Show inequality involving capacities of the union and intersection of cuts.

Question

Let $(S,\overline S)$, $(T,\overline T)$ be cuts of a network $G$. I've alredy showed que $(S\cup T,\overline{S\cup T})$ and $(S\cap T,\overline{S\cap T})$ are cuts for $G$. Now, I have left to prove that $$cap(S\cup T,\overline{S\cup T})+cap(S\cap T,\overline{S\cap T})\leq cap(S,\overline S)+cap(T,\overline {T}),$$ where $cap$ denotes the capacity of an edge (its weight).

What I've tried so far is that if we take $e\in E(G)$ such that $e=uv$, with $u\in S\cup T$ and $v\in \overline{S\cup T}$, then it sums $cap(e)$ on the LHS and sums at least once $cap(e)$ on the RHS. We have a same argument for an edge $e=uv$ with $u\in S\cap T$ and $v\in\overline{S\cap T}$. But what I need to conclude the inequality is that each edge $e$ cannot sum $cap(e)$ in both $cap(S\cup T,\overline{S\cup T})$ and $cap(S\cap T,\overline{S\cap T})$.

Answer 1

Decompose into disjoint unions as follows: \begin{align} S &= (S \setminus T) \cup (S \cap T) \\ \overline{S} &= (\overline{S} \setminus \overline{T}) \cup (\overline{S} \cap \overline{T}) \