5 ms·
No. x -> y && !x does not give !y, it just gives nothing. He proved y under some set of assumptions, and then later !y under another set of assumptions. So he
by nightcracker 7y ago
No. x -> y && !x does not give !y, it just gives nothing.
He proved y under some set of assumptions, and then later !y under another set of assumptions. So he proved a -> y && b -> !y.
- xamuel 7y agoBut, given x->y, if you prove !y, you can conclude !x, so in a sense kibibu is right (possibly requiring "proved" to be read as "indirectly proved"), and we should give them the benefit of the doubt.
- pdonis 7y ago> given x->y, if you prove !y, you can conclude !x But given x -> y, if you prove !x, you can't conclude anything. And kibibu said !x, not !y.
- kibibu 7y agoYeah, I interpreted: > quantum fields near the horizon can violate some of the assumptions that are required for the area theorem as !x