Maverick Philosopher

Footnotes to Plato from the foothills of the Superstition Mountains

Definitions and Axioms of Classical Mereology

Is a wall or a brick house a whole of its parts?  Obviously — that's a pre-analytic datum.  But is it a sum of its parts?  I have been arguing, with no particular originality, in the negative.  I have been arguing that it is a big mistake to assume  that, just because y is a whole of the xs, that y is a sum of the xs. But it depends on what exactly is meant by 'sum.'  My point is well-taken if 'sum' is elliptical for 'classical mereological sum.'  But what does that mean?  Since 'classical mereological sum' is a technical term, it has all and only the meaning conferred upon it by the definitions and axioms of classical mereology.  I will now present what I take to be the essentials of classical mereology.  I will use 'sum' as short for 'classical mereological sum.'  Later we will look at neoclassical variants that result from tampering with the classical definitions and axioms.

If anything in what follows is original, it is probably a mistake on my part.  Feel free to correct me — but only if you know the subject matter.

I will take proper parthood and identity as primitives.  To simplify the exposition I will drop universal quantifiers.  They are there in spirit if not in letter.

D1. x is a PART of y =df x is a proper part of y or x = y.

D2. x OVERLAPS y =df there is a z such that z is part of x and z is part of y.

D3. x is DISJOINT from y =df it is not the case that x overlaps y.

D4. y is a SUM of the xs =df z overlaps y iff z overlaps one of the xs.

A1. Asymmetry of Proper Parthood.  If x is a proper part of y, then y is not a proper part of x.

A2. Transitivity of Proper Parthood.  If x is a proper part of y, and y is a proper part of z, then x is a proper part of z.

A3. Supplementation of Proper Parthood.  If x is a proper part of y, then there is a z such that z is a proper part of y and z is disjoint from x.

A4. Uniqueness of Summation.  If u is a sum of the xs and v is a sum of the xs, then u = v.

A5. Unrestricted Summation.  For any xs, there is a y such that y is a sum of the xs.

When I used the word 'sum' in previous posts, I intended that its meaning be not merely the meaning assigned to it by (D4), but the meaning assigned to it by (D4) in conjunction with the rest of the definitions and the axioms (not to mention the theorems that follow as logical consequences of the definitions and axioms). 

Extensionality is a feature of classical mereology.  I leave it as an exercise for the reader to derive Extensionality of Parthood  — if x and y are sums with the same proper parts, then x = y — as a theorem from the above.

 

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Bill Vallicella

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5 responses to “Definitions and Axioms of Classical Mereology”

  1. David Brightly Avatar
    David Brightly

    Hello Bill,
    and thank you for pulling these definitions and axioms together.
    (1) We might rephrase the V/VI definition of sum as

    D4*. y is a sum of the xs =df each of the xs is a part of y and every part of y overlaps at least one of the xs.

    To continue our discussion we need to establish that D4 and D4* are equivalent. I was having some difficulty doing this, mainly because D4 is couched in terms of overlap and D4* in terms of parthood, and it wasn’t clear how the two are related. But, reading Varzi once again, I found his

    (55) Pxy ↔ ∀z(Ozx → Ozy).

    Varzi says this is a theorem ‘of every extensional mereology’. If we allow ourselves this result (I’d like to see a proof) then it’s straightforward to show that D4 and D4* are indeed equivalent.
    (2) Let CM denote the system D1–D4, A1–A5. If CM is applicable to the examples we have been discussing—the bricks, the TT pieces, the book—then D4 requires us to say that BrickHouse, TTH, the book, are all sums. The only way to avoid this is to deny that CM is applicable. But this is not unreasonable. We could take any distant pair of bricks, say, as the xs and deny that there is a y consisting of just those two bricks. We could I think take as objects any connected set of bricks. These objects would satisfy A1, A2, and A4, but not A5 nor A3 (the supplement of a course of bricks partway up the house is disconnected.) So the system CM would not apply to these objects and hence we need not feel obliged to call BrickHouse a sum. Is this what you meant by saying “If you take ‘sum’ in this full sense, then I insist that TTH [or BrickHouse] is not a sum”?
    (3) Regarding the homework problem: Doesn’t this follow directly from A4, Uniqueness of Summation?

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  2. Bill Vallicella Avatar
    Bill Vallicella

    David,
    Doesn’t (55) strike you as intuitively correct?
    I think we are on the same page. I am denying that CM is applicable to things like books and brick houses.
    Yes, Extensionality of Parthood follows from A4.

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  3. David Brightly Avatar
    David Brightly

    Hello Bill,
    Yes (55) certainly looks intuitively correct and the left-to-right implication is easy enough to prove. It’s the right-to-left I’m stuck on. That it’s a theorem of *extensional* mereology suggests sums are involved.
    Regarding the applicability of CM, we are in agreement.

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  4. Bill Vallicella Avatar
    Bill Vallicella

    It’s the right-to-left implication of (55) that you are stuck on, thus:
    (Ozx –> Ozy) –> Pxy
    For x to overlap y is for them to have a part in common. This can happen in four ways. Either x = y, or y is a proper part of x or x is a proper part of y or there is a z which is a proper part of x and a proper part of y.
    Now the only way z’s overlapping of x can imply z’s overlapping of y is if x = y or x is a proper part of y. Either way it follows that x is a part of y. QED
    Is this right?

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  5. David Brightly Avatar
    David Brightly

    Hello again Bill,
    I have made a little progress with this, again with Varzi’s help. He says that our problematic theorem, viz, the right to left half of

    (55) Pxy ↔ ∀z(Ozx → Ozy)

    holds in all *extensional* mereologies. In his section 3.2 he says that an extensional mereology is one in which ‘strong supplementation’ holds. This axiom is

    (P5) ¬Pxy → ∃z(Pzx ∧ ¬Ozy).

    And indeed from this one can deduce (55).
    Proof. Suppose for contradiction ∀z (Ozx → Ozy) but ~Pxy. Then we have ∀z (Ozx → Ozy) ∧ ∃w(Pwx ∧ ¬Owy). Substitute this w for the z in the universally quantified expression on the left:

    ∃w((Owx → Owy) ∧ Pwx ∧ ¬Owy).

    But Pwx implies Owx and hence Owy, so this reduces to ∃w(Owy ∧ Pwx ∧ ¬Owy). Contradiction, as desired.
    Varzi says that his P5 is strictly stronger than his

    (P4) PPxy → ∃z(Pzy ∧ ¬Ozx)

    which is a slightly weaker version of your A3. Indeed, the model in his figure 3 satisfies your A3 but makes (55) false. Every part overlaps both x and y so LHS of (55) is true, but RHS is false.
    Conclusion: The system you quote less the summation axioms A4 and A5 is not strong enough to rule out the non-extensional model of figure 3 in which x and y have the same parts but are distinct, and not strong enough to prove the intuitive result (55). So presumably it’s the inclusion of A4 and A5 that sufficiently strengthens the system. Extensionality comes in by fiat and the model of figure 3 in which x and y are distinct sums of the zs is ruled out, for example. But I’m still at a loss to prove (55) from the axioms you give. Other intuitive results such as xP(x+y) are also problematic, though this follows easily from (55). The difficulty perhaps stems from a lack of theorems of the form …x…y… → Pxy. Or maybe I’m just missing an essential mereological proof technique.

    Reply

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