Going from output of D/C to input of D/C $$y_c[n] {\longrightarrow} \boxed{\text{D/C}} {\longrightarrow} y(t)$$
If the following equation describes the IDEAL D/C converter:
$$y_c(t) =  \sum \limits_{n=-\infty}^{\infty} y_d[n]~ \text{sinc}\left(\frac{\pi (t - nT)}{T}\right)\tag{1}$$
$$Y_c(\Omega) = \begin{cases} T~ Y_d(\omega) \Big|_{\omega=\Omega T} &amp; |\Omega| &lt; \left(\Omega_r = \frac{\pi}{T_s}\right) \\0 &amp; \text{otherwise}\end{cases}$$
How to prove that we can find $y_c[n]$ from $y(t)$ using this formula:
$$y_d[n] = y_c(nT)\tag{2}$$
It doesn't seem entirely obvious to me by rearrange equation $(1)$, that it equals $(2)$.

conceptually, if function $f$ represents a D/C converter, and function $g$ represents a C/D converter, and $f$ and $g$ are inverse functions of each other. then:
$$y(t) = f\Big( y_n[n]\Big)$$
$$g\Big( y(t) \Big) = g\bigg(f\Big(y_n[n]\Big)\bigg)$$
$$g\Big( y(t) \Big) = y_n[n]$$
You won't get the discrete $y_d[m]$ by re-arranging the equations, but by sampling the $y_c(t)$ at integral multiples of $T$. When you sample the RHS in equation1 exactly at $t=mT$ instances, you get $y_d[m]$. You need to remember that $sinc(m-n) = 1$ for $m=n$ and $0$ otherwise, and (BTW in equation 1, in $sinc$ argument $\pi$ won't be there). Anyway, take RHS and evaluate the expression at $t = mT$ where $m$ are integers.
$$y_c(t) \big|_{t=mT} = \sum^{\infty}_{n=-\infty}y_d[n] \, \mbox{sinc} \left( \frac{t-nT}{T} \right) \Big|_{t=mT} = y_d[n]$$