= Idempotent-ultrafilter star-set lemma
{title2=$A^*=\{n\in A:A-n\in\mathcal U\}$}
For $A\in\mathcal U$ and an <idempotent ultrafilter>, put $A^*=\{n\in A:A-n\in\mathcal U\}$. Then $A^*\in\mathcal U$, and $A^*-n\in\mathcal U$ for every $n\in A^*$. For the second assertion, apply idempotence to $A-n$ and intersect its resulting good-translation set with $A-n$. This allows each new finite-sums generator to be chosen from finitely many translation constraints, proving the <Idempotent-ultrafilter proof of Hindman's theorem>.
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