Okay, hello everybody. I think this is the kind of quiz that we may actually want to

go over, right? Okay. Before I do, who managed to register for the exam? On your own? Okay.

Everyone who managed to register for the exam on their own without having to like, okay.

So everyone who's not yet registered for the exam and still needs to do so, please ping

Florian Raber because he's the person who actually has access to the system where you

do all the exam related things. I have no idea what's going wrong. Neither does he,

but in the worst case what we do is we just register you manually by literally entering

your matricle number, right? So if you want to register for the exam and you haven't yet,

send your matricle number to Florian Raber either in the Matrix channel or like in a

private message on Matrix or via email, anything like that. In the meantime, we will like see

whether we can figure out what's wrong. Well, I can't, but like Florian will still try to

figure out what's wrong. Yeah. Either way, ping Florian, then you're on the safe side.

Okay. Then let's maybe look at those. Yeah, sure.

Now this is, I think, the easier one, right? So incompleteness theorem tells us that any

extension of Peano arithmetic, there is some formula, sorry, any computable extension of

Peano arithmetic, there is some formula phi such that neither P A proves phi nor P A proves

not phi, right? Which means there is some closed first order proposition phi such that

neither P A nor P A, such that neither P A proves phi nor P A proves not phi. Yeah, that's

what it means for P A to be incomplete. And so that is correct. There are infinitely many

models of Peano arithmetic that all satisfy different propositions. That is a direct corollary

of that because since every extension or every computable extension of Peano arithmetic is

incomplete, that means every time you find one such a proposition, you can just add that

to Peano arithmetic, which gives you a new theory to which the incompleteness theorem

applies, which again gives you another one of those formulas which you can add and so

on and so forth, and you can do that infinitely often. That's probably the least obvious option

in this entire quiz. But yeah, so therefore this one is correct. There's some super set

of Peano arithmetic that is complete. Yeah, there is. You just take the standard natural

numbers, take every formula that is true in there, put them all in one theory and that gives you a

complete theory. That theory is not computable. Okay, so you don't know which formula is actually

in that theory, but there is no algorithm that tells you. But you can construct a complete

extension of Peano arithmetic, which is there is some computable super set of Peano arithmetic.

That one is wrong. That's exactly the point of basically of the incompleteness theorem.

Every computable extension of Peano arithmetic suffers from the same flaw, namely it's just

necessarily incomplete. But if you just don't care about computability, you can just easily

just take all the statements that are true about the natural numbers and you get a complete

theory. There's some subset of Peano arithmetic that proves that Peano arithmetic is consistent.

No, Peano arithmetic cannot prove its own consistency. That's the second Gödel incompleteness

theorem. And if Peano arithmetic can, definitely no subset of Peano arithmetic can. There's some

super set of Peano arithmetic that proves that Peano arithmetic is consistent. I mean, it does

say correct here. Yeah, sure. You just need to add the statement Peano arithmetic is consistent

to Peano arithmetic and then gives you a super set of Peano arithmetic that proves that Peano arithmetic...

Sorry? Can prove its own consistency. Yes, but it can prove the consistency of Peano

arithmetic alone. Just take the axiom Peano arithmetic is consistent, add it to Peano

arithmetic. That gives you a new theory that proves that Peano arithmetic is consistent,

which itself cannot prove its own consistency. For a more trivial example, naive set theory

proves that Peano arithmetic is consistent. Or any set theory proves that Peano arithmetic

is consistent because it gives you a direct model of the natural numbers. Peano arithmetic

is inconsistent. Not necessarily true. I mean, you can sort of like epistemologically claim

that we don't know because we can't prove that it's consistent unless you assume something

that is stronger than Peano arithmetic, blah, blah, blah. But definitely we don't know