Maybe the final part of the proof I gave is a bit dense, so here are some additi... | Hacker News

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		vog on July 6, 2016 \| parent \| context \| favorite \| on: Mental Models I Find Repeatedly Useful 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.

asQuirreL on July 6, 2016 [–]

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.

vog on July 6, 2016 | [–]

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

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