You’d think that since P(Q1|~∀xQx) = 1⁄2 and P(Q1|∀xQx) = 1, observing Q1 is evidence in favor of ∀xQx.
And it is, but the hidden catch is that this depends on the implication that ∀xQx->Q1, and that implication is exactly the same amount of evidence against ∀xQx.
It’s also an amusing answer to the end of part 1 exercise.
This is counterintuitive in an interesting way.
You’d think that since P(Q1|~∀xQx) = 1⁄2 and P(Q1|∀xQx) = 1, observing Q1 is evidence in favor of ∀xQx.
And it is, but the hidden catch is that this depends on the implication that ∀xQx->Q1, and that implication is exactly the same amount of evidence against ∀xQx.
It’s also an amusing answer to the end of part 1 exercise.