On atomic composition as identity

Abstract

In this paper I address two important objections to the theory called ‘(Strong) Composition as Identity’ (‘CAI’): the ‘wall-bricks-and-atoms problem’ (‘WaBrA problem’), and the claim that CAI entails mereological nihilism. I aim to argue that the best version of CAI capable of addressing both problems is the theory I will call ‘Atomic Composition as Identity’ (‘ACAI’) which consists in taking the plural quantifier to range only over proper pluralities of mereological atoms and every non-atomic entity to be identical to the (proper) plurality of atoms it fuses. I will proceed in three main steps. First, I will defend Sider’s (in: Baxter D, Cotnoir A (eds) Composition as identity. Oxford University Press, Oxford, pp 211–221, 2014) idea of weakening the comprehension principle for pluralities and I will show that (pace Calosi in Philos Q 66(263):219–235, 2016a) it can ward off both the WaBrA problem and the threat of mereological nihilism. Second, I will argue that CAI-theorists should uphold an ‘atomic comprehension principle’ which, jointly with CAI, entails that there are only proper pluralities of mereological atoms. Finally, I will present a novel reading of the ‘one of’ relation that not only avoids the problems presented by Yi (J Philos 95:163–190, 1999a, in: Baxter D, Cotnoir A (eds) Composition as identity. Oxford University Press, Oxford, pp 169–191, 2014) and Calosi (Log Log Philos 25(3):429–443, 2016b, Am Philos Q 55(3):281–292, 2018) but can also help ACAI-theorists to make sense of the idea that a composite entity is both one and many.

Appendix

1.1 ACAI entails (MP1-3)

ACAI entails (L1):

(L1):

$\forall x\left( \left( { \sim Ax \wedge \forall z\left( {z \le x \leftrightarrow \phi z} \right)} \right) \to {\mathbb{P}}^{x} = IW.\forall z\left( {Wz \leftrightarrow \left( {Az \wedge \phi z} \right)} \right) \right)$

Proof

Suppose that x is not an atom and that something is a $\phi$-er if and only if it is a part of x. By (AT), every $\phi$-er has some atomic part. By (ACP), there is, thus, a plurality W such that something is one of the Ws if and only if it is an atomic $\phi$-er. Each of the Ws is a $\phi$-er and is, thus, part of x. Therefore, every atomic $\phi$-er is part of x. By (AT), every part of x has atomic parts. By transitivity, every atomic part of a part of x is an atomic part of x and is, thus, an atomic $\phi$-er. Therefore, every part of x overlaps an atomic $\phi$-er. By the definition of fusion, x fuses the Ws. By ACAI, x is identical to the Ws. By (ATI-2), ${\mathbb{P}}^{x}$ is identical to W. □

(T1) is a theorem of classical mereology that follows from the definition of mereological sum and Strong Supplementation (which is itself a theorem of classical mereology):

(Strong Supplementation):

$\forall x\forall y\left( {x{ \nleqslant }y \to \exists z\left( {z \le x \wedge \sim Ozy} \right)} \right)$

(T1):

$\forall x\forall y\forall z\left( {x = y + z \to \forall w\left( {w \le x \leftrightarrow \forall k\left( {k \le w \to \left( {Oky \vee Okz} \right)} \right)} \right)} \right)$

Proof

Suppose that x is the sum of y and z. (a) Let w be an arbitrary part of x. By the definition of sum, every part of x overlaps either y or z. By the transitivity of parthood, every part of w overlaps either y or z. (b) Let w be an entity such that every part of w overlaps either y or z and suppose that w is not part of x. Since y and z are parts of x it follows that every part of w overlaps x. By Strong Supplementation, w is part of x. Contradiction! Therefore, something is part of x if and only if all of its parts overlap either y or z. □

The left-to-right directions of (MP1-3) can be, thus, proved as follows (x, y and z are thought of as ranging over non-atomic entities):

(MP1-lr):

