$\newcommand{R}{\mathcal{R}}\newcommand{S}{\mathcal{S}}\newcommand{dom}{\operatorname{dom}}$
Let $\R,\S\colon A\to A$ i.e. $\R,\S\subseteq A\times A$. Prove or disprove: $$\dom(R\cap S)=\dom(R)\cap\dom(S).$$ Note. If $\mathcal{M}\subseteq A\times B$ then $\dom(\mathcal{M})$ is defined as the set $\{x\in A\mid\exists y\in B\wedge(x,y)\in \mathcal{M}\}$.
True. Proof:
$\subseteq)$ Let $x\in\dom(\R\cap\S)$. Then by definition of domain, there exists an $y\in A$ such that $(x,y)\in\R\cap\S$. By definition of intersection, it means $(x,y)\in\R$ and $(x,y)\in\S$. Again by definition of domain, we have $x\in\dom(\R)$ and $x\in\dom(\S)$, and by definition of intersection, we end up with $x\in\dom(\R)\cap\dom(\S)$.
$\supseteq)$ Let $x\in\dom(\R)\cap\dom(\S)$. By definition of intersection, $x\in\dom(\R)$ and $x\in\dom(\S)$. By definition of relation, there exists an $y\in A$ such that $(x,y)\in\R$ and $(x,y)\in\S$. By definition of intersection, $(x,y)\in\R\cap\S$, so by definition of domain, we end up with $x\in\dom(\R\cap\S)$.
Is it correct?
Thanks!!
Consider $A=\{1,2,3\}, R=\{(1,2),(2,2)\}, S=\{(1,3),(2,2)\}$
$$\text{dom}(R\cap S) = \text{dom}\{(2,2)\} = \{2\}$$ $$\text{dom}(R)\cap \text{dom}(S) = \{1,2\}\cap\{1,2\} = \{1,2\}$$