Showing that the flow value is well defined.

Solution 1:

You should sum over all vertices.

For simplicity, I adopt the notation $x(\delta^+(v))=\sum_{e\in \delta^+(v)}x(e)$ and analogously for $\delta^-(v)$ for any $v\in V$. We have

$$\sum_{v\in V}x(\delta^+(v))=\sum_{v\in V}x(\delta^-(v))$$ because every edge is ingoing exactly once and outgoing exactly once (i.e. every edge is counted exactly once on both sides).

Then essentially subtract what you have calculated:

$$\sum_{v\in V}x(\delta^+(v))-\sum_{v\neq s,t}(\delta^+(v))=\sum_{v\in V}x(\delta^-(v))-\sum_{v\neq s,t}(\delta^-(v))$$

This holds because $\sum_{v\neq s,t}(\delta^+(v))=\sum_{v\neq s,t}(\delta^-(v))$ by the second property. But this just means

$$\delta^+(s)+\delta^+(t)=\delta^-(s)+\delta^-(t)$$

which yields the result

$$\delta^+(s)-\delta^-(s)=\delta^-(t)-\delta^+(t)$$

by a simple arrangement.