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May 12, 2021 at 17:42 comment added Peter LeFanu Lumsdaine @MikeShulman: and indeed the D–G–M proof can usefully be factored as “If all subsets of A satisfy choice, then A has decidable equality” (which doesn’t need general quotients/suspension, just propositional truncation) and “if all sets have decidable equality, then LEM holds” (using set quotients/suspension as you say).
May 12, 2021 at 15:51 comment added Mike Shulman Note that $\Sigma A$ is the same set that appears in the Diaconescu-Goodman-Myhill proof that the axiom of choice implies excluded middle.
May 12, 2021 at 15:46 history answered aws CC BY-SA 4.0