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I don't see how the conclusion must be that he's the Pope. It containing just him and the Pope but actually only just him is a contradiction that could just as

I don't see how the conclusion must be that he's the Pope. It containing just him and the Pope but actually only just him is a contradiction that could just as easily be resolved by saying "it contains me and the Pope but actually just me so there is no Pope." Even if I accept the false proposition, the conclusion doesn't necessarily follow.

_ZeD_ 6y ago
it's not that difficult to follow: let B = {set made of "Bertrand Russell" }, P = {set made of the pope} let BP = B ∪ P |B| = 1, |P| = 1 (trivials) |BP| may be 1 or 2 if |BP| = 1 it means B=P -> Bertrand Russell is the Pope if |BP| = 2, and 2=1, it means B=P -> Bertrand Russell is the Pope
- rembrandt 6y ago
  But since |BP| is both 1 and 2 it means Bertrand Russell is both the Pope, and not the Pope.
  - unanswered 6y ago
    ... yes. That is indeed what is meant by ex falso quodlibet. It does not make the example invalid.
    - DangitBobby 6y ago
      The point was merely that it cannot be used to prove "anything" since the actual conclusions do not necessarily follow, leaving the original claim that you can prove anything with false axioms unproven.
      - unanswered 6y ago
        The conclusion "Bertrand Russell is the Pope" does necessarily follow from 1=0. I'm not sure why you think it doesn't. The conclusion "Bertrand Russell is not the Pope" also follows. Yes, this is a contradiction. So is 1=0.
        DangitBobby 6y ago
        The claim was that if you accept a false premise, you can prove any false thing to be true. What we've actually seen is random claiming of contradictory things that supposedly follow by people trying to appear clever. It so obviously doesn't follow that it's a little strange we are having this conversation. If the conclusions themselves can be seen to be contradictory, anyone can dismantle your proof and show that it's bogus, even if you do accept the original premise. So the claim is bogus, and it's probably not a real story anyway.
        unanswered 6y ago
        The most wonderful thing about mathematics (and by extension logic) is that we don't have to give a shit what random people like yourself want to guess on the internet. We have actual proof checkers. They can show the correctness of proof terms for exfalso from more basic, more obviously correct principles. Those proof terms can't be "dismantled"; they're proved correct no matter how much you whine that you don't like them.
- DangitBobby 6y ago
  I didn't have a difficult time following it, it's just not correct.
tgfjsdr 6y ago
It's not the conclusion, it's a conclusion: remember that one can prove everything, so yours is just as (in)valid as his.
rubenfiszel 6y ago
The conclusion is that if the set containing "the pope and bertrand russell" has only 1 member, then it must follow that the pope and bertrand russell are the same person/member. An implicit axiom is that, since the pope is in a set, it exists (or at least in that set), so that other conclusion of yours does not follow directly.
- DangitBobby 6y ago
  Another implicit axiom is that if there is a set of at least two people which contains him and the Pope, one of them is not the Pope. And if the set is reduced to just Bertrand Russell then he must not be the Pope.

_ZeD_ 6y ago

it's not that difficult to follow: let B = {set made of "Bertrand Russell" }, P = {set made of the pope} let BP = B ∪ P |B| = 1, |P| = 1 (trivials) |BP| may be 1 or 2 if |BP| = 1 it means B=P -> Bertrand Russell is the Pope if |BP| = 2, and 2=1, it means B=P -> Bertrand Russell is the Pope

rembrandt 6y ago
But since |BP| is both 1 and 2 it means Bertrand Russell is both the Pope, and not the Pope.
- unanswered 6y ago
  ... yes. That is indeed what is meant by ex falso quodlibet. It does not make the example invalid.
  - DangitBobby 6y ago
    The point was merely that it cannot be used to prove "anything" since the actual conclusions do not necessarily follow, leaving the original claim that you can prove anything with false axioms unproven.
    - unanswered 6y ago
      The conclusion "Bertrand Russell is the Pope" does necessarily follow from 1=0. I'm not sure why you think it doesn't. The conclusion "Bertrand Russell is not the Pope" also follows. Yes, this is a contradiction. So is 1=0.
      - DangitBobby 6y ago
        The claim was that if you accept a false premise, you can prove any false thing to be true. What we've actually seen is random claiming of contradictory things that supposedly follow by people trying to appear clever. It so obviously doesn't follow that it's a little strange we are having this conversation. If the conclusions themselves can be seen to be contradictory, anyone can dismantle your proof and show that it's bogus, even if you do accept the original premise. So the claim is bogus, and it's probably not a real story anyway.
        unanswered 6y ago
        The most wonderful thing about mathematics (and by extension logic) is that we don't have to give a shit what random people like yourself want to guess on the internet. We have actual proof checkers. They can show the correctness of proof terms for exfalso from more basic, more obviously correct principles. Those proof terms can't be "dismantled"; they're proved correct no matter how much you whine that you don't like them.
DangitBobby 6y ago
I didn't have a difficult time following it, it's just not correct.

tgfjsdr 6y ago

It's not the conclusion, it's a conclusion: remember that one can prove everything, so yours is just as (in)valid as his.

rubenfiszel 6y ago

The conclusion is that if the set containing "the pope and bertrand russell" has only 1 member, then it must follow that the pope and bertrand russell are the same person/member. An implicit axiom is that, since the pope is in a set, it exists (or at least in that set), so that other conclusion of yours does not follow directly.

DangitBobby 6y ago
Another implicit axiom is that if there is a set of at least two people which contains him and the Pope, one of them is not the Pope. And if the set is reduced to just Bertrand Russell then he must not be the Pope.