## Composition as identity, part II

December 11, 2008

Aside from Leibniz’s law, there are various other constraints identity must obey. For example, every object is identical with at most one thing, so every object presumably is identical* to at most one plurality. But here we have a disanalogy with the relation “x is the fusion of the yy’s”, for x is the fusion of many pluralities. If there may be more than one plurality *identical to x, then our notation for pluralizing x, x*, isn’t justified: * isn’t a function.

A way of sharpening this problem, pointed out by Jeff in the comments, is that you’d want (*)identity(*) to be transitive. For example, my upper and lower body are *identical to me, and I’m identical* to my arms, legs head and torso. Does that mean my lower and upper body are *identical* to my arms, legs, head and torso? How do you state that they’re different pluralities?

So there appear to be two ways you can go. One is to say that every object is really identical* to only one plurality, and the other is to say that every plurality is *identical to exactly one object. I’ll call the two approaches (a) unique decomposition, and (b) unique composition.

The first seems to be the most natural. By way of analogy, note that the pluralities, over a domain of objects, have many of the formal properties of mereology. (i) there’s no null fusion/no empty plurality, (ii) pluralities are closed under ‘unioning’ and ‘non-empty intersecting’ (fusion and products) (iii) they’re closed under complements (supplementation.)

In fact, they form a complete Boolean algebra under ‘subplurality’, and thus model the standard mereological axioms. However, there are some drawbacks. Firstly, you can form pluralities of mereological objects, in standard mereological theories (that’s how unrestricted composition is usually stated.) However, on this picture, you can’t, for it would amount to forming pluralities of pluralities – which is nonsense.

You might think this is not too much of a cost; after all, you can always talk about superpluralities when the standard mereologist talks of pluralities.

So this seems to answer our original problem, which was to ensure that many-one identity really associated each object with one plurality. What we have is unique decomposition: there is a unique plurality associated with each object, and that plurality fuses to that object (is *identical to it.) The way we have achieved unique decomposition in this case is by identifying xx with x’s atoms.

There may be other ways to achieve unique decomposition, but it seems they’ll all fall to the following problem. There are some situations where unique decomposition can’t be achieved, at least according to the standard mereologist. One of these is the possibility of a gunky world: a world where everything has a proper part. Formally, we have an atomless Boolean algebra. But if the points in our algebra are pluralities, what are they pluralities of? There cannot be any singleton pluralities, and if there can’t be singleton pluralities, there can’t be objects for there to be pluralities of.

[Side note: by Stone’s representation theorem, any gunky world a standard mereologist can conjure up may be represented by a ‘Henkin’ model of plural logic. Thus, you may feel like you’re in a gunky world – but only because your plural quantifiers are restricted. You’re failing to quantify over all pluralities (in particular, the singletons.)]

The other approach was what I labelled unique composition. Every plurality is *identical to exactly one object. In particular, the tables legs and surface are *identical to exactly one object, x, and the tables atoms are *identical to exactly one object, y. Since they two pluralities aren’t *identical*, neither is x and y. But now we should be worried: this seems to mean we must be able to uniquely assign one object to every plurality in the domain. Since we already have a condition for plurality identity, namely $xx ^*\!\!=^* yy \leftrightarrow \forall z(z\prec xx \leftrightarrow z\prec yy)$ we get the following:

• $\forall x\forall y(x=y \leftrightarrow \forall z(z\prec xx \leftrightarrow z\prec yy))$

This is essentially Frege’s infamous Basic Law V, which entails that there is exactly one object. (Actually, in Frege’s logic it entailed a contradiction, but he allowed there to be empty pluralities.)

In Frege’s system you could derive this, via Russell’s paradox, and I (probably) haven’t written out enough axioms for you to be able to derive the paradox formally. But the problem is still there in the form of Cantor’s theorem, which says you cannot uniquely assign an object to each plurality.

(Note: I never said this, but $\forall x$ can bind $xx$ and vice versa.)

1. The last section where you talk about “you cannot uniquely assign an object to each plurality” is explained in some detail in Agustin Rayo’s /Word and Objects/. There he calls it Bernays’ Principle (but it’s really just a generalization of Cantor’s Theorem).

I’m assuming in your $\forall x \forall y (x = y \leftrightarrow \forall z (z \prec xx \leftrightarrow z \prec yy))$ x is the unique object identical to xx and y is the unique object identical to yy. Is that what is meant by the final note?

