As discussed, I have a new version which preserves the proof structure, but weakens the premises about as much as possible.

A1. The collection of all entities is a set E, with a causal relation C and a partial order P, such that x P y if and only if x is a part of y.

Note: This merges the assumption that P is a partial order into the overall set-up; that feature of P now gets used earlier in the argument.

A2. The set E can be well-ordered.

This ensures we can apply Zorn's Lemma when considering chains in E, but is not as strong as the full Axiom of Choice. If the set E is finite or countable, for instance, then A2 applies automatically.

Definitions: We define a relation C such that x C y iff there are entities v, w such that v P x, y P w and v C w.

Note: This gives a broader causal relation which automatically satisfies "if x C y and x P z then z C y" as well as "if x C z and y P z then x C y", loosely "anything which is caused by a part is caused by the whole" and "anything which causes the whole, causes the part". So we don't need to state those as extra premises.

We then define a further relation <= such that x <= y iff x = y, or there are finitely many entities x1, ..., xn such that x1 = x, xn = y and xi C* xi+1 for i=1.. n-1.

Note: This construction ensures that <= is a pre-order on E.

Say that a subset S of E is a "chain" iff for any x, y in S we have x <= y or y <= x. Say that S is an "endless chain" iff for any x in S there is some y in S distinct from x with y <= x. We shall say that y is "uncaused" if and only if there is no z in E distinct from y with z C* y (this of course implies there is no z distinct from y with z C y, but it also implies that y isn't part of anything which is caused by something distinct from y). Say that x is a proper part of y iff x is distinct from y but x P y.

A3. Let S be any endless chain in E; then there is some z in E such that every x in S is a proper part of z.

Lemma 1: For any chain S in E, there is an entity x in E such that x <= y for every y in S.

Proof: Suppose S is not endless. Then there is some x in S such that for no other y in S is y <= x. By the chain property we must have x <= y for every member y of S. Alternatively, suppose that S is endless, then by A3, there is some z in E of which every x in S is a part. Now consider any y in S. There is some x not equal to y in S with x <= y, so there are entities x = x1... xn = y with each xi C xi+1 for i=1..n-1. Further, as x P z we have z C x2 and hence z <= y.

Lemma 2: For any x in E, there is some y in E such that y <= x, and for every z <= y, we must have y <= z.

Proof: This follows from Zorn's Lemma applied to pre-orders.

Theorem 3: For any x in E, there is some uncaused y in E such that y <= x.

Proof: Take a y as given by Lemma 2 and consider the set S = {s: s <= y}. By Lemma 2, y <= s for every member of S, and if S has more than one element, then S is an endless chain. So by A3 there is some z of which every s in S is a proper part, which implies that z is not in S. But by the proof of Lemma 1, z <= y, which implies z is in S: a contradiction. So it follows that S = {y}, which completes the proof.

We now partition E into three subsets. I are the "inert" entities, which do not cause anything and have no causes themselves. (Note that the new version allows there to be some of these, unlike the previous version; you can think of them as abstract entities like numbers, sets, propositions and so on, if you want to). Formally I = {x in E: there is no y distinct from x with x C y or y C x}. U are the "uncaused causes" - formally U = {x in E: there is no y distinct from x with y C x, but there is z distinct from x with x C z}. O are all the "other", caused entities, so that formally O = {x in E: there is some y distinct from x with y C* x}.

B1. If S is any subset of U such that for any x, y in S we have x P y or y P x, (call such an S a "chain of parts"), then there is some entity z of which all members of S are parts.

B2. Suppose that y <= x and z <= x. Then there is some entity w such that: w <= x; w <= y or y P w; w <= z or z P w.

(EDIT: Restated to ensure that Theorem 4 properly follows.) Informally, the idea is that y and z can't independently cause x without any further causal explanation. So there must be some common cause, however each of them may be part of that common cause.

Definition: Say that entities x and y are causally-connected if and only if x=y or there are finitely-many entities x=x1,..,xn=y with each xi C xi+1 or xi+1 C xi for i=1..n-1.

B3. Any two entities x, y in O are causally-connected.

Informally, O doesn't "come apart" into disconnected components, such as a bunch of isolated universes. Premises B1-B3 turn out to be necessary for Theorem 6 to hold, as well as sufficient (see below). So they can't be made any weaker!

Theorem 4: For any x in O, there is a unique entity f(x) in U such that: f(x) <= x, and any other y in U with y <= x satisfies y P f(x).

Proof: For any x in O, define a subset U(x) = {y in U: y <= x}; this is non-empty by Theorem 3. Consider any chain of parts S that is a subset of U(x). If it has at least two members, then by B1 there is some z in E of which all members of S are parts, and such a z must be in U. (If not, then note any w C z would also satisfy w C s for each member s of S, which would require them all to be equal to w). Also since y <= x for any member of S and y P z we have z <= x. So z is also a member of U(x). Or if S is a singleton - say {z} - then clearly all members of S are parts of z, and z is also in U(x). By application of Zorn's Lemma to U(x), there is a P-maximal element f(x) in U(x) such that there is no other y in U(x) with f(x) P y. By B2, for any other y in U(x) there must be some z in U(x) with f(x) P z and y P z; given f(x) is maximal we have z = f(x) and so y P f(x). This makes f(x) the unique maximal element of U(x).

Theorem 5: For any x, y in O, f(x) = f(y) if and only if x and y are causally-connected.

Proof: It is clear that if f(x) = f(y) then x is causally-connected to y (just build a path backwards from x to f(x) and then forward again to y). Conversely, consider any two x, y in O:

a) If x C y, then f(x) is in U and satisfies f(x) <= y so we have f(x) P f(y). Since x is not f(x), we have f(x) C x2 <= x for some x2, and hence f(y) C* x2 <= x i.e. f(y) <= x which means f(y) P f(x), and so f(x) = f(y).

b) If z is in U with z C x and z C y, then z P f(x) so f(x) <= y and f(x) P f(y); similarly, f(y) P f(x) so that f(x) = f(y).

The result now follows by induction on the length of the causal path connecting x to y.

Theorem 6: If O is non-empty, then there is a single entity g in U such that: f(x) = g for every x in O, and y P g for every y in U.

Proof: Assuming O is non-empty, take any element y in O, and set g = f(y); then the result that f(x) = g for any x in O follows from Theorem 5 and B3; further, for any y in U, there is some x in O with y C* x, so by Theorem 4, y P f(x). If there are no elements of O (meaning there are none in U either) then the Theorem is trivial.

Finally, note that B1, B2 and B3 are entailed by the statement of Theorem 6. For B1, we can just take g as the relevant z. For B2, we can take g as the relevant w. B3 follows using using the first part of Theorem 5 (just track from x back to g, then forward to y again).

I'm just about done now, so unless there are errors in the above proof will leave it. What are the residual weak points? Well, B2 and B3 have been weakened a bit, but are still basically unjustifiable (we can imagine them being false without absurdity) and the above re-work shows they are needed for the uniqueness conclusion (Theorem 6). Also, we have the weakness of not deriving anything else useful about g.