Atomic ontology

Abstract

The aim of this article is to offer a method for determining the ontological commitments of a formalized theory. The second section shows that determining the consequence relation of a language model-theoretically entails that the ontology of a theory is tied very closely to the variables that feature in that theory. The third section develops an alternative way of determining the ontological commitments of a theory given a proof-theoretic account of the consequence relation for the language that theory is in. It is shown that the proof-theoretic account of ontological commitment does not entail that the ontological commitments of a theory depend on the variables of that theory. The last section of the article discusses how this account of ontological commitment can be used in other philosophical projects such as Wright’s (Frege’s conception of numbers as objects, Aberdeen University Press, Aberdeen, 1983) abstractionism. The article concludes with a discussion of the upshots of adopting the proof-theoretic account of ontological commitment for ontology generally.

Appendix

Lemma 1

If $\Delta \Rightarrow \Lambda \in \triangledown ( \Gamma \Rightarrow \Sigma )$, then $\Gamma \subseteq \Delta$ and $\Sigma \subseteq \Lambda$.

Proof

Let $\Gamma \Rightarrow \Sigma , \Delta _1 \Rightarrow \Lambda _1, \ldots$ be an open branch in $\tau (\Gamma \Rightarrow \Sigma )$. Since the root of this branch is $\Gamma \Rightarrow \Sigma$, $\Gamma \subseteq \Gamma \cup \bigcup _i \Delta _i$ and $\Sigma \subseteq \Sigma \cup \bigcup _i \Lambda _i$. Since that branch was arbitrary this holds for every open branch in $\tau (\Gamma \Rightarrow \Sigma )$ and thus for every position in $\triangledown (\Gamma \Rightarrow \Sigma )$. $\square$

Theorem 1

Let $\Gamma$ be a theory. $\Gamma$ is committed to there being entities of which K is true iff for any model M if $M \vDash \Gamma$ then for some $\sigma \in \Sigma _M$ and x that is free for K.

Proof

For the left to right direction let $\Gamma$ be committed to there being entities of which K is true. It follows that there is no model M such that $M \vDash \Gamma$ and K is empty in M. Let M be a model of $\Gamma$, i.e. $M \vDash \Gamma$. Since K is not empty in M there is variable assignment $\sigma$ and variable x such that x is free for K and .

For the right to left direction let it be that for any model M if $M \vDash \Gamma$ then for some $\sigma \in \Sigma _M$ and X that is free for K. Suppose that $\Gamma$ is not committed to there being entities of which K is true. It follows that there is a model M such $M \vDash \Gamma$ and K is empty in M. Since K is empty in M there is no variable assignment $\sigma \in \Sigma _M$ and variable x that is free for K such that . But by the assumption there is such a $\sigma$ and variable. $\square$

Theorem 2

The proof-theoretic and model-theoretic accounts of quantification agree, i.e. $\Gamma \vDash \Sigma$ iff $\vdash \Gamma \Rightarrow \Sigma$.

Proof

This is a familiar result for first-order classical logic. For a proof of this see Ebbinghaus et al. (1980). $\square$

Theorem 3

For any sentence $\varphi$ that does not contain elements of W, it holds that $\vdash \Gamma \Rightarrow \varphi , \Sigma$ iff for every $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \Sigma )$, $\varphi \in \Delta$.

Proof

For the left to right direction suppose that $\vdash \Gamma \Rightarrow \varphi , \Sigma$. Let $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \Sigma )$ be such that $\varphi \in \Lambda$ and $\beta$ be the open branch of $\tau (\Gamma \Rightarrow \Sigma )$ from which $\Delta \Rightarrow \Lambda$ is generated. By Lemma 1$\Gamma \subseteq \Delta$ and $\Sigma \bigcup \{\varphi \} \subseteq \Lambda$. Since $\varphi \in \Lambda$ there is an n such that $\Delta ' \Rightarrow \Lambda '$ in $\beta$ and $\varphi \in \Lambda '$. Let $\delta \vdash \Gamma \Rightarrow \varphi , \Sigma$. The following deduction establishes that $\vdash \Delta ' \Rightarrow \Lambda '$ and thus that $\beta$ is not an open branch in $\tau (\Gamma \Rightarrow \Sigma )$ contradicting the assumption that it was.