$x = y + z \to {\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$

Proof

Suppose that x is the sum of y and z. By (T1), something is part of x if and only if all of its parts overlap either y or z. Letting $\phi$ be ‘$\forall w\left( {w \le v \to \left( {Owy \vee Owz} \right)} \right)$’ it follows from (L1) and the definition of mereological sum that ${\mathbb{P}}^{x}$ is the plurality of atoms overlapping either y or z, and thus the plurality of the atomic parts of either y or z. Therefore, by the definitions of ‘${\mathbb{P}}^{x}$’ and ‘$\mathop \cup \nolimits$’, it follows that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$. □

(MP2-lr):

$x = y - z \to {\mathbb{P}}^{x} = {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$

Proof

Suppose that x is identical to $y - z$ and let $\phi$ be ‘$v \le y \wedge \sim Ovz$’. It follows from (L1) and the definition of mereological difference that ${\mathbb{P}}^{x}$ is the plurality of the atomic parts of y that don’t overlap z. Therefore, ${\mathbb{P}}^{x}$ is the plurality of the atomic parts of y that are not part of z. By the definitions of ‘${\mathbb{P}}^{x}$’ and ‘$-$’, it follows, thus, that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$. □

(MP3-lr):

$x = y \times z \to {\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z}$

Proof

Suppose that x is identical to $y \times z$ and let $\phi$ be ‘$v \le y \wedge v \le z$’. It follows from (L1) and the definition of mereological product that ${\mathbb{P}}^{x}$ is the plurality of atoms that are parts of both y and z. Therefore, by the definitions of ‘${\mathbb{P}}^{x}$’ and ‘$\mathop \cap \nolimits$’, it follows that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z}$. □

From the definition of ‘${\mathbb{P}}^{x}$’ and (${\mathbb{P}}$A)⁴²

(${\mathbb{P}}$A):

$\forall x\left( { \sim Ax \to \forall z\left( {{\mathbb{P}}^{x} z \leftrightarrow \left( {Az \wedge z \le x} \right)} \right)} \right)$

it is possible to derive (L2) which says that, for every non-atomic x and y, x is part of y if and only if the ${\mathbb{P}}^{x}$s are among (‘$\subseteq$’) the ${\mathbb{P}}^{y}$s:

($\subseteq$):

$X \subseteq Y =_{\text{df}} \forall z\left( {Xz \to Yz} \right)$

(L2):

$\forall x\forall y\left( {\left( { \sim Ax \wedge \sim Ay} \right) \to \left( {y \le x \leftrightarrow {\mathbb{P}}^{y} \subseteq {\mathbb{P}}^{x} } \right)} \right)$

Proof

Left-to-right. Suppose that (i) x and y are not atomic, (ii) y is part of x, and (iii) w is one of the ${\mathbb{P}}^{y}$s. By (${\mathbb{P}}$A), w is an atomic part of y. Therefore, since y is part of x, w is also an atomic part of x. By generalisation, the ${\mathbb{P}}^{y}$s are among the ${\mathbb{P}}^{x}$s. Right-to-left. Suppose that (i) x and y are both not atomic, (ii) ${\mathbb{P}}^{y} \subseteq {\mathbb{P}}^{x}$, and (iii) y is not part of x. By Strong Supplementation, there is a part w of y that doesn’t overlap x. Therefore, since, by (AT), w is either an atom or has atomic parts, there must be an atomic part v of y that doesn’t overlap x and is, thus, not part of x. Since v is an atomic part of y, it is one of the ${\mathbb{P}}^{y}$s. But we are assuming that each of the ${\mathbb{P}}^{y}$s is also one of the ${\mathbb{P}}^{x}$s. Contradiction! Therefore, y is part of x. □

The right-to-left directions of (MP1-3) can be proved by means of (L2) (x, y and z are thought of as ranging over non-atomic entities):

(MP1-rl):

