Some Reduction Theorems on "Why?" Questions

Geez, Peter Lipton is a moron.

He goes on and on about explaining P alone, as opposed to explaining P-rather-than-Q, even though in the previous sentence he's admitted that he has no story about what he even means by explaining P alone (211).  Presumably he means giving a factor statistically relevant to P; he speaks of certain "intuitive evidence" about our explanations of P (simpliciter) -- where we simply ask "Why P?"

He's wrong to think there is such a thing as giving an explanation of P -- as opposed to a contrastive explanation (of P-rather-than-Q).  Consider some indirect questions:

(1) I wonder whether Rex will see the chick flick or the kung fu movie.

(2) I wonder whether Rex will see the chick flick.

Now while not a "Why?" question, (1) is clearly considering two possibilites: Rex seeing the chick flick, and Rex seeing the kung fu movie; and I am wondering which of the two will be the case [hence assuming the  exclusive reading of "or"].  (2) doesn't seem to offer two different options, in the way (1) does.  But that appearance is mistaken:  (2) is asking which of these two possibilities will be the case:

(a) Rex will see the chick flick.
(b) Rex will not see the chick flick.
This is easy to see;  for another way of asking (2) would be 2*:

(2*) I wonder whether or not Rex will see the chick flick.

Moral: when we ask "whether P," with no second option, we are asking "whether P or not-P".  When there's no second option, the negation of the first is the default.

Same with "Why?" questions [examples]

(3) I wonder why Rex saw the chick flick rather the kung fu movie.
(4) I wonder why Rex saw the chick flick.

So all "Why?" Q's are contrastive, contra Lipton.

Now note: in such cases -- asking "Why P (rather than  not-P)," VF's favoring collapses into positive relevance.

Notice also Temple's suggestion (which Lipton objects to) that to ask "Why P (rather than Q)?" is just to ask "Why P and not Q?" (which is prima facie very nice, since the two "Why?" Q's do seem synonymous).  Temple's suggestion is that we take the topic (P) and conjoin to it the negation of each of the remaining members of the contrast class.   Once we eliminate Lipton's distinction between contrastive and non-contrastive "Why?" questions, we can test it out on the default case -- "Why P?".  On my reading of this (as using the default CC "rather than not-P," how will the question come on under Temple-transformation?  Like this:

(5) Why P and ~~P?

which logically reduces to just 

(6) Why P?

Ironically, Lipton's erroneous postulation of non-contrastive "Why?", and his erroneous denial of Temple's rephrasing of the contrast class, have cancelled each other out, ending up with (my reconstruction of) his original view: there can, after all, be such things as "Why P?" simplicter.

Theorems following from this model.

(T1) The question "Why is there something rather than nothing" reduces to "Why is there something?"

(T2) On my introduction of the default contrast class, we see a link bet. VF's favoring and simple positive relevance..  On VF's favoring, the explanandum makes the explanas (topic) likelier than the other members of the CC.  [But then favoring P over ~P is just increasing the probability of P (over 50%). (True?)]

So X favors E over ~E iff:

P(E&X) > P(~E&X)

Then: 

(T3) Favoring P over Q, R is just positive relevance for P&~Q&~R.  

But now keep in mind it has to be the cause favoring the effect w/respect to other members of the CC.  Doubts about the above results might be dispelled in working this out.
(VF, 1980: 27):

P(X, Y) = P(X&Y)/P(Y)

X is positively relevant to E iff:
P(E, X) > P(E)

Hence 

X is positively relevant to E iff:

P(E&X) > P(E) x P(X)

X favors E over F iff: P(E, X) > P(F, X)

So X favors E over F iff:

P(X, Y) = P(X&Y)/P(Y)

P(E&X)/P(X)  >  P(F&X)/P(X)

that is,

X favors E over F iff:

P(E&X) > P(F&X)
(though P(E&X) may not be > P(E). )

Now according to VF, if X favors E over F, then X tells me why E rather than F ; and according to my reduction principle, if X tells me E rather than F, then X   is positively relevant to E&~F.  So:
if X favors E over F, then X is positively relevant to   E&~F.

We can state that formally, and see if it's true:

(i) X favors E over F: P(E&X) > P(F&X)
(ii) X is positively relevant to E&~F: 
P(E&~F, X) > P(E&~F),
that is,
P(E&~F&X) > P(E&~F) x P(X)

Now in the abstract, 
P(Y&Z) = P(Y) x P(Z)
and 
P(~F) = 1-P(F)

Prove it both ways:
Suppose (1) P(E&X) > P(F&X)
[Get: (2) P(E&~F&X) > P(E&~F) x P(X)
That is,
(2*) P(E&~F&X)/P(X) > P(E&~F)
That is, 
(2**) P(E&~F, X) > (P(E&~F)]
Reductio: Suppose not; then either 
(i) P(E&~F&X) = P(E&~F) x P(X)
or 
(ii) P(E&~F&X) < P(E&~F) x P(X)

Let P(E&~F&X) = P(E) x P(~F) x P(X)
Let P(E&X) = P(E) x P(X)
Let P(

Suppose (1) P(E&X) > P(F&X)
[Get: (2) P(E&~F&X) > P(E&~F) x P(X)
From (1): P(E) x P(X) > P(F) x P(X)
Then P(E) > P(F)
(i)If P(F) = P(~F), then P(E) > P(~F) and P(E) > .5
(ii)If P (F) > P(~F) then P(E) > P(~F) and P(E)> .5
(iii)If P (F) < P(~F) then P(~F) > .5 , we don't know its relation to P(E) .
(Suppose P(E) = .6 and P(F) = .45; then P(E) > P(~F)
Suppose P(E) =.5 and P(F) = .2; then P(E) < P(~F)
Suppose P(E) = .6 and P(F) = .4; then P(E) = P(~F))

But we know that either P(E) > .5 or P(~F) > ,5.
That is, (iiia) P(E) > P(~F) and 2P(E) > 1 
or P(iiib) (F) < P(~F) and 2P(~F) > 1

(iiia) P(E) > P(~F) and 2P(E) > 1
Then 2P(E) x P(X) > P(X)
Then 2P(E) x P(X) x P(~F) > P(X) x P(~F)
And since P(E) > P(~F), 
2P(E) > P(E) x P(~F)
2P(E) x P(X) > P(E) x P(~F) x P(X)
And by hypothesis, P(E) > P(F)
So P(E) x P(X) > P(F) x P(X)
And from 2P(E) > 1, we get P(E) > P(~E)
So P(E) > [1 - P(E)]
So 1 > [1/P(E) - 1]

Then P(E) < [1 - P(~F)]
Then P(E) x P(X) < P(X) x [1 - P(~F)]
Then P(E) x P(X) < P(X) - [P(X) x P(~F)]
Then [P(E) x P(X)] + [P(X) x P(~F)] < P(X)
Then [P(E) + P(~F)] x P(X) < P(X)
Then P(E) + P(~F) < 1
Cases: 

[Axiom: P(E) = P(E&F) +P(E&~F)
Axiom: P(E&F) = P(E) x P(F)]