Let’s consider loops A->B->C->A->B->C and D->E->F->D->E->F.
Let’s say, further, that B is a cause of E and D is a cause of A. Then each loop has an external cause.
Then there are also a few other loops possible:
A->B->E->F->D->A->B->E->F->D (external cause: C)
A->B->E->F->D->A->B->C->A->B->E->F->D… huh. That includes all of them, in a sort of double-loop with no external cause. I guess that would be the super-loop.
Finally, I rather liked your thought that causality may be so loopy that everything is a cause of everything else. The only way to get a first cause out of that mess is to treat the entire “super-duper-loop” of all things as a single uncaused entity, and if you insist on calling that “God”, you’re a pantheist.
Better yet; no matter what causality looks like, you can still always combine everything into a single giant, uncaused entity. You don’t need to assume away loops or infinite chains without external causes if you do that.
I’ve been doing a bit more “stir in fancy set theory” over the weekend, and believe I have an improved recipe! This builds on the idea to treat chains and loops as a single “entity” and look for a cause of that entity. It is a lot subtler than just throwing every entity together into one super-duper-entity.
Here are a bunch of premises that I think will do the trick:
A1. The collection of all entities is a set E, with two relations C and P on E, such that: x C y if and only if x is a cause of y; x P y if and only if x is a part of y.
Note: 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.
A3. If x C y and x P z then z C y.
Informally, “anything caused by a part is caused by the whole”.
Definitions: We define ⇐ such that x ⇐ y if and only if x = y or there are finitely many entities x1, …, xn such that x1 = x, xn = y and xi is a cause of xi+1 for i=1.. n-1. Say that a set S is a “chain” in E iff for any x, y in S we have x ⇐ y or y ⇐ x. Say that such an S is an “endless chain” iff for any x in S there is some y not equal to x in S with y ⇐ x. Say that an entity y is “uncaused” if and only if there is no z distinct from y with z ⇐ y. Also say that x is a “proper part” of y iff x is not equal to y but x P y.
Note: These definitions ensure that ⇐ is a pre-order on E. Note that an endless chain may be an infinite chain of distinct elements, or a causal loop.
A4. 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 element x of E with x ⇐ y for every y in S.
Proof: Suppose S has an end (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 A4, there is some z in E such that every x in S is a part of z. Now consider any y in S. There is some x not equal to y in S with x ⇐ y, so there are x = x1… xn = y with each xi C xi+1 for i=1..n-1. Further, by A3, as x C x2, 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 any z ⇐ y, y ⇐ z.
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 A4 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.
I’ve also got some premises for aggregating multiple uncaused entities into a single entity. This gives another approach to “uniqueness”. More on my next comment, if you’re interested.
For uniqueness, we build on the idea of all uncaused causes being part of a whole. The following premises look interesting here:
B1. If x P y and y P z then x P z; x = y if and only if x P y and y P x.
This states that P is a partial order, which is reasonable for the “part of” relation.
B2. If S is any chain of parts, such that for any x, y in S we have x P y or y P x, then there is some z in E of which all members of S are parts.
This states that E is inductively ordered by the “part of” relation.
B3. If x C z and y P z then x C y.
Informally, “a cause of the whole is a cause of any part”.
B4. Suppose that y ⇐ x and z ⇐ x and both y, z are uncaused. Then y P z or z P y, or there is some w of which both y and z are proper parts.
Informally, two uncaused y and z can’t independently conspire to cause x unless they are parts of a common entity.
Definition: Say that entities x and y are causally-connected if and only if x = y, or there are entities x=x1,..,xn=y with either xi C xi+1 or xi+1 C xi for each i=1..n-1.
B5. Any two entities in E are causally-connected.
Informally, E doesn’t “come apart” into completely disconnected components, such as a bunch of isolated universes.
Theorem 4: For any x in E, there is a unique entity f(x) in E such that: f(x) is uncaused, f(x) ⇐ x, and any other uncaused y with y ⇐ x satisfies y P f(x).
Proof: For any x, define a subset E’ = {y in E: y ⇐ x, y is uncaused}. Consider any chain of parts S in E’ with at least two elements. By B2 there is some z in E of which all members of S are parts. By B3, z must be uncaused (or else some w C z would also be a cause of all the members of S, which would require them all to be equal to w, so S would be a singleton), and by A3, z ⇐ x. So z is also a member of E’. By application of Zorn’s Lemma to E’, there is a P-maximal element f in E’ such that there is no other y in E’ with f P y. But then, by B4, for any y in E’ we must have y P f; this makes f unique.
