The following is taken from Algebra Vol 2 by Luthar and Passi
Background
$\textbf{Definition 6.6.}$ The ring $(A-\mathfrak{p})^{-1}A$ is a local ring and is called the $\textit{localization of $A$ at its prime ideal}$ $\mathfrak{p}$ be denoted by $A_\mathfrak{p}$, and it is defined as $$A_\mathfrak{p}=\{a/s\mid a\in A, s\not\in\mathfrak{p}\}$$
If $\mathfrak{a}$ is an ideal of $A$, it is customary to denote the extension of $\mathfrak{a}A_\mathfrak{p}=(A-\mathfrak{p})^{-1}\mathfrak{a}$ of $\mathfrak{a}$ in $A_\mathfrak{p}$ by $\mathfrak{a}_\mathfrak{p}$. Thus $\mathfrak{a}_\mathfrak{p}$ consists of elements of the form $a/s$, $a$ in $\mathfrak{a}$ and $s\not\in \mathfrak{p}$.
$\textbf{2.1.1 Proposition.}$ The ring $A_\mathfrak{p}$ is a local ring with maximal ideal $\mathfrak{p}A_\mathfrak{p}$. As $\mathfrak{q}$ runs once through the prime ideals of $A$, contained in $\mathfrak{p}$, the ideal $\mathfrak{q}A_\mathfrak{p}$ of $A_\mathfrak{p}$ runs once through the prime ideals of $A_\mathfrak{p}$. For any ideal $\mathfrak{a}$ of $A$, contained in $\mathfrak{p}$, we have
$$A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p}\cong (A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}\quad(*)$$.
In particular, $A_\mathfrak{p}/\mathfrak{p}A_\mathfrak{p}$ is isomorphic to the field of fractions of $A/\mathfrak{p}$.
Question
For 2.1.1 Proposition above, would the two sets $A_\mathfrak{p}/\mathfrak{a}_\mathfrak{p},(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}$ in the isomorphism of the proposition in set builder notation in terms of coset representation should be as follows:
let $m,s\in A$, and $\bar{(\frac{m}{s})}=\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}$
$\frac{A_\mathfrak{p}}{\mathfrak{a}_\mathfrak{p}}=\{\bar{(\frac{m}{s})}\mid m\in A, s\not\in P\}=\{\frac{m}{s}+\mathfrak{a}_{\mathfrak{p}}\mid m\in A, s\not\in P\}$
and since $(\frac{A}{\mathfrak{a}})_{\frac{\mathfrak{p}}{\mathfrak{a}}}$ denotes the localization of the quotient ring $\frac{A}{\mathfrak{a}}$ at the ideal $\frac{\mathfrak{p}}{\mathfrak{a}}$. Then,
let $\bar{m}, \bar{s}\in \frac{A}{\mathfrak{a}}$ where $\bar{m}=m+\mathfrak{a}, \bar{s}=s+\mathfrak{a}$. So
$(\frac{A}{\mathfrak{a}})_{\frac{\mathfrak{p}}{\mathfrak{a}}}=\{\frac{\bar{m}}{\bar{s}}\mid \bar{s}\not\in \frac{\mathfrak{p}}{\mathfrak{a}}\}=\{\frac{m+\mathfrak{a}}{s+\mathfrak{a}}\mid s+\mathfrak{a} \not\in \frac{\mathfrak{p}}{\mathfrak{a}} \}$
Thank you in advance.