undefined | Better HN

0 pointsvog9y ago0 comments

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.

0 comments

2 comments · 1 top-level

asQuirreL9y ago· 1 in thread

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.

vogOP9y ago

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

j / k navigate · click thread line to collapse