Below shows the two-channel filter bank. The first half of the filter bank (before $Y_0(z)$ and $Y_1(z)$) is known as an analysis filter bank. The second half of the filter bank (after $Y_0(z)$ and $Y_1(z)$) is known as a synthesis filter bank.
If you go thru each step of the bank, you will find that the $$ \begin{eqnarray*} V(z) &=& \frac{1}{2} X(z) \left[ H_0(z) G_0(z) + H_1(z) G_1(z) \right] \\ &~& + \frac{1}{2} X(-z) \left[ H_0(-z) G_0(z) + H_1(-z) G_1(z) \right] \end{eqnarray*} $$
If we want to use the analysis filter bank to transform data into a new domain that we can manipulate, we need to return back from that domain without loss. This is known as the perfect reconstruction condition.
The aliasing canceling condition is defined by: $$ \begin{eqnarray*} H_0(z) G_0(z) + H_1(z) G_1(z) &=& 2 \\ H_0(-z) G_0(z) + H_1(-z) G_1(z) &=& 0 \end{eqnarray*} $$
In the frequency domain, this is equivalent to $$ \begin{eqnarray*} H_0(\omega) G_0(\omega) + H_1(\omega) G_1(\omega) &=& 2 \\ H_0(\omega-\pi) G_0(\omega) + H_1(\omega-\pi) G_1(\omega) &=& 0 \end{eqnarray*} $$
In the time domain, this is equivalent to $$ \begin{eqnarray*} h_0[n] \ast g_0[n] + h_1[n] \ast g_1[n] &=& 2 \delta[n] \\ \left( (-1)^n h_0[n] \right) \ast g_0[n] + \left( (-1)^n h_1[n] \right) \ast g_1[n] &=& 0 \end{eqnarray*} $$
In the orthogonal filter bank condition, we define $$ \begin{eqnarray*} H_0(z) &=& G_0(z^{-1}) \\ H_1(z) &=& G_1(z^{-1}) \end{eqnarray*} $$ and we assume that $X(z)$ can be defined by $$ X(z) = \alpha G_0(z) + \beta G_1(z) $$ where $\alpha$ and $\beta$ are scalar constants.
Under this condition, the output of the two-channel filter bank can be expressed as: $$ \begin{eqnarray*} V(z) &=& \frac{1}{2} \left[ \alpha G_0(z) + \beta G_1(z) \right] \left[ G_0(z^{-1}) G_0(z) + G_1(z^{-1}) G_1(z) \right] \\ &~& + \frac{1}{2} \left[ \alpha G_0(-z) + \beta G_1(-z) \right] \left[ G_0(-z^{-1}) G_0(z) + G_1(-z^{-1}) G_1(z) \right] \end{eqnarray*} $$ By clustering all of the $\alpha$ terms and $\beta$ terms together, this can be rearranged into $$ \begin{eqnarray*} V(z) &=& \frac{1}{2} \alpha \left( G_0(z) \left[ G_0(z) G_0(z^{-1}) + G_0(-z) G_0(-z^{-1}) \right] + G_1(z) \left[ G_0(z) G_1(z^{-1}) + G_0(-z) G_1(-z^{-1}) \right] \right) \\ &~& + \frac{1}{2} \beta \left( G_0(z) \left[ G_0(z) G_0(z^{-1}) + G_1(-z) G_0(-z^{-1}) \right] + G_1(z) \left[ G_1(z) G_1(z^{-1}) + G_1(-z) G_1(-z^{-1}) \right] \right) \end{eqnarray*} $$ The orthogonal filter bank conditions are derived from this expression by forcing $$ V(z) = \alpha G_0(z) + \beta G_1(z) = X(z) $$
The orthogonal filter bank condition is defined by: $$ \begin{eqnarray*} G_0(z) G_0(z^{-1}) + G_0(- z) G_0(- z^{-1}) &=& 2 \\ G_1(z) G_1(z^{-1}) + G_1(- z) G_1(- z^{-1}) &=& 2 \\ G_0(z) G_1(z^{-1}) + G_0(- z) G_1(- z^{-1}) &=& 0 \\ G_1(z) G_0(z^{-1}) + G_1(- z) G_0(- z^{-1}) &=& 0 \end{eqnarray*} $$ The last two equations are equivalent conditions since transforming $z$ into $z^{-1}$ gives the other expression. Hence, we only need three equations in the condition: $$ \begin{eqnarray*} G_0(z) G_0(z^{-1}) + G_0(- z) G_0(- z^{-1}) &=& 2 \\ G_1(z) G_1(z^{-1}) + G_1(- z) G_1(- z^{-1}) &=& 2 \\ G_0(z) G_1(z^{-1}) + G_0(- z) G_1(- z^{-1}) &=& 0 \end{eqnarray*} $$
In the frequency domain, this is equivalent to $$ \begin{eqnarray*} G_0(\omega) G_0(-\omega) + G_0(\omega - \pi) G_0(- \omega - \pi) &=& 2 \\ G_1(\omega) G_1(-\omega) + G_1(\omega - \pi) G_1(- \omega - \pi) &=& 2 \\ G_0(\omega) G_1(-\omega) + G_0(\omega - \pi) G_1(- \omega - \pi) &=& 0 \end{eqnarray*} If the time-domain filter coefficients are real (i.e., $G(-\omega)=G^*(\omega)$), this is equivalent to: \begin{eqnarray*} |G_0(\omega)|^2 + |G_0(\omega - \pi)|^2 &=& 2 \\ |G_1(\omega)|^2 + |G_1(\omega - \pi)|^2 &=& 2 \\ G_0(\omega) G_1^*(\omega) + G_0(\omega - \pi) G_1^*(\omega - \pi) &=& 0 \end{eqnarray*}
In the time domain, this is equivalent to $$ \begin{eqnarray*} g_0[n] \ast g_0[-n] + \left( (-1)^n g_0[n] \right) \ast \left( (-1)^n g_0[-n] \right) &=& 2 \delta[n] \\ g_1[n] \ast g_1[-n] + \left( (-1)^n g_1[n] \right) \ast \left( (-1)^n g_1[-n] \right) &=& 2 \delta[n] \\ g_0[n] \ast g_1[-n] + \left( (-1)^n g_0[n] \right) \ast \left( (-1)^n g_1[-n] \right) &=& 0 \end{eqnarray*} $$ So the condition states that the sum of the auto-correlation of the filters and their transformed versions must equal 2, and the sum of cross-correlation must equal 0.
It can be shown that this is equivalent to: $$ \begin{eqnarray*} \sum_{n=-\infty}^{\infty} g_0[n] g_0^*[n-2k] &=& \delta[k] \\ \sum_{n=-\infty}^{\infty} g_1[n] g_1^*[n-2k] &=& \delta[k] \\ \sum_{n=-\infty}^{\infty} g_0[n] g_1^*[n-2k] &=& 0 \\ \end{eqnarray*} $$