The
Dude, if you have one group doing something and another doing something, they are doing something. I didn't know you'd play a game that's idiotic — Gregory
If logic has an inherent contradiction, and math is based on logic, than math is self refuting — Gregory
The barber (B), a philosopher (T) who doesn't shave himself, and a mathematician (M) who does.
We have all and only men who shave all and only men who don't shave themselves.
1. M is never a member of R because he shaves a man who shaves himself.
2. P can't be a member either because he doesn't shave himself, so he'd have to shave himself to be a member, but he doesn't.
3. What about B? He would have to shave P and not M. No problem. If he shaves himself, he'd be out, like M, but if he doesn't, he'd be out like P. So B can't be a member no matter what he does.
So R = { }. No one shaves all and only men who do not shave themselves, therefore the barber does not shave all and only men who do not shave themselves. The three cases are exhaustive, in fact: no one can be a member of R whether they shave themselves or not.
— Srap Tasmaner"
Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.