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Monday, September 03, 2012

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Does a cracker work as an analogy?

A cracker is a number of molecules taken collectively.
Each of these molecules are contingent.
It does not followw that because of this the cracker is contingent.
But nevertheless the cracker is contingent.

Does that work?

That works.


Dear Bill,

I have some comments on your exposition.

First: If your subtractive argument were valid, it would show that, after all, id does follow that if all members of a collection are contingent then the collection is contingent. Therefore, either your argument works or your claim that the contingency of the collection does not follow from the contingency of each of its member is false.

More detail: Let us assume what you had assumed, viz. that it does not follow that Universe is contingent if each of its members is contingent. That is to say that is is possible that the antecedent is true but the consequent is false. Moreover, since we are talking about S5-modalities and all truths about S5 modalities are S5-necessary, the fact that it is possible that the Universe is not contingent although each of its member is amounts to the fact that the Universe actually is not contingent, even assuming each of its members are (if the Universe is non-continget in a possible world, then it is non-contingent in all possible worlds). From that follows that your subtractive procedure cannot be gone through. There must be some point at which the subtraction of a one single member is impossible (although we are still assuming that all memebers are contingent).

Is this absurd? No. Even if a member is contingent in itself, it may be necessarily implied by some other memeber(s) and therefore incapable to be consistently "removed" in isolation.

However, further reflection shows that the "subtractive" argument can be amended to show evidently that a collection of which all members are contingent must be contingent as well. At each point of the subtractive sequence it holds that either it is possible to subtract a chosen memeber m or not. If it is not, then either because the non-existence of m is impossible in itself or incompatible with some necessarily existing being(s), or because the non-existence of m is incompatible with some contingently existing being(s). But the former alternative is excluded by the assumption that m is contingent, which would not be the case if the non-existence of m were impossible in itself or incompatible with something necessary. So the only reaosn why m may be unsubtractible can be that its non-existence would be incompatible with some contingent being(s). In that case, it will always be possible to subtract m plus all its ontological necessitants. Note that I do not need to iterate the reasoning at this point and argue to the impossibility of infinite regress. Given that the relation of ontological necessitation is transitive, all the second-order ontological necessitants of the first-order ontological necessitants of m are ontological necessitants of m as well and therefore are considered in the first step already, regardless whether they are finite or infinite in number. The bottom line is: If all memebers of the Universe are contingent, then there exists a series of possible worlds ordered by the described relation of "subtraction"such that the last member of this series is the world where the Universe does not exist (have members). In other words, the non-existence of the Universe is possible given that all its members are contingent.

We have proved that from the fact that all members of the Univerese are contingent it follows that the Universe is contingent. But what about the "fallacy of composition" objection? There is a sublte mistake in that objection. It is right that it would be false to claim that "For any property P and any collection C, if all the members of C are P then C is P" is a valid form of reasoning. But likewise it would be false to claim that for any P it holds that "For any C, if all members of C are P then C is P" is an invalid form of reasoning. For example, let us substitute "to be in this room" for P. Then "For any C, if all memebers of C are in this room, then C is in this room" is a valid form of reasoning. The same holds, I claim, for the property of contingency.

Finally, one comment concerning your metaphilosophical maxim. Clearly, the maxim cannot be strictly proven, if it is true. It seems therefore that its justification can be merely of inductive character. As such, however, it can hardly be used take into question the epistemical force of deductive arguments, since it will be always more dubious than these arguments themselves.

Lukas

Hi Lukas,

I don't understand your opening paragraph. My claim was that it does not follow just from the fact that each being is contingent that the collection is contingent. That would be the fallacy of composition. But saying that is consistent with saying that there is a valid argument from the contingency ofthe members to the contingency of the collection.

Perhaps you could clarify . . . .

I must be missing something. If there is a valid argument form A to B, then B follows just from A, doesn't it?

No. The inference to B may be mediated by auxiliary premises. There is an important difference between immediate and mediate inference.

Immediate: p; therefore p v q

Mediate: p v q; ~p; therefore q.

If additional premises are required, then I would say that the argument from A to B is not valid, as such. But (to address the real issue) my point is that in case of the argument in question, no additional premises are required (except perhaps some necessary prinicples which follow from anything, and so need not be expressly assumed).
Note that I am speaking about the (semantic) relation of entailment, not (necessarily) about '(syntactic) derivability in some (or any) formal axiomatic system.

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