3

Sorry for the somewhat dumb question! Please do see if you can make sense of the latter, and put it in formal or whatever terms.

If a chariot is equal to its parts then the chariot is not its "being" equal to its parts. Else changing the parts of a chariot would be impossible – and would (?) that mean that every referent of “chariot” is composed of the same axle and wheel etc..

[P1 P2 P3 ↔ C] ← C as opposed to P1 P2 P3 ↔ C ← C The chariot is (let's suppose) just the chariot's parts, but that fact is not something about the chariot.

3
  • 1
    Think of a molecule of water. Is it identical to a mere collection of an oxygen atom and two hydrogen atoms? (Hint: think of isomers). The obvious 'no' might be applicable to this question as well. Not uncontroversial, I admit. I think Shane recently contributed an answer to the Ship of Theseus puzzle, so I'm hoping he, and others, will be able to make sense of this question. I'm not sure I interpreted it properly. Good luck. Commented Jun 2, 2014 at 7:02
  • hi, notice the garbled C/P thing..? i think this is about modality - maybe? if C having such and such parts is equivalent to C existing, then (it seems to me) every existing C has those parts. but if C is not its having those parts, that does not hold... QED (?) ?
    – user6917
    Commented Jun 2, 2014 at 8:40
  • @user3293056 not really sure how the P and C would change what Hunan is telling you here. At the simplest level a structured list is more than the elements on the list precisely because of the structure.
    – virmaior
    Commented Jun 4, 2014 at 11:57

3 Answers 3

3

At minimum, composition is the specific composite parts plus their configuration, i.e. their position relative to each other and/or the way they are joined.

Additionally (and much more interestingly), you can have emergence of qualities that the composition exhibits.

1
  • hi this makes sense... but i'm interested in proofs of your first statement and / or formal ways of stating it.
    – user6917
    Commented Jun 2, 2014 at 9:55
2

The answer to your question depends on how one spells out certain supplementation intuitions about composites. This can be done in precise terms in mereology, a theory (or rather class of theories) dealing with the parthood relation.

Intuitions first: The supplementation intuition says that a composite, i.e. a whole having proper parts (parts distinct from the whole) cannot consist of a single proper part. In other words proper parts are always 'supplemented' by certain other proper parts.

This intuition can be formalized in a first-order theory whose signature only contains = and the two place relation symbol P representing the parthood relation. The first three axioms capture the intuition that parthood is a partial order:

  1. ∀xPxx
  2. ∀xyz(Pxy & Pyz → Pxz)
  3. ∀xy(Pxy & Pyx → x = y)

Now one quite strong way to represent the supplementation intuition is by adding the axiom

  1. ∀xy(¬Pyx → ∃z(Pzy ∧ ¬Ozx))

where Oxy := ∃z(Pzx & Pzy) ('x and y overlap' in the sense of sharing some part). This axiom says that in case y is no part of x there is some part of y such that this part and x do not overlap. Lets call 1.-4. extensional mereology (EM). Now it is easily seen that 4. and so EM entails

  1. ∀xy(PPxy → ∃z(Pzy ∧ ¬Ozx))

where PPxy := Pxy & x ≠ y. 5. seems to be a proper rendering of the supplementation intuition. But EM also entails that composites having exactly the same proper parts cannot be distinguished:

  1. ∀xy(∃zPPzx ∨ ∃zPPzy) → (x=y ↔ ∀z(PPzx ↔ PPzy)).

So, as long as your chosen mereology contains 4. it is extensional in the sense that composites are nothing more than their proper parts.

4
  • i'm not sure if i will be able to make sense of it at all easily; so may i ask / double check: have you included formal proof that the whole is the property of "being" a whole? tho i am not sure the question makes sense. perhaps i am only asking if wholes defined in that way essentially have the parts they do: if the whole always has the same parts.
    – user6917
    Commented Jun 2, 2014 at 15:31
  • OK I thought you wanted to know: Are there formal conditions being sufficient for a composite to be determined by its proper parts? What I did is providing a formal framework (first-order logic), wherein it is possible to prove that: Given a certain formally expressible condition, namely strong supplementation (the fourth axiom), composites are determined by their parts. I didn't spell out the simple proof, for otherwise my reply would have been even longer.
    – sequitur
    Commented Jun 2, 2014 at 21:01
  • ok, i should have asked: does this "formally expressible condition" entail anything problematic, like every whole with parts p1 p2 p3 being composed of the same parts?
    – user6917
    Commented Jun 4, 2014 at 12:16
  • @sequitur I've taken a metaphysics course, but didn't have the opportunity to explore this fascinating area in detail. I've come across Simons 2000 and Varzi's SEP article, but I was wondering if you recommend some further things to read on mereology. I could go through their bibliographies, but I'd love to hear your personal advice on this. Thanks in advance. Commented Jun 4, 2014 at 22:54
0

Much more often than not the sum of parts is much more than than the collection of parts.

A house is made of bricks; but its much more than a collection of bricks.

A novel is made of letters from the alphabet; but its much more than the mere collection of letters.

A molecule of water is made of two hydrogen atoms and one oxygen atom; but again its not just the collection of those atoms.

Only in simple arithmetic do we get 1+1=2; that is the sum is the collection. This simplicity is vital to its utility; so much so that many parts of mathematics implicitly use this or try to generalise it so thats its true. This is what is called the principle of linearity - and is the key idea in the theory of Linear Algebra - and its many avatars. Quantum Mechanics, strangely enough given its bizarre behaviour is a linear theory. The differential calculus, or its geometric avatar - differential geometry - is linear in the small.

You must log in to answer this question.