Help Wanted – Part 4

A key premise in Swinburne’s (deductive) argument in defense of his inductive version of the Cosmological argument (TCA) goes like this:

(TCA9) The probability that there will be a complex physical universe given that God does not exist is low. (EOG, p.151)

Swinburne has provided an explanation of his reasoning in support of this premise (in email dated 10/24/11), and I am working my way through that explanation.P(e &~h&~c&k;) will be the probability that a complex physical universe exists without an explanation. By the calculus this equals P(eI~h&~c&k;) P(~h&~c&k;).
[Swinburne:] Let c be ‘there is a personal creator other than God’. Then (given the sentence on p.149, ‘e could not, as we have seen..’), with k as a mere tautology,

See my previous post:

https://secularoutpost.infidels.org/2011/11/help-wanted-part-3_06.html

[Swinburne:] I have argued that c is much less probable than h. If (see p.109) P(hIk) is very low, P(cIk) will be even lower.

Swinburne argues that the a priori probability of the existence of God (a person with infinite power, infinite knowledge, and unlimited freedom) is greater than the a priori probability of the existence of one or more limited deities, because of the greater simplicity of the hypothesis of God as compared with the hypothesis of one or more limited gods.

God’s existence is a contingent fact, and the a priori probability of the existence of anything that is contingent is very low according to Swinburne. Thus, the a priori probability of God’s existence is “very low”. As a general rule, in trying to clarify Swinburne’s reasoning, I interpret “very low” probability to mean: greater than 0 but less than .2:

0 < P(hIk) < .2

Since Swinburne says “c is much less probable than h” his statement that “P(cIk) will be even lower” should be understood as meaning significantly lower. I suggest that this be quantified as
‘P(cIk) is greater than 0 but no more than half of P(hIk)’ :

0 < P(cIk) ≤ .5 x P(hIk)

So, for example, if P(hIk) = .1, then at most P(cIk) = .05.

[Swinburne:] In that case P(~h&~c&k;) will be close to 1, and so
P(e&~h&~c&k;) will equal approx P(eI~h&~c&k;)…

The claim that “P(~h&~c&k;) will be close to 1” seems plausible, but needs clarification and a proof.
The validity of the inference that Swinburne makes from this assumption can be partly confirmed by proving the following conditional statement:

If P(~h&~c&k;) = 1, then P(e&~h&~c&k;) = P(eI~h&~c&k;).

One can prove a conditional statement by assuming the antecedent and deriving the consequent.

1. [suppose] P(~h&~c&k;) = 1…………………..supposition for conditional derivation
2. P(AIB) = P(A&B;) / P(B)………………………….conditional probability formula
3. P(eI~h&~c&k;) = P(e&~h&~c&k;) / P(~h&~c&k;)…2, an instance of the formula
4. P(eI~h&~c&k;) = P(e&~h&~c&k;) / 1…………..1, 3 substitution of equals
5. a/1 = a…………………………………………………………see proof below
6. P(e&~h&~c&k;) / 1 = P(e&~h&~c&k;)………….5, an instance of the formula
7. P(eI~h&~c&k;) = P(e&~h&~c&k;)………………..4,6 transitivity of equality
8. P(e&~h&~c&k;) = P(eI~h&~c&k;)………………..7, symmetry of equality
9. If P(~h&~c&k;) = 1, then P(e&~h&~c&k;) = P(eI~h&~c&k;)…1-8 conditional derivation

Show: a/1 = a

1. c/d x d = c…………………………………already proved (see previous post).
2. a/1 x 1 = a……………………………………..1, an instance of the formula
3. a x 1 = a…………………………………………identity element for multiplication
4. a/1 x 1 = a/1…………………………………..3, an instance of the formula
5. a/1 = a/1 x 1…………………………………..4, symmetry of equality
6. a/1 = a……………………………………………5,2 transitivity of equality

To be continued…

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h: God exists.
e: There is a complex physical universe.
c: There is a personal creator other than God.
k: [tautological background knowledge]