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You are right, they are different questions, but the straw man was not intentional, I thought the original phrasing was ambiguous enough that it could be interpreted in both ways ;)

In other words, it was unclear to me what the answer to the question "Are the assumptions part of the hypothesis?" was. If, as I did, we assume that "yes, they are" then I don't think it follows that the probabilities will both be `1`, because we do not have logical proofs for the claims, the implication could only be true in the model (they are not necessarily entailments).

The waters are muddied further still when the hypothesis itself is phrased as an implication.

EDIT

It also strikes me that for your line of reasoning to hold, it is not sufficient that Pd1 = Pd2 are small, but instead `Pd1 = 0 = Pd2`, in order to justify this line:

    > = Pd1 + P((A1 & A2 & ... & Ak) & H1)
    > = Pd1 + P(A1 & A2 & ... & Ak)
Which is tantamount to saying

    (A1 & A2 & ... & Ak) <-> H1
  & (A1 & A2 & ... & Ak & ... & An) <-> H2
Is it not?

EDIT (2)

Ignore that, it is not tantamount, it is a weaker condition.




Maybe the final part of the proof I gave is a bit dense, so here are some additional notes.

First of all, if you know that

    (A1 & A2 & ... & Ak) -> H1
then the following two terms are logically equivalent:

    A1 & A2 & ... & Ak
    (A1 & A2 & ... & Ak) & H1
Also, for the proof which I gave it is sufficient that Pd1 = Pd2. It does not need them to be zero.


Ah yes, I see. I guess I was looking for a place where the fact that `Pd1` and `Pd2` were small, but I guess that's not necessary.


No, the "both are small" was just meant to be a justification for assuming Pd1=Pd2.




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