Louis Wasserman

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That might actually be easier to prove by working directly with classical-Fin and constructing a new equivalence, rather than the preexisting product-Fin...

I'm uncertain. I am mostly planning to go from Bishop's proof in Foundations of Constructive Analysis, and I don't see any choice principles in that proof yet; he accepts countable...

I see now that it uses dependent choice, actually.

I'm excited to see this work; I attempted once or twice to tackle the commutative semiring of cardinalities but didn't get very far.

Is this theorem on the 100 Theorems list? If not, I feel less urgency about the classical formulation.