I think if you were going to take this unique composition line, you could require that we can’t take pluralities of objects identical to pluralities. But I’m not entirely sure how that would go, and it would seem to have all the same problems as the unique decomposition approach.

I definitely like the unique decomposition line better. I think there is another route too. You could allow pluralities to be identical to their fusion wrt to a cover. Thus, table = atoms’ wrt to the cover {atoms}, but not wrt another {legs, top}; but legs, top = table’ wrt to the cover {legs, top} but not {atoms}. [Actually, strictly speaking the covers will have as its members sets of atoms grouped differently, but I haven’t said enough to spell that out.]

On this view, transitivity fails but only when you switch covers, which shouldn’t be all that surprising.

Anyway, I think you’re right that this sort of idea, too, will require atomism and rule out gunk. That’s unfortunate.

2. Thanks Aaron,

Yeah, it was kind of built into the notation that x was the unique object denoted by xx. So there’s an ambiguity: when you write something like $\forall xxF*xx$ have you quantified over every plurality, or just the pluralities identical to objects. For the unique decomposition theorist it make’s a difference, and it’s the latter, so you you’ll need to introduce a new device to achieve real quantification over everything (and it will probably involve superplural quantification.)

I have a question about the cover proposal. If ‘table = {{table atoms}}’ and ‘table = {{leg atoms}, {top atoms}}’ (that’s the proposal right?), then what is the condition for an object to be identical with a plurality w.r.t a cover. Won’t table =* the atoms under every cover?

Also, what happens when a sentence is true wrt some covers but not others? Do you supervaluate? I’m just worried about Leibniz’s law again, because you might be able to get a version of Evans argument running. Depending on how you do things, you’ll pretty much always get sentences like ‘the F’s are four’ coming out indeterminate…

Lastly, it seems you get the same problems as the unique decomposition theorist anyway, because you only have pluralities of atoms (w.r.t various covers.)

3. Sorry, I haven’t been very clear about the actual proposal. I was trying to put things in some your terminology, but it just made it messy.

The idea is complicated. It involves what I call carvings of the domain, they serve as stand-ins for non-atomic individuals (these are achieved via superplurals). [Incidently, covers are also treated superplurally and not using sets.] And so I should probably spell out the full idea in a post of my own.

But the basic idea is that the domain includes only atoms. Plural quantification over atoms achieves the expression resources of an atomic mereology. We’ll also have plural predicates, say F, G, H. Some theories of plural predication (including yours, as I read it) have different predicates F and F* for distributive and collective readings. “They carried the chair” is ambigous between the two readings. [There are also intermediate readings for some sentences, but let’s not get too detailed.] Is “carried the chair” really two distinct predicates “carried_c” and “carried_d”? I say no. The trick is to interpret the predicate wrt a cover. That yields the differences between the collective and distributive readings.

So, a plural term denotes some things (which are just some atoms). The predicate “is a table” is true of them wrt to the cover {{atoms}}. The predicate “are atoms” is true of them wrt cover {{a},{b}, …}. You see the pattern. So really, there are no ‘individuals’ identical to pluralities in the domain. The many-one identity is really just many-many identity, and we get all the predicates true of tables, chairs, and ordinary objects.

I do not supervaluate over covers. What is true or false of some things depends on the cover, and that’s that. (This is given a better motivation in the paper, but it’s not done yet…)

Leibniz’s law? Well we have that $\forall xx \forall yy (xx=yy \rightarrow (\varphi(xx) \leftrightarrow \varphi(yy))$ so long as the predicate phi is interpreted wrt to the same cover. We needed this already (independent of composition as identity), and so the restriction on Leibniz’s law comes as no surprise.

Anyway. I wasn’t really thinking about gunk. I just assumed atomism for simplicity. I was — to tell you the truth — in the back of my mind assuming that gunk would be fine and wouldn’t bring any problems. But your post has shown that that’s not so. I didn’t really want a theory that assumed gunk is impossible, but … oh well.

So, yes, the proposal does have all the problems of your unique decomposition idea. I mentioned it only because you said in the original post “There may be other ways to achieve unique decomposition, but it seems they’ll all fall to the following problem.” And so I’m just confirming that, yes, there is another way, and it does fall to the same problem.