Suppose there are at least two distinct individuals, Alex and Bob, and that Alex is part of Bob.

Ground mereology has the two-place parthood relation as reflexive, i.e. everything is part of itself: the maximal part. The relation is not, however, symmetric as it is intuitively false that if x is part of y, then y is thereby part of x. There are at least two ways to reject symmetry: asymmetry or antisymmetry. On the former, any x part of y entails y is not part of x. On the latter, if x and y are parts of each other, they are identical. The first is too strong, since inconsistent in the presence of reflexivity:

1.      ∀x P(x,x)                                 Premise
2.      ∀x∀y P(x,y) -> ~P(y,x)          Premise
3.            SHOW   !                           DD
4.                  P(a,a)->~P(a,a)         2, ∀ Instantiation
5.                  P(a,a)                          1, ∀ Instantiation
6.                  ~P(a,a)                       4,5 MP
7.                    !                                 5,6 !

Ground mereology accepts instead the weaker antisymmetry. Additionally, parthood is taken to be transitive as it is plausible any part x of y which is part of z entails x is also part of z.

Useful definitions can be constructed from this characterization of parthood. Two individuals are said to overlap if they share a part in common, are discrete if they do not overlap, and an individual is said to overlap the complement of another, if the first shares a part with the complement of the second.

Remarks in hand, return to Alex and Bob, denoting the first with “a” and the second with “b”, and the parthood between them as “P(a,b)”. Observe, parthood entails overlap for these individuals:

1.      P(a,b)                                    Premise
2.      SHOW ∃x P(x,a) & P(x,b)     DD
3.             P(a,a)                             Reflexivity
4.             P(a,a) & P(a,b)               1,3, CI
5.             ∃x P(x,a) & P(x,b)          4, ∃ Introduction

Line 5 reflects that “a” and “b” share a part in common. Hence, if Alex is part of Bob, then Alex overlaps Bob. The converse does not hold (Exercise: Find a countermodel).

Observe next, to say Alex and Bob are discrete, is to deny they overlap. Equivalently, it is to claim they have no common parts. Moreover, we would be saying of, say, Alex, that Alex overlaps some part of the complement of Bob, and vice versa. For Alex to overlap Bob’s complement is for there to be a part of Alex that is not a part of Bob.

1.      ~∃x P(x,a) & P(x,b)                           Premise
2.           SHOW∃x P(x,a) & ~P(x,b)           DD
3.                 ∀x P(x,a) -> ~P(x,b)              1, Substitution
4.                 P(a,a)                                     Reflexivity
5.                 P(a,a) -> ~P(a,b)                   3, ∀ Instantiation
6.                 ~P(a,b)                                   4,5 MP
7.                 P(a,a) & ~P(a,b)                     4,6 CI
8.                 ∃x P(x,a) & ~P(x,b)                7, ∃ Introduction

Hence, if Alex is discrete from Bob, then Alex overlaps the complement of Bob. Since both overlap and complement overlap are symmetric, we can say the same for Bob (Exercise: Prove overlap and complement overlap are symmetric).

Our brief foray into the mereology wilderness permits, given the assumptions with which we began, a square of individuals (cp. Square of Opposition). As is well-known, blindly translating categorical sentences into first-order notation undermines logical relations of the traditional square, as classical logic permits conditionals which are vacuously true. We avoid the problem of existential import by sticking with individuals. Hence, our square parallels the tradition:

Implication holds whenever the first is true then the second must be, and if the second is false so must the first be. Contraries are sentences which may both be false but which may not both be true. Contradictory sentences require that if one is true the other is false, and if one is false the other is true. Subcontraries are sentences which may both be true, but which may not both be false (Exercise: Verify the remaining corners).