{1} X_{1}(v_{1}) \rightarrow \exists v_{2} (FU(v_{2}, X_{1}))\).
Proof
Let \(R \in {D^{1}_{M}}\). Again, there are two cases. Case one: \(\bigsqcup _{a \in S} \bigsqcup _{g \in N} R(g) \sqcap g(a) = 0\). Then for any g ∈N,a ∈S, R(g) ⊓g(a) = 0. This case can be proven easily by unpacking the definitions. Case two: \(\bigsqcup _{a \in S} \bigsqcup _{g \in N} R(g) \sqcap g(a) > 0\). Then define fR ∈N: for any a ∈S, let \(f(a) = \bigsqcup _{g \in S^{B}} R(g) \sqcap g(a)\). We will show that \(\llbracket FU(f^{R}, R) \rrbracket = \llbracket \forall v_{2} (R(v_{2}) \rightarrow v_{2} \precsim f^{R}) \land \forall v_{3} (v_{3} \precsim f^{R} \land E(v_{3}) \rightarrow \exists v_{4} (R(v_{4}) \land v_{3} \circ v_{4})) \rrbracket = 1\).
\(\llbracket \forall v_{2} (R(v_{2}) \rightarrow v_{2} \precsim f^{R}) \rrbracket = \bigsqcap _{h \in S^{B}} R(h) \Rightarrow (\bigsqcap _{a \in S} h(a) \Rightarrow f^{R}(a))\). Fix some h ∈N. Then \(-R(h) \sqcup (\bigsqcap _{a \in S} -h(a) \sqcup (\bigsqcup _{g \in S^{B}} R(g) \sqcap g(a))) = \bigsqcap _{a \in S} -(R(h) \sqcap h(a)) \sqcup \bigsqcup _{g \in S^{B}} R(g) \sqcap g(a)) \geqslant \bigsqcap _{a \in S} -(R(h) \sqcap h(a)) \sqcup (R(h) \sqcap g(h)) = 1\).
\(\llbracket \forall v_{3} (v_{3} \precsim f^{R} \land E(v_{3}) \rightarrow \exists v_{4} (R(v_{4}) \land v_{3} \circ v_{4})) \rrbracket = \bigsqcap _{g \in S^{B}} (\llbracket g \precsim f^{R} \rrbracket \sqcap \llbracket E(g) \rrbracket ) \Rightarrow (\bigsqcup _{h \in S^{B}} (R(h) \sqcap \llbracket h \circ g \rrbracket ))\). Fix some g ∈N. \(\bigsqcup _{h \in S^{B}} (R(h) \sqcap \llbracket h \circ g \rrbracket ) = \bigsqcup _{h \in S^{B}} R(h) \sqcap \bigsqcup _{a \in S} h(a) \sqcap g(a) = \bigsqcup _{a \in S} \bigsqcup _{h \in S^{B}} R(h) \sqcap h(a) \sqcap g(a) = \bigsqcup _{a \in S} f^{R}(a) \sqcap g(a) = \llbracket f^{R} \circ g \rrbracket \). But \(\llbracket f^{R} \circ g \rrbracket = \llbracket \exists v_{1} (E(v_{1}) \land v_{1} \precsim f^{R} \land v_{1} \precsim g) \rrbracket = \bigsqcup _{t \in S^{B}} \llbracket E(t) \rrbracket \sqcap \llbracket t \precsim f^{R} \rrbracket \sqcap \llbracket t \precsim g \rrbracket \geqslant \llbracket E(g) \rrbracket \sqcap \llbracket g \precsim f^{R} \rrbracket \). □
Lemma E.3.1
Let f ∈N. \(\mathfrak {S}^{S}_{V} \models At(f)\) just in case {f(a)|a ∈S} is a maximal antichain in B.
Proof
Using the same proof as in Lemma D.4.1 we can show that for any f ∈N, \(\mathfrak {S}^{B}_{V} \models \forall v (E(v) \rightarrow \neg v \precnsim f)\) just in case {f(a)|a ∈S} is an antichain in B.
Recall that \(At(f) = E(f) \land \forall v (E(v) \rightarrow \neg v \precnsim f)\). Suppose ⟦At(f)⟧ = 1. Then {f(a)|a ∈S} is an antichain. Also, since \(\bigsqcup _{a \in S} f(a) = \llbracket E(f) \rrbracket = 1\), {f(a)|a ∈S} is a maximal antichain. Similarly, suppose {f(a)|a ∈S} is a maximal antichain, then \(\llbracket E(f) \rrbracket = \bigsqcup _{a \in S} f(a) = 1\). Also, \(\llbracket \forall v (E(v) \rightarrow \neg v \precnsim f) \rrbracket = 1\). Hence ⟦At(f)⟧ = 1. □
Theorem E.4 (Atomicity)
\(\mathfrak {S}^{B}_{V} \models \forall v_{1} (E(v_{1}) \rightarrow \exists v_{2} (At(v_{2}) \land v_{2} \precsim v_{1}))\).
