# Tag Info

## New answers tagged axioms

1

ZF without the Axiom of Infinity is still strong enough to derive Peano's axioms PA. So we cannot prove its consistency without proving the consistency of PA.

0

Suppose that $R$ is the well-order on $\bigcup_{i\in I}A_i$, then $f(i)=\min_R A_i$ is a neat way of writing that. Another would be: $$\varphi(i,a):=a\in A_i\land(\forall b\in A_i)(\langle a,b\rangle\in R\lor a=b)$$ Of course if you want to be strict about it, then $\{A_i\mid i\in I\}$ and $R$ are parameters in the formula.

1

Generally, in set theory, one works with "pure" sets - that is sets whose every element is itself a set. So in your example, we would have to choose set-theoretic definitions of $aab$, $+$, and $50$. For example, natural numbers can be defined set-theoretically by the following recursive definition: $0 = \emptyset, n+1 = n \cup \{n\} = \{0, \ldots, n\}$, so ...

2

I think what your professor had in mind is the following: one can prove "by induction" that $f'(a)=f''(a)$ for all $a\in A$, just by noting that if the statement holds for all $b<a$, then it holds as well for $a$, by definition. What you did is in fact more: you justified that such a reasoning is indeed correct, by considering the smallest $a$ such that ...

0

$\emptyset$ is not a member of $A\cap B$. It is $A\cap B$. So $\emptyset$ represents the set of common elements of $A$ and $B$ but is not a common element itself. Also you got the axiom wrong as explained in comments.

0

While I agree that faith is not required to accept the rules of certain game in order to play that game, I am pretty sure that, if the rules of the game, and the consequences of playing by those rules, were entirely arbitrary, a lot of mathematicians wouldn't be able to earn a living by doing mathematics. For ex. according to some recent stats, ...

0

"We know that: ((A∧B≡A≡B)≡(A∨B)) is true since it is an axiom(golden rule). But does this mean ((A∨B)≡(A∧B≡A≡B)) is a valid as a corollary since it is equivalent?" Yes. A quick way to verify this comes as to look at the truth table for ≡. ≡ False True False True False True False True Now suppose that (p≡q) is true and "p" is true also. ...

2

The so-called Golden Rule is "typical" of Equational Logic; thus I'll assume that your question must be "interpreted" in this context. I'll refer to the textbook by George Tourlakis, Mathematical Logic (2008), which is devoted to the exposition of mathematical logic according to the "equational" style. The axioms of the system are listed at page 42-43, and ...

1

To prove that what you've described is a bijection, you have to show that it's one-to-one and onto. To show that it's one-to-one, we let $x$ and $y$ be vectors that both map to the same vector $x + v$, and we want to show that $x$ and $y$ must be equal. Since $y \longrightarrow y + v$, we have that $x + v = y + v$. It follows from the inverse axiom that ...

0

It is onto since given any $y \in V_2$, we have $y-v \mapsto y$. It is 1-1 since $x+v = y+v \implies x=y$. Another way is to note that the inverse map is $x \mapsto x-v$.

Top 50 recent answers are included