parsed completely in isolation, so compilation starts by parsing all
In general, there probably isn't a simple, mechanical rule to prove that a
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This new approximation is vulnerable to the same fate as the a1 / b1 solution it replaced; that’s to say, we can keep incrementing K to conjure as many distinct 2-good pairs as we want. The proof doesn’t guarantee that it’s going to happen on any specific cadence, but it says that it will if we try long enough.