Timeline for Why would the existence of a sufficiently strong PRNG prove P=BPP?

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Jul 9 at 10:36	comment	added	rus9384		But in BPP every instance has >50% chance to be correct. A BPP machine is not required to return the same result on the same input every time, unlike a P machine. We already know that $\mathsf{BPP\subset P}/poly$, which means there exists a polynomially (in dependence on $n$) sized set of random seeds that solves the problem correctly. The issue is computing this set without advice.
Jul 8 at 19:56	comment	added	Mike Battaglia		I agree it isn't meaningful - that's the point, that being in $P$ is a much stronger statement. Your latter statement seems to be the exact thing I am talking about in the last two paragraphs of my question, right?
Jul 8 at 19:52	history	edited	Mike Battaglia	CC BY-SA 4.0	latex formatting
Jul 8 at 18:53	comment	added	rus9384		I am saying that even $\mathsf{REG_{\text{only correct on 51%/99%/etc of instances}}=ALL}$. That is, it's not really meaningful. Anyway, you can look at $\mathsf{BPP}$ as a machine/RNG that selects one of the exponentially many $\mathsf P$ machines, to solve a problem. And an exponential number of these $\mathsf P$ machines have to be correct for any given instance. Now, a strong enough PRNG would allow you to replace the $\mathsf{BPP}$ selector with a $\mathsf P$ selector.
Jul 8 at 18:05	history	edited	Mike Battaglia	CC BY-SA 4.0	correct typo involving equality vs subset inclusion
Jul 8 at 2:47	answer	added	D.W.♦		timeline score: 1
Jul 7 at 15:33	history	asked	Mike Battaglia	CC BY-SA 4.0