Proof
The same proof as in Theorem D.5, using the previous lemma. □
Theorem E.5 (NoZero is false.)
Proof
This can be proven by showing two things. First, has value 1 . Let f1,f2 ∈N be such that for some a ∈S, f1(a) = 1 and f2(a) = 0. Then f1(a) ⊓−f2(a) = 1. Second, ⟦¬∃v3¬(E(v3))⟧ has value 0. Define fp ∈SB to be the constant function that takes every a ∈S to p, where 0 < p < 1, and f−p ∈N to be the constant function that takes every a ∈S to −p. Then ⟦E(fp)⟧ = p and ⟦E(f−p)⟧ = −p. Hence \(\llbracket \neg \exists v_{3} \neg (E(v_{3})) \rrbracket \geqslant \llbracket E(f^{p}) \rrbracket \sqcap \llbracket E(f^{-p}) \rrbracket = 0\). □
Corollary E.5.1
\(\mathfrak {S}^{B}_{V}\) is a model of MACM, but not a model of ACM−.
Appendix F: Identity and Anti-Symmetry
Recall that an atomic Boolean model is a VI model if it is SEVI or VEVI, and similarly is a TI model if it is SETI or VETI.
Proposition F.1
In any VI model, Anti-Symmetry has value 1.
Proof
Directly follows from Vague-Identity: for any f1,f2 ∈M/N, \(\llbracket f_{1} = f_{2} \rrbracket = \bigsqcap _{a \in S} f_{1}(a) \Leftrightarrow f_{2}(a) = \llbracket f_{1} \precsim f_{2} \rrbracket \sqcap \llbracket f_{2} \precsim f_{1} \rrbracket \). □
Proposition F.2
In any TI model, Anti-Symmetry has value 0.
Proof
Define \(f_{1}: S \rightarrow B\) as follows: for some a ∈S, f1(a) = p, where 0 < p < 1; for any b ≠ a ∈S, f1(b) = 1. Define \(f_{2}: S \rightarrow B\) as follows: f2(a) = −p and for any b ≠ a ∈S, f2(b) = 1. Define \(f: S \rightarrow B\) as follows: for any c ∈S, f(c) = 1. It is easy to see that \(f_{1}, f_{2} \in M \subseteq N\).
It is also easy to see that \(\llbracket f_{1} \precsim f \rrbracket = \llbracket f_{2} \precsim f \rrbracket = 1\). And \(\llbracket f \precsim f_{1} \rrbracket = 1 \Rightarrow p = p\), \(\llbracket f \precsim f_{2} \rrbracket = 1 \Rightarrow -p = -p\). Also, since f,f1,f2 are different functions, ⟦f1 = f⟧ = ⟦f1 = f⟧ = 0.
Hence \(\llbracket f_{1} \precsim f \land f \precsim f_{1} \rightarrow f = f_{1} \rrbracket = (1 \sqcap p) \Rightarrow 0 = -p\). And \(\llbracket f_{2} \precsim f \land f \precsim f_{2} \rightarrow f = f_{2} \rrbracket = (1 \sqcap -p) \Rightarrow 0 = p\). Hence \(\llbracket \forall v_{1} \forall v_{2} (v_{1} \precsim v_{2} \land v_{2} \precsim v_{1} \rightarrow v_{1} = v_{2}) \rrbracket \leqslant p \sqcap -p = 0\). □
Corollary F.0.1
In any SEVI model, Transitivity, Supplementation, Fusion, Atomicity, NoZero and Anti-Symmetry all have value 1.
Corollary F.0.2
In any SETI model, Transitivity, Supplementation, Fusion, Atomicity and NoZero all have value 1, but Anti-Symmetry has value 0.
Corollary F.0.3
In any VEVI model, Transitivity, Supplementation, Fusion, Atomicity, and Anti-Symmetry all have value 1, but NoZero has value 0.
Corollary F.0.4
In any VETI model, Transitivity, Supplementation, Fusion and Atomicity all have value 1, but NoZero and Anti-Symmetry have value 0.
Rights and permissions
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.
About this article
Cite this article
Wu, X. Boolean Mereology. J Philos Logic 52, 731–766 (2023). https://doi.org/10.1007/s10992-022-09686-0
Received:
Accepted:
Published:
Issue Date:
DOI: https://doi.org/10.1007/s10992-022-09686-0