Shannon reconstruction formula: who can prove alternative form? I am familiar with the standard formulation of the Shannon-Whittaker reconstruction formula which is 
$$s(t) = \sum\nolimits_{n =  - \infty }^\infty  {s(nT){\mathop{\rm sinc}\nolimits} (B(t - nT))}$$
where B = 1/T.
However, in the standard book "Simulation of Communication Systems, 2nd Ed" by Michel C. Jeruchim et al., the following formula is given for the reconstruction of $s(t-\tau)$, where $\tau$ is a delay parameter (real number) in Chapter 9, eq. 9.1.30:
$$s(t-\tau) = \sum\nolimits_{n =  - \infty }^\infty  {s(t-nT){\mathop{\rm sinc}\nolimits} (B(\tau - nT))}$$
How does s(.) in the summation become a function of $t$ (in addition to the sample timing offsets $nT$) and the $\mathop{\rm sinc(.)}$ devoid of the continuous variable $t$? So in essence, instead of samples of $s(t)$,  samples of $\mathop{\rm sinc()}$ are being used to reconstruct the (delayed) signal ??
In the case $\tau$ = 0, the formula reduces to 
$$s(t) = \sum\nolimits_{n =  - \infty }^\infty  {s(t-nT){\mathop{\rm sinc}\nolimits} (B(-nT))} = \sum\nolimits_{n =  - \infty }^\infty  {s(t-nT){\mathop{\rm sinc}\nolimits} (B(nT))}$$ since $\mathop{\rm sinc()}$ is an even function.
Thanks for any help clarifying this!
 A: First note that in the formula (9.1.30), $t$ is a fixed value parameter, whereas $\tau$ is the independent variable (taking arbitrary values). However, I will consider below the common notation where $t$ is the independent time variable and $\tau$ is a (fixed) delay. You can translate the results by simply switching the notation.
From the standard sampling theorem expansion, it holds 
$$
\tag{1}
s(t)=\sum_n s(nT) \text{sinc}(B(t-nT)) , t\in \mathbb{R}.
$$ 
Now, consider another signal $y(t)=s(t+\tau)$, which is also bandlimited. Applying the sampling theorem to $y(t)$ gives a more general version of (1) as
$$
\tag{2}
s(t+\tau)=\sum_n s(\tau+nT) \text{sinc}(B(t-nT)),t\in \mathbb{R}, \tau\in \mathbb{R}.
$$
From (2), we have
$$
\begin{align}
s(-t+\tau)&=\sum_n s(\tau+nT) \text{sinc}(B(-t-nT))\\
&\stackrel{(a)}=\sum_m s(\tau-mT) \text{sinc}(B(-t+mT))\\
&\stackrel{(b)}=\sum_m s(\tau-mT) \text{sinc}(B(t-mT))
\end{align}
$$
where $(a)$ follows by changing the summation index variable to $m=-n$ and $(b)$ follows by noting that $\text{sinc}(\cdot)$ is a symmetric function, i.e., $\text{sinc}(x)=\text{sinc}(-x)$.
This is formula (9.1.30) with the notations $t$-$\tau$ exchanged.
