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0

We have found a proof and formalized it using the Coq proof assistant, the details are here: https://hal.inria.fr/hal-01332044


5

Recall that $x\in L$ if and only if there is some $\alpha$ such that $x\in L_\alpha$. And that is exactly the content of the axiom $V=L$: For every $x$, there exists an ordinal $\alpha$ such that $x\in L_\alpha$. But wait, what is $L_\alpha$ anyway? Ah, well, luckily, it is the unique set $X$ such that there exists a function $f$ with domain $\alpha+1$,...


3

If by "what does it mean" you mean "what does it mean" I have to ask what do you mean, what does it mean? Saying that ZF has infinitely many axioms means exactly what it says. There are infinitely many primes. There are only finitely many natural numbers less than $3$. There are infinitely many axioms of ZF. Probably you actually understand that, and you're ...


0

The full statement of the axiom schema is as follows: given $\varphi(x,y,u)$, $$\forall u(\forall x(\forall a(a\in x\rightarrow\exists y(\varphi(a,y,u)\land\forall z(\varphi(a,z,u)\rightarrow y=z)))\rightarrow\exists w(\forall y(y\in w\leftrightarrow\exists a(a\in x\land\varphi(a,y,u)))))$$ Namely, if $\varphi$ is a formula which defines a function on $x$, ...


1

The idea is that you are defining $f(y)$ to be the unique $x$ such that $P(y,x)$ is true. If it said simply ${\forall}y({\exists}x:P(y,x))$ then you would be right that it needs to say $\exists!$ instead of just $\exists$. But notice that it says not just that there is an $x$ such that $P(y,x)$ but that there is an $x$ such that $P(y,z)\Leftrightarrow z=x$ ...


2

No. The axiom already implies that there's only one such $x$. Given $y$, let $x$ and $x'$ be elements with ${\forall}z(P(y,z){\iff}(x=z))$ and ${\forall}z(P(y,z){\iff}(x'=z))$. Using $z=x$ in the first statement yields $P(y,x)\iff(x=x)$, and thus $P(y,x)$; then using $z=x$ in the second statement yields $P(y,x)\iff(x'=x)$, and thus $x'=x$.


0

One possibility is to "vectorize" the ordinary Booleans. So your set will consist of all triples of Booleans, and the NOT and OR operations will apply in parallel to the three elements; i.e., $$ \neg(b_1,b_2,b_3)=(\neg b_1,\neg b_2, \neg b_3) $$ and $$ (b_1,b_2,b_3) \vee (b_1', b_2', b_3')=(b_1\vee b_1',b_2\vee b_2', b_3\vee b_3'). $$ Clearly this set ...


1

I interpret your question about group algebras to really be something like the following: What extra structure do group algebras have that allows you to "remember" that they come from groups, and in particular how do you see the group theory axioms in terms of this structure? The answer is that group algebras have the additional structure of a Hopf ...


3

What extensionality actually says is that if $x$ and $y$ are such that $z\mathbin{E} x\Leftrightarrow z\mathbin{E} y$ for all $z$, then $x=y$. In your example it is true that $z\mathbin{E} b\Leftrightarrow z\mathbin{E} c$ for all $z\in D$ (since the only $z$ such that $z\mathbin{E} b$ is $z=a$ and similarly for $c$). Since $b\neq c$, this means ...


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The attached article (What is Mathematics - an Overview) discusses a good deal about "Axioms". Abstract: Mathematics is based on deductive reasoning though man's first experience with mathematics was of an inductive nature. This means that the foundation of mathematics is the study of some logical and philosophical notions. We elaborate in simple terms that ...



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