${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z} \to x = y + z$

Proof

Suppose that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$. (a) Clearly, ${\mathbb{P}}^{y} \subseteq {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$ and ${\mathbb{P}}^{z} \subseteq {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$. By (L2), both y and z are parts of x. (b) Suppose that w is part of x. If w is an atom, then w is one of the ${\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$ and so it clearly overlaps either y or z. If w is not atomic, we have by (L2) that ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} \mathop \cup \nolimits {\mathbb{P}}^{z}$, so that all the atomic parts of w are atomic parts of either y or z. Therefore, w overlaps either y or z. Either way, w and thus, by generalization, every part of x overlaps either y or z. Therefore, x is the sum of y and z. □

(MP2-rl):

${\mathbb{P}}^{x} = {\mathbb{P}}^{y} - {\mathbb{P}}^{z} \to x = y - z$

Proof

Suppose that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$.

Part I. Let w be part of x. (a) Suppose that w is not atomic. Then, by (L2), ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{x}$ and, so ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$. Clearly, ${\mathbb{P}}^{y} - {\mathbb{P}}^{z} \subseteq {\mathbb{P}}^{y}$. Therefore, ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y}$, and by (L2), w is part of y. Suppose w overlaps z. Then some v is part of both w and z. By (L2), we have that ${\mathbb{P}}^{v} \subseteq {\mathbb{P}}^{w}$ and ${\mathbb{P}}^{v} \subseteq {\mathbb{P}}^{z}$. Therefore, each of the ${\mathbb{P}}^{v}$s is both one of the ${\mathbb{P}}^{w}$s and one of the ${\mathbb{P}}^{z}$ s, thus contradicting ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$ (by the definition of ‘$-$’). Therefore, every non-atomic part of x is a part of y that doesn’t overlap z. (b) Suppose that w is an atom. Then w is one of the ${\mathbb{P}}^{x}$s and, by the definition of plural difference, also one of the ${\mathbb{P}}^{y}$s without being one of the ${\mathbb{P}}^{z}$s. Suppose w overlaps z. Since w is an atom, w is an atomic part of z and, thus, one of the ${\mathbb{P}}^{z}$s. Contradiction! Therefore, also every atomic part of x is a part of y that doesn’t overlap z.

Part II. Let w be a part of y that doesn’t overlap z. (a) Suppose that w is not an atom. By (L2), we have, thus, that ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y}$ and that none of the ${\mathbb{P}}^{w}$s is one of the ${\mathbb{P}}^{z}$s. Therefore, ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} - {\mathbb{P}}^{z}$ and thus ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{x}$. By (L2), w is part of x. (b) Suppose that w is an atom. w is, thus, an atomic part of y and, thus, one of the ${\mathbb{P}}^{y}$s. Since w is an atom that doesn’t overlap z, w is not an atomic part of z and is, thus, not one of the ${\mathbb{P}}^{z}$s. Therefore, w is a member of ${\mathbb{P}}^{y} - {\mathbb{P}}^{z}$ and, thus, one of the ${\mathbb{P}}^{x}$s. By (L2), w is part of x. Every part of y that doesn’t overlap z is, thus, part of x.

Therefore, x is the mereological difference between y and z. □

(MP3-rl):

${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z} \to x = y \times z$

Proof

Suppose that ${\mathbb{P}}^{x} = {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z}$.

Part I. Let w be part of x. (a) Suppose that w is not an atom. By (L2), ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z}$. By the definition of ‘$\mathop \cap \nolimits$’, we have both ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y}$ and ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{z}$. By (L2), w is, thus, part of both y and z. (b) Suppose that w is an atom. w is, thus, one of the ${\mathbb{P}}^{x}$s. By the definition of ‘$\mathop \cap \nolimits$’, w is both one of the ${\mathbb{P}}^{y}$s and one of the ${\mathbb{P}}^{z}$s. Therefore, w is part of both y and z.