For the right to left direction suppose that $\varphi \in \Delta$ for every $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \Sigma )$ and for reductio that

• (H) $\nvdash \Gamma \Rightarrow \varphi , \Sigma$

Let $\varphi$ be the nth sentence. Suppose that at stage n of the construction there is a branch $\beta \in \tau (\Gamma \Rightarrow \Sigma )$ with an open leaf $\Delta ' \Rightarrow \Lambda '$ such that $\nvdash \Delta ' \Rightarrow \varphi , \Lambda '$. At this stage $\beta$ is at least extended to $\beta '$ with open leaf $\Delta ' \Rightarrow \varphi , \Lambda '$. But in this case there is a branch $\beta ''$ that passes through $\beta '$ when the construction of $\tau (\Gamma \Rightarrow \Sigma )$ is completed such that the position $\Delta \Rightarrow \Lambda$ is the maximal leaf of $\beta ''$ and $\varphi \in \Lambda$. Since $\vdash \Delta ' \Rightarrow \varphi , \Lambda '$$\varphi \not \in \Delta$, contradicting the assumption. This establishes

• (B) There is no open leaf $\Delta ' \Rightarrow \Lambda '$ at stage n such that $\nvdash \Delta ' \Rightarrow \varphi , \Lambda '$.

Claim 1

There is no stage i such that there is an open leaf $\Pi \Rightarrow \Theta$ in the construction of $\tau (\Gamma \Rightarrow \Sigma )$ where $\nvdash \Pi \Rightarrow \varphi , \Theta$ but at $i+1$ there is no open leaf $\Pi ' \Rightarrow \Theta '$ such that $\nvdash \Pi ' \Rightarrow \varphi , \Theta '$.

Suppose that Claim 1 is true. It follows that for any stage i if there is an open leaf $\Pi \Rightarrow \Theta$ at a stage i such that $\nvdash \Pi \Rightarrow \varphi ,\Theta$ then there is an open leaf $\Pi ' \Rightarrow \Theta '$at $i+1$ such that $\nvdash \Pi ' \Rightarrow \varphi , \Theta '$. By (H) at stage 0, there is an open leaf $\Gamma \Rightarrow \Sigma$ such that $\nvdash \Gamma \Rightarrow \varphi , \Sigma$. This contradicts (B).

This leaves only the proof of Claim 1. It is proved by reductio. Let i be a stage such that there is an open leaf $\Pi \Rightarrow \Theta$ in the construction of $\tau (\Gamma \Rightarrow \Sigma )$ where $\nvdash \Pi \Rightarrow \varphi , \Theta$ but at $i+1$ there is no open leaf $\Pi ' \Rightarrow \Theta '$ such that $\nvdash \Pi ' \Rightarrow \varphi , \Theta '$. To be specific let $\Pi \Rightarrow \Theta$ be the position under consideration at $i\cdot j\cdot k$. Either there is a step $i\cdot j\cdot k+l$ with position $\Pi ' \Rightarrow \Theta '$ such that $\vdash \Pi ' \Rightarrow \varphi , \Theta '$ or not.

Case 1

(There is a step $i\cdot j\cdot k+l$ with position $\Pi ' \Rightarrow \Theta '$ such that $\vdash \Pi ' \Rightarrow \varphi , \Theta '$) In this case $\Pi ' \Rightarrow \Theta '$ is $\Pi , \psi [w/x] \Rightarrow \Theta$. It follows that $\exists x \psi _j \in \Pi$, w does not appear in $\Pi \bigcup \Theta$, $\nvdash \Pi , \exists x \psi _j \Rightarrow \Theta$. It was assumed that no witnesses occur in $\varphi$ so w does not appear in $\varphi$. By assumption $\vdash \Pi , \exists x \psi _j, \psi [w/x] \Rightarrow \varphi , \Theta$ and $\nvdash \Pi , \exists x \psi _j \Rightarrow \varphi , \Theta$. However, L$\exists$ read from bottom to top guarantees that if $\nvdash \Pi , \exists x \psi _j \Rightarrow \varphi , \Theta$ and w does not occur in $\Pi \bigcup \Theta \bigcup \{\varphi \}$ then $\nvdash \Pi , \exists x \psi _j, \psi [w/x] \Rightarrow \varphi , \Theta$ contradicting the assumption.

Case 2

(There is no step $i\cdot j\cdot k+l$ with position $\Pi ' \Rightarrow \Theta '$ such that $\vdash \Pi ' \Rightarrow \varphi , \Theta '$) Without loss of generality let there be no existential steps between i and $i+1$. If there are then there is a point at which Case 1 applies. There are then three sub-cases for how i is extended.