Theorem 5: For any x, y in E, 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, suppose that x C y, then f(x) is uncaused and satisfies f(x) ⇐ y so we have f(x) P f(y). This implies f(x) = f(y). By a simple induction on n we have that if x is causally-connected to y, then f(x) = f(y).
Corollary 6: There is a single entity g in E such that f(x) = g for every entity x in E.
(Huh. One of the ancestors to this comment—several levels up—has been downvoted enough to require a karma penalty. I wonder if there should be some statute of limitations on that; whether, say, ten levels of positive-karma posts can protect against a higher-level negative-karma post?)
A4. 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.
An interesting assumption. Necessary for theorem 3, but I suspect that it’ll mean that the original cause described in theorem 3 will then very probably be an entity z that is the earliest cause.
I also note that, while z consists of all the parts in the endless chain, there is no guarantee that any of the elements in the chain, even those that cause other elements in the chain, is in any way a cause of z. In fact, the way that z is defined, z may well be causeless (or, then again, z may have a cause). While I can’t actually find anything technically invalid in theorem 3, or in assumption 4, I get the general feeling of wool being pulled over my eyes in some way.
When I consider B3, it becomes even more important to note that z as a whole is not necessarily caused by any element that is a proper part of z. The cause of a part may or may not be the cause of the whole.
Hmmm… B4 appears to be pretty much just shoehorning monotheism in. It seems a questionable assumption; if I decide to get into my car and drive, and you decide to get into your car and drive, and we drive into each other, then we both are causes of the resultant accident but we are not the same. (We are not causeless, either, so it’s not quite a counterexample, just an explanation of why I don’t think B4 is justified,) B5 is unsupported, but I can prove that all entities that I will ever observe evidence of are causally connected (i.e. they are connected to the effects on my actions of having observed them) so it will look true whether it is or not.
Though I can raise questions about your assumptions, I can’t find anything wrong with your logic from then on. So congratulations; you have a very convincing argument! …as long as you can persuade the other person to accept your assumptions, of course.
Let’s consider loops A->B->C->A->B->C and D->E->F->D->E->F.
Let’s say, further, that B is a cause of E and D is a cause of A. Then each loop has an external cause.
Then there are also a few other loops possible:
A->B->E->F->D->A->B->E->F->D (external cause: C) A->B->E->F->D->A->B->C->A->B->E->F->D… huh. That includes all of them, in a sort of double-loop with no external cause. I guess that would be the super-loop.
Better yet; no matter what causality looks like, you can still always combine everything into a single giant, uncaused entity. You don’t need to assume away loops or infinite chains without external causes if you do that.
I’ve been doing a bit more “stir in fancy set theory” over the weekend, and believe I have an improved recipe! This builds on the idea to treat chains and loops as a single “entity” and look for a cause of that entity. It is a lot subtler than just throwing every entity together into one super-duper-entity.
Here are a bunch of premises that I think will do the trick:
A1. The collection of all entities is a set E, with two relations C and P on E, such that: x C y if and only if x is a cause of y; x P y if and only if x is a part of y.
A2. The set E can be well-ordered
Note: 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.
A3. If x C y and x P z then z C y.
Informally, “anything caused by a part is caused by the whole”.
Definitions: We define ⇐ such that x ⇐ y if and only if x = y or there are finitely many entities x1, …, xn such that x1 = x, xn = y and xi is a cause of xi+1 for i=1.. n-1. Say that a set S is a “chain” in E iff for any x, y in S we have x ⇐ y or y ⇐ x. Say that such an S is an “endless chain” iff for any x in S there is some y not equal to x in S with y ⇐ x. Say that an entity y is “uncaused” if and only if there is no z distinct from y with z ⇐ y. Also say that x is a “proper part” of y iff x is not equal to y but x P y.
Note: These definitions ensure that ⇐ is a pre-order on E. Note that an endless chain may be an infinite chain of distinct elements, or a causal loop.
A4. 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 element x of E with x ⇐ y for every y in S.
