2 - Knowledge Representation for Mathematical Theories [ID:57073]
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Okay, you all know what you have to do, right?

Or rather in two minutes, I guess.

Sorry?

Sorry?

No, it's just about the stuff we did last week.

Yeah, which is also why there's only four problems and the problems are rather short,

because we didn't do much.

Yeah, that's because Michael has less confidence in the system than I do.

Also, their test didn't work this morning.

Mine does.

I've tested it.

Okay, so it's definitely not the HDMI thing.

I've now switched to USB-C and the beamer still goes off.

Okay, so it's definitely the beamer, not me.

That's good to know at least.

Bad news, that means I can do nothing about it.

Okay, and that was it.

I'm not sure how to interpret that.

Should we maybe actually go over the questions just to make sure we're all on the same page?

Like, I'm definitely not going to do this every week, but we might as well do it this time, I think.

Okay, so maybe just to recap first.

There we go.

Okay, so we're currently at the topic of talking about mathematical foundations first,

because mathematical foundations allow us to make precise what we're actually talking about when we're talking about mathematical objects.

And the primary foundation we should be aware of is naive set theory,

because to some extent that is basically what the vast majority of mathematics happens to be in to some extent or another.

We know that naive set theory is contradictory.

That's what Russell's paradox tells us.

So if you want to be precise, you would say that actually like most of informant mathematics happens in Sermilo-Frenkel set theory

with the axiom of choice and potentially a club of inaccessible cardinals, blah, blah, blah, blah, blah.

We are going to talk about that at some point.

But also to be fair, most mathematicians don't necessarily even know what that means.

So for most mathematicians, the only thing that they care about is we have some kind of set theory where we can do various things.

And naive set theory is the simplest attempt to describe what those things that we're allowed to do actually are.

So we have two axioms. We have the extensionality axiom, which tells us that two sets are equal if and only if they have the same elements.

And we have the comprehension axioms, which tells us that for any property phi, there is a set such that that particular set contains exactly those elements that satisfy the property.

That gives us the answer to that particular question.

So we have one set which contains a set X and it contains a set X, Y, and it contains a set Y, X.

You might notice X, Y, Y, X immediately have the same elements, so we can get rid of one of them because they're equal anyway.

So what we have here is the set that contains the set X and the set that contains X and Y.

So the first one, is that equal to that one? Sorry? No. Why not?

Yeah, there is one more set of braces surrounding this, which means in particular this set has exactly one element, namely this set, which means they're not equal.

This one has two elements, this one has one element. Next one, X and XY, are those equal? Yes, they're equal, exactly because we've just gotten rid of this Y, X, which is redundant because we already have it here.

X and Y, X, yeah, same reason. Again, the order doesn't matter, so that one is equal to S.

This one, XY, yeah, no, we've gotten rid of X, which we're not allowed to and also we've added additional braces around it.

So again, this one has one element, this one has two elements, that doesn't match. XY, also not equal.

This one has two elements, one of which contains X and Y, this one has two elements, one of which is Y, so that also doesn't work.

XY, XY, YX. No, this has the additional Y, which isn't in there. XX, XY, YX. Yes, we've just doubled the X, which we're allowed to do because they're equal anyway, and XY, YX are equal also.

So, yeah, so much for that one. We haven't talked about Kowalchowsky pairs yet.

So axioms, which of the followings are axioms of naive set theory? For all XY, X equals Y if and only if for all Z, Z in X and Z in Y.

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01:22:49 Min

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2025-05-06

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2025-05-07 01:29:07

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