Derivation of the energy function of a hopfield network
I believe what you want to show is that the energy function is monotonically decreasing from time $t$ to time $t+1$ given the state update rules. Since there is only a finite number of states, this means the state must converge to a equilibrium under the given dynamics.  

To this end let $\mathbf{s}(t) \in \{0,1\}^n$ be the state of the network at time $t$. We update the network state by picking a unit $s_i(t)$ and updating the state according to
$$
s_i(t+1) = \left\{\begin{array}{r l} 1:&amp; \sum_{j} w_{ij} s_j(t) &gt; 0\\ 0:&amp; \text{otherwise}\end{array}\right.
$$
and define the energy function as 
$$
E(\mathbf{s}(t)) = -\frac{1}{2}\sum_i\sum_jw_{ij}s_i(t)s_j(t),
$$
where $w_{ij}$ is the weight between unit $i$ and $j$, $w_{ij} = w_{ji}$ and $w_{ii} = 0$ for al $i$.

WLOG assume we will update the $i^{th}$ unit. First, because of the restrictions on the weights, we may rewrite $E(\mathbf{s}(t))$ as the difference of 2 terms
$$
E(\mathbf{s}(t)) = -\frac{1}{2}\sum_{j\neq i}\sum_{k\neq i}w_{jk}s_j(t)s_k(t) - s_i(t)\sum_jw_{ij}s_j.
$$
Let $c_t = -\frac{1}{2}\sum_{j\neq i}\sum_{k\neq i}w_{jk}s_j(t)s_k(t)$ be the part of the energy function that does not depend on $s_i(t)$. Now if $s_i(t+1) = s_i(t)$ then clearly we have $E(\mathbf{s}(t+1)) = E(\mathbf{s}(t))$, so we only need to consider the 2 other cases.

Case 1: If $s_i(t+1) = 1$ and $s_i(t) = 0$ then it must be that $\sum_{j} w_{ij} s_j(t) = \Delta_t &gt; 0$. So 
$$
E(\mathbf{s}(t+1)) = c_{t+1} - \Delta_t = c_t - \Delta_t &lt; c_t = E(\mathbf{s}(t)).
$$
Case 2: If $s_i(t+1) = 0$ and $s_i(t) = 1$ then it must be that $\sum_{j} w_{ij} s_j(t) = \Delta_t \le 0$. So 
$$
E(\mathbf{s}(t+1)) = c_{t+1} = c_t \le c_t - \Delta_t = E(\mathbf{s}(t)).
$$
Having considered each case we can conclude that $E(\mathbf{s}(t+1)) \le E(\mathbf{s}(t))$ and is therefore monotonically decreasing.