In a paper I was reading an ordered pair had a slightly different definition $\langle a,b \rangle = \{a,\{a,b\}\}$ instead the normal Kuratowski definition which is that $\langle a,b \rangle = \{\{a\},\{a,b\}\}$. notice in the first one the $a$ is without braces. now I know that it is wrong , ...

Assume ZF-infinity and assume that there is a set $X$ such that $\varnothing\in X$ and for all $t\in X$, $\{t\}\in X$. How to deduce axiom of infinity from these new axioms?

Definitions $(a,b) := \{\{a\},\{a,b\}\}$ $\Bbb N := \{n \in \infty \mid \forall A (0 \in A \land \forall n (n \in A \to n \cup \{n\} \in A) \to n \in A) \}$, where $\infty$ is the set guaranteed to exist by the axiom of infinity. $\Bbb Z := \Bbb N \times \Bbb N / \sim$ where $(a,b) \sim (c,d) \...

Let $\mathbf{ZFC}$ denote the theory of ZFC. Let $\mathbf{ZF-R}$ denote the theory of ZF without replacement. The motivation is that the remaining axioms do not require ordered pairs. Let $p(x,y,z)$ be a valid encoding for ordered pairs, where $p$ is a formula over the language of ZFC with $x...

The following is a classically valid deduction for any propositions $A,B,C$. $\def\imp{\rightarrow}$ $A \imp B \lor C \vdash ( A \imp B ) \lor ( A \imp C )$. But I'm quite sure it isn't intuitionistically valid, although I don't know how to prove it, which is my first question. If my conje...

A set $Y\subseteq X$ is computable iff its characteristic function $f_Y:X\rightarrow\{0,1\}$ defined as $$f_Y(x)=\begin{cases} 1, & x\in Y \\ 0, & x\notin Y \end{cases}$$ is computable. In principle $Y$ need not be a subset of $\mathbb N$, it may be e.g. a subset of the Baire space $\mathbb N^{\o...

how can i proved that polynomial ring Z[x] is not an integeral domain ? i was thinking that Z[x] is not a field so it is will not form integral domain as every finite integral domain is a fied but here Z[x] contain infinitely many element so ,,it will not form field ..... Is my thinking i...

What you may be looking for in your formal system is variously called full abbreviation power or definitorial expansion. Basically, it comprises rules that allows you to create on the fly new symbols extending the original language. We need one type of rule for each kind of symbol: $\def\eq{\left...

The only way to prove that $\Bbb Z[X]$ is not an integral domain is to prove that ZFC is inconsistent, because (in ZFC) $\Bbb Z[X]$ is actually an integral domain: Let $f,g \in \Bbb Z[X]$ with $fg=0$. Convince yourself (prove via induction) that if the leading coefficient of $f$ is $m$ and th...