Ah, yes, that is almost correct. You need redundancy over TWO distinct observables (i.e. the subset must be at least size two), not just one, but otherwise yes. With just one observable, you don’t have two branches to dangle a Λ′ off of in the graphical proof, so we can’t get Λ between two copies of Λ′.
Ah, yes, that is almost correct. You need redundancy over TWO distinct observables (i.e. the subset must be at least size two), not just one, but otherwise yes. With just one observable, you don’t have two branches to dangle a Λ′ off of in the graphical proof, so we can’t get Λ between two copies of Λ′.