Case 2.1 (The branch with leaf $\Pi \Rightarrow \Theta$ is expanded according to (b) in the construction of $\tau (\Gamma \Rightarrow \Sigma )$) In this case $\nvdash \Pi , \psi _i \Rightarrow \Theta$ and $\nvdash \Pi \Rightarrow \psi _i, \Theta$. By assumption there are no open leaves $\Pi ' \Rightarrow \Theta '$ at this stage such that $\vdash \Pi ' \Rightarrow \varphi ,\Theta '$, so there are deduction $\delta$ and $\delta '$ such that $\delta \vdash \Pi , \psi _i \Rightarrow \varphi , \Theta$ and $\delta ' \vdash \Pi \Rightarrow \psi _i, \varphi , \Theta$. It was also assumed that $\nvdash \Pi \Rightarrow \varphi , \Theta$. The following deduction contradicts this assumption.

Case 2.2 (The branch with leaf $\Pi \Rightarrow \Theta$ is expanded according to (c) in the construction of $\tau (\Gamma \Rightarrow \Sigma )$) In this case there is a $\delta$ such that $\delta \vdash \Pi , \psi _i \Rightarrow \Theta$ but $\nvdash \Pi \Rightarrow \psi _i, \Theta$. By assumption there is no open leaf $\Pi ' \Rightarrow \Theta '$ at this stage such that $\nvdash \Pi ' \Rightarrow \varphi , \Theta '$. It follows that there is a $\delta '$ such that $\delta ' \vdash \Pi \Rightarrow \psi _i, \varphi , \Theta$. The following deduction contradicts the assumption that $\nvdash \Pi \Rightarrow \varphi , \Theta$.

Case 2.3 (The branch with leaf $\Pi \Rightarrow \Theta$ is expanded according to (d) in the construction of $\tau (\Gamma \Rightarrow \Sigma )$) In this case there is a $\delta$ such that $\delta \vdash \Pi \Rightarrow \psi _i , \Theta$ but $\nvdash \Pi , \psi _i \Rightarrow \Theta$. By assumption there is no open leaf $\Pi ' \Rightarrow \Theta '$ at this stage such that $\nvdash \Pi ' \Rightarrow \varphi , \Theta '$. It follows that there is a $\delta '$ such that $\delta ' \vdash \Pi , \psi _i \Rightarrow \varphi \Rightarrow \Theta$. The following deduction contradicts the assumption that $\nvdash \Pi \Rightarrow \varphi , \Theta$.

$\square$

Theorem 4

A set of sentences $\Gamma$ has a model-theoretic ontological commitment to K iff the position $\Gamma \Rightarrow \quad$ has a proof-theoretic commitment to K.

Proof

For the left to right direction suppose that $\Gamma$ has a model-theoretic commitment to K. By Theorem 1 for any model M such that $M \vDash \Gamma$ there is a $\sigma \in \Sigma _M$ and variable x that is free for K such that . It follows that . Since this holds for any model M it follows that . By Theorem 2. It follows from Theorem 3 that for any $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \quad )$, . Let $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \quad )$ that is the maximal leaf of a branch $\beta$. Since there is a node $\Pi \Rightarrow \Theta$ in $\beta$ such that was considered at stage n.j of the construction of $\tau (\Gamma \Rightarrow \quad )$. Let be the kth sentence. At stage n.j.k a sentence the node directly above $\Pi \Rightarrow \Theta$ is for a witness w. By Theorem 1. Since $\Delta \Rightarrow \Lambda$ was selected arbitrarily it holds that for any $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \quad )$ there is a sentence of the form . It follows that $\Gamma \Rightarrow \quad$ has a proof-theoretic commitment to K.

For the right to left direction suppose that $\Gamma \Rightarrow \quad$ has a proof-theoretic commitment to K. It follows that for any $\Delta \Rightarrow \Lambda \in \triangledown (\Gamma \Rightarrow \quad )$ there is a sentence of the form . The sentence is closed and contains no witnesses. Since . If it were then $\Delta \Rightarrow \Lambda$ would not be coherent. Since $\Delta \Rightarrow \Lambda$ is maximal . By Theorem 3. By Theorem 2. Let M be a model of $\Gamma$. It follows that and that there is a $\sigma \in \Sigma _M$ such that . So K is not empty in M. Since M was arbitrary it holds that there are no models of $\Gamma$ for which K is empty. $\Gamma$ has a model-theoretic commitment to K. $\square$