A=Aa+e A-Aa=e A(1-a)=e A=e/(1-a) not e/(e-a) | Hacker News

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		emilfihlman on Nov 12, 2018 \| parent \| context \| favorite \| on: Derivatives of Regular Expressions (2007) A=Aa+e A-Aa=e A(1-a)=e A=e/(1-a) not e/(e-a)

dataflow on Nov 12, 2018 [–]

Nope, I'm pretty sure it's not wrong. I believe ϵ is the multiplicative identity ("1") here. Note that ϵ is the empty string, and concatenating it with anything results in the same thing, so Aϵ = A. Hence we have Aϵ - Aa = A(ϵ - a) = ϵ.

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