# In mereology, how are the product and sum axioms derived from the axiom schema of fusion or unrestricted composition?

I am stuck with a simple problem in mereology (the theory of parthood), where $$P x y$$ stands for $$x$$ is part of $$y$$ and $$O x y$$ stands for $$x$$ and $$y$$ overlap. From the axiom schema $$(\exists x) \phi x \rightarrow (\exists z)(\forall y)(O y z \leftrightarrow (\exists x)(\phi x \wedge O y x))$$, known as fusion or unrestricted composition, I am trying to derive the following, known as the sum and product axioms respectively:

$$(\exists z) (Pxz \wedge Pyz) \rightarrow (\exists z) (\forall w)(O w z \leftrightarrow (O w x \vee O w y))$$

$$(\exists z) (Pzx \wedge Pzy) \rightarrow (\exists z) (\forall w) (P w z \leftrightarrow (Pwx \wedge Pwy))$$.

I suppose it should be obvious, but what substitutions should I use for $$\phi$$?

It's assumed that $$P$$ is reflexive, antisymmetric and transitive, and that $$O x y \leftrightarrow (\exists z)(P z x \wedge P z y)$$. I shouldn't need to use the axiom of strong supplementation, $$\neg P y x \to (\exists z) (P z y \wedge \neg O z x)$$. But an answer which does use this axiom would be very helpful too.

The formulations are from Achille Varzi and Roberto Casati's Parts and Places (Cambridge, Mass: MIT Press, 1999), pp. 33-47. See also https://plato.stanford.edu/entries/mereology/

• What is mereology ? Dec 15 '17 at 6:43
• Mereology is an axiomatic theory of parthood.
– Ben
Dec 15 '17 at 7:10
• I think that Fusion must be: $(∃x)ϕx→(∃z)(∀y)(Oyz↔(∃x)(ϕx∧Oyx))$ Dec 16 '17 at 10:11

To answer my own question, to prove the sum axiom, let $\phi x$ be $x = a \vee x = b$. Then we have:

1. $(\exists x) (x = a \vee x = b) \rightarrow (\exists z)(\forall y)(O y z \leftrightarrow ((x = a \vee x = b) \wedge O y x))$

2. $(\exists z)(\forall y)(O y z \leftrightarrow ((x = a \vee x = b) \wedge O y x))$ from (1), because the antecedent is a tautology.

3. $(\exists z)(\forall y)(O y z \leftrightarrow (O a y \vee O by))$ from (2), since $(x = a \vee x = b) \wedge O y x)$ implies $Oya \vee Oyb$.

And this is the same as the sum axiom, except with different letters.

For the second question, let $\phi x$ be $P x a \wedge P x b$. Then we have:

1. $(\exists x) (P x a \wedge P x b) \rightarrow (\exists z)(\forall y)(O y z \leftrightarrow ((P x a \wedge P x b) \wedge O y x))$

which says that if $a$ and $b$ overlap there is a sum of all the things that are part of both of them. The problem is then just to show that this sum is also the product of $a$ and $b$. But doing this does require strong supplementation.

The proof turns out to be quite lengthy, but can be found in  p. 204, which also notes that the claim that the claim that the product axiom can be derived from weak supplementation alone is a common misconception in the literature.

 Carsten Pontow (2004) "A note on axiomatic theories of parthood" Data and Knowledge Engineering 50(2): 195-213.