Key focus: As per Nyquist ISI criterion, to achieve zero intersymbol interference (ISI), samples must have only one non-zero value at each sampling instant.
Consider an equivalent baseband communication system model with baud rate $latex F_{sym}=1/T_{sym}$ ($latex T_{sym}$ is the symbol period) shown in Figure 1, where the transmitter, channel and the receiver are represented as band-limited filters.

The entire combination of filters can be represented as the following Fourier transform pair

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Given the noise term $latex n(t)$, the output $latex b(t)$ of the receiving filter is given as
$latex b(t)=\sum_{k=-\infty}^{\infty}a_k h(t-kT_{sym})+n(t) \quad\quad\quad (2) &s=2$
where, $latex h(t)=h_T(t) \ast h_C(t) \ast h_R(t)$ is the overall impulse response of the system due to an impulse at its input. Normalizing $latex h(0)=1$ (where the signal of interest lies), at the symbol sampling instants $latex t=mT_{sym}$,
$latex b(mT_{sym})=a_m + \underbrace{ \sum_{ \substack{k=-\infty \ k \neq m} }^{\infty} a_k h(mT_{sym}-kT_{sym})}_{\text{ISI terms}} + n(mT{sym}) \quad\quad\quad\quad\quad\quad (3) &s=2$
where, $latex a_m$ is the symbol of interest at $latex mT_{sym}$ sampling instant, $latex n(mT_{sym})$ is the noise at that sampling instant and the remaining terms are contributions from other symbols – representing intersymbol interference. For nullifying the ISI terms, with an impulse of unit value applied at $latex t=0$ to the combined filters $latex h(t)$, the samples of the $latex h(t)$ at the output of the filter combination should be 1 at the sampling instant $latex t=0$ and zero at all other sampling instants $latex kT_b$ $latex (k \neq 0)$. This is called Nyquist criterion for zero ISI. In frequency domain, the frequency-shifted replicas of the overall system transfer function $latex H(f)=H_T(f)H_C(f)H_R(f)$ should add up to a constant value.
$latex \sum_{k=-\infty}^{\infty} H \left(f+\frac{k}{T_{sym}} \right)= T_{sym} \quad\quad \forall f \quad\quad (4) &s=2$
It can be readily seen that, in time domain, the simplest signal that naturally avoids ISI is the rectangular pulse of width $latex T_{sym}=1/F_{sym}$ but it consumes infinite bandwidth. On the other hand, the signal that avoids ISI with the least amount of bandwidth is a sinc pulse of bandwidth $latex F_{sym}/2$. The next section walks thorough the various choices of pulse shapes that are available for avoiding or mitigating ISI.
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where is the proof for eq 4 ?
You may check standard text books for proof. Thanks.