I have two edge cuts $A,B$. An edge cut is a set of edges, whose removal from $G$ makes the graph disconnected (there exist some vertices $x,y$ such that there's no path connecting them). I need to prove, that the symmetric difference of those two sets is too an edge cut. Let $Z=(A \cup B)/(A \cap B)$
So there are a couple of cases to consider.
$|A \cap B|=0$, then obviously the symmetric difference is an edge cut.
$G-Z$ is disconnected, so it is an edge cut.
$G-Z$ is for some reason connected. That would mean, that for every $x,y$ there exists a path between those two vertices, an it goes through some edges in $A \cap B$. An there probably lies a contradiction here somewhere, but where exactly I do not know...
This exercise is from one of the old exams I'm doing in preparations for my exam next week. Is there any possibility, that the exercise is just wrong? I mean take a look at this drawing.
$1$ shows edge cut $A$, that separates the middle left vertex, $2$ shows edge cut $b$ that separates two bottom vertices from the left, and $3$ is their symmetrical difference that hardly does anything.