Proof: Suppose S has an end (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 A4, there is some z in E such that every x in S is a part of z. Now consider any y in S. There is some x not equal to y in S with x ⇐ y, so there are x = x1… xn = y with each xi C xi+1 for i=1..n-1. Further, by A3, as x C x2, 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 any z ⇐ y, y ⇐ z.
Proof: This is the version of 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 A4 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.
I’ve also got some premises for aggregating multiple uncaused entities into a single entity. This gives another approach to “uniqueness”. More on my next comment, if you’re interested.
For uniqueness, we build on the idea of all uncaused causes being part of a whole. The following premises look interesting here:
B1. If x P y and y P z then x P z; x = y if and only if x P y and y P x.
This states that P is a partial order, which is reasonable for the “part of” relation.
B2. If S is any chain of parts, such that for any x, y in S we have x P y or y P x, then there is some z in E of which all members of S are parts.
This states that E is inductively ordered by the “part of” relation.
B3. If x C z and y P z then x C y.
Informally, “a cause of the whole is a cause of any part”.
B4. Suppose that y ⇐ x and z ⇐ x and both y, z are uncaused. Then y P z or z P y, or there is some w of which both y and z are proper parts.
Informally, two uncaused y and z can’t independently conspire to cause x unless they are parts of a common entity.
Definition: Say that entities x and y are causally-connected if and only if x = y, or there are entities x=x1,..,xn=y with either xi C xi+1 or xi+1 C xi for each i=1..n-1.
B5. Any two entities in E are causally-connected.
Informally, E doesn’t “come apart” into completely disconnected components, such as a bunch of isolated universes.
Theorem 4: For any x in E, there is a unique entity f(x) in E such that: f(x) is uncaused, f(x) ⇐ x, and any other uncaused y with y ⇐ x satisfies y P f(x).
Proof: For any x, define a subset E’ = {y in E: y ⇐ x, y is uncaused}. Consider any chain of parts S in E’ with at least two elements. By B2 there is some z in E of which all members of S are parts. By B3, z must be uncaused (or else some w C z would also be a cause of all the members of S, which would require them all to be equal to w, so S would be a singleton), and by A3, z ⇐ x. So z is also a member of E’. By application of Zorn’s Lemma to E’, there is a P-maximal element f in E’ such that there is no other y in E’ with f P y. But then, by B4, for any y in E’ we must have y P f; this makes f unique.
Theorem 5: For any x, y in E, 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, suppose that x C y, then f(x) is uncaused and satisfies f(x) ⇐ y so we have f(x) P f(y). This implies f(x) = f(y). By a simple induction on n we have that if x is causally-connected to y, then f(x) = f(y).
Corollary 6: There is a single entity g in E such that f(x) = g for every entity x in E.
Proof: This follows from Theorem 5 and B5.
Done!
(Huh. One of the ancestors to this comment—several levels up—has been downvoted enough to require a karma penalty. I wonder if there should be some statute of limitations on that; whether, say, ten levels of positive-karma posts can protect against a higher-level negative-karma post?)
An interesting assumption. Necessary for theorem 3, but I suspect that it’ll mean that the original cause described in theorem 3 will then very probably be an entity z that is the earliest cause.
I also note that, while z consists of all the parts in the endless chain, there is no guarantee that any of the elements in the chain, even those that cause other elements in the chain, is in any way a cause of z. In fact, the way that z is defined, z may well be causeless (or, then again, z may have a cause). While I can’t actually find anything technically invalid in theorem 3, or in assumption 4, I get the general feeling of wool being pulled over my eyes in some way.
When I consider B3, it becomes even more important to note that z as a whole is not necessarily caused by any element that is a proper part of z. The cause of a part may or may not be the cause of the whole.
Hmmm… B4 appears to be pretty much just shoehorning monotheism in. It seems a questionable assumption; if I decide to get into my car and drive, and you decide to get into your car and drive, and we drive into each other, then we both are causes of the resultant accident but we are not the same. (We are not causeless, either, so it’s not quite a counterexample, just an explanation of why I don’t think B4 is justified,) B5 is unsupported, but I can prove that all entities that I will ever observe evidence of are causally connected (i.e. they are connected to the effects on my actions of having observed them) so it will look true whether it is or not.
Though I can raise questions about your assumptions, I can’t find anything wrong with your logic from then on. So congratulations; you have a very convincing argument! …as long as you can persuade the other person to accept your assumptions, of course.