There probably exists a necessary, external cause to the universe

Thanks Pro! I hope you enjoy this debate!First for the introductory definitions. i) S = everything that exists or has existed. ii) --> on S = if some y-->x then y is a cause of x. iii) x is at least one of the causes of x for all x in S iv) :. --> is transitive. v) If x-->y and y-->x then x=y. vi) :. --> is a partial ordering of S. vii) An External, Necessary cause is defined as an element G of S with the property that if X is another element of S such that X-->G, it must be the case that X=G. (G has no cause and cannot not be) viii) Creation occurs when G-->X. Now, onto the argument. Premiss 1: The set S is a set in a universe of sets that satisfies the axiom of choice Premiss 2: The set S is inductively ordered. Theorem: If Premisses 1 and 2 hold, then an external, necessary cause of S exists. Defence of Premise 1 This premise is uncontroversial, as it would be hard to deny that S exists, and it would be equally as hard to deny that the axiom of choice to hold, which states that the cartesian product of a collection of non-empty sets is non-empty - i.e., for every set whose elements are a disjoint (have no element in common) collection of non-empty sets, it is possible to choose another set which contains precisely one element from each of the sets in the original set. Why should we believe this? Well, it seems self-evident. Imagine there is a set of boxes, each with 100 items in them, and there is a non-empty subset in each box (each box has more than one item). The axiom of choice will say that it is metaphysically possible to pick exactly one item out of each box. Simple, right? Now, simply apply this principle to S, and premise 1 is established. Defence of premise 2 This states that in some chain C in S, there exists some X in S such that X is the cause of everything in C. Some chain C in S means that there is a set in S such that if x and y are distinct elements of C, then either x-->y or y-->x. Thus, C is totally ordered by causation. In other words, imagine that a causes b, b cases c, c causes d and so on in an ordered chain. If this chain is inductively ordered, then a will be the cause of everything else in the chain in a causal sequence. [1] Proof: The most basic view of dependence states that a is the cause of b iff if a were not to exist, then neither would b. [2] If so, then in any ordered chain (i.e., the universe), if a were not to exist, then neither would b,c,d...., quod erat demonstrandum. But even if we reject this view of dependency (but I see no reason to), I fail to see any other view of dependence which does not imply the same principle. So, this premise must be accepted. Proof of Theorem Lemma 1 (Zorn's Lemma) Every partially ordered set P has the property that every chain (i.e. totally ordered subset) has an upper bound. Then the set P contains at least one maximal element. Now, replace P with the inductively ordered (partially ordered) set S in premise 1. Then, S contains at least one maximal element. In this case, the maximal element must cause every other member of S. Moreover, to avoid a vicious causal regress, this must result in G-->G and thus G is necessary. So we have a necessary being who is the cause of all members of S. Proof of Lemma 1: For a reductio, assume Zorn's Lemma is false. Then, there exists a partially ordered set or poset such that for every totally ordered subset, every element has an upper bound ad infinitum. Therefore, we need to define each element a1<a2<a3<a4..... In the set. But this will include not only all natural numbers but also all ordinals (a proper class) in the set, and consequently, there will be more ordinals than elements in the set and thus a contradiction arises. [3] Conclusion I have shown that, via certain axioms of maths, that the existence of a necessary first cause must be admitted. For if there exists an inductively ordered set where the axiom of choice applies, then via Zorn's Lemma, there must exist a maximal element of that set. This is equivalent to an external, necessary cause of S, which is equivalent to the universe.Sources: [1] http://mathworld.wolfram.com...[2] http://plato.stanford.edu...[3] http://en.wikipedia.org...'s_lemma