Part II. Let w be part of both y and z. (a) Suppose that w is not an atom. By (L2), we have both ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y}$ and ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{z}$ and, thus, that ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{y} \mathop \cap \nolimits {\mathbb{P}}^{z}$. By (L2), w is part of x. (b) Suppose that w is an atom. w is, thus, both one of the ${\mathbb{P}}^{y}$s and one of the ${\mathbb{P}}^{z}$s. By the definition of ‘$\mathop \cap \nolimits$’, w is also one of the ${\mathbb{P}}^{x}$s. w is, thus, part of x.

Therefore, x is the product of y and z. □

1.2 ACAI entails (PQ-SF)

In what follows x is thought of as ranging over non-atomic entities:

(PQ-SF-lr):

$xSFu\phi \to \forall z\left( {{\mathbb{P}}^{x} z \leftrightarrow \left( {\left( {Az \wedge \phi z} \right) \vee \exists w\left( {\phi w \wedge {\mathbb{P}}^{w} z} \right)} \right)} \right)$

Proof

Suppose that x S-fuses everything that $\phi$s.

Left-to-right. Let z be one of the ${\mathbb{P}}^{x}$s. z is, thus, an atomic part of x. Since x S-fuses everything that $\phi$s, it follows there is some w such that w is a $\phi$-er and z overlaps w. z is, thus, an atomic part of w. Therefore, if w is a composite entity, z is one of the ${\mathbb{P}}^{w}$s. If, instead, w is an atom, then z is identical to w and is, thus, a $\phi$-er. In either case, it follows that z, and thus, by generalization, each of the ${\mathbb{P}}^{x}$s is either an atomic $\phi$-er or it is one of the ${\mathbb{P}}^{w}$s for some w that $\phi$s.

Right-to-left. (a) Suppose that w is a $\phi$-er and that z is one of the ${\mathbb{P}}^{w}$s. Since x S-fuses everything that $\phi$s, w is part of x. By (L2), ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{x}$, so that z is also one of the ${\mathbb{P}}^{x}$s. (b) Suppose instead that z is an atomic $\phi$-er. Then, since x S-fuses everything that $\phi$s, it follows that z is an atomic part of x and, thus, that z is one of the ${\mathbb{P}}^{x}$s. □

(PQ-SF-rl):

$\forall z\left( {{\mathbb{P}}^{x} z \leftrightarrow \left( {\left( {Az \wedge \phi z} \right) \vee \exists w\left( {\phi w \wedge {\mathbb{P}}^{w} z} \right)} \right)} \right) \to xSFu\phi$

Proof

Suppose that, for every z, z is one of the ${\mathbb{P}}^{x}$s if and only if, either z is an atomic $\phi$-er or there is some w such that w is a $\phi$-er and z is one of the ${\mathbb{P}}^{w}$s.

Part I. Let w be a $\phi$-er. Suppose that w is a composite entity. Then each of the ${\mathbb{P}}^{w}$s is one of the ${\mathbb{P}}^{x}$s. It follows by (L2) that w is part of x. Suppose that w is, instead, an atom. Therefore, w is an atomic $\phi$-er and, thus, one of the ${\mathbb{P}}^{x}$s. It follows that w is part of x. Therefore, every $\phi$-er is part of x.

Part II. Let w be part of x. Suppose that w is a composite entity. By (L2), ${\mathbb{P}}^{w} \subseteq {\mathbb{P}}^{x}$. Suppose that z is one of the ${\mathbb{P}}^{w}$s. z is, thus, also one of the ${\mathbb{P}}^{x}$s and, therefore, either an atomic $\phi$-er or one of the atomic parts of a $\phi$-er. Either way, z is also part of a $\phi$-er. Therefore, w overlaps a $\phi$-er. Suppose, instead, that w is an atom. Then, w is one of the ${\mathbb{P}}^{x}$s, and so either an atomic $\phi$-er or one of the atomic parts of a $\phi$-er. In both cases, w overlaps a $\phi$-er. Therefore, every part of w overlaps a $\phi$-er.

We have, thus, proved that (i) every $\phi$-er is part of x, and that (ii) every part of x overlaps a $\phi$-er, and so that x S-fuses everything that $\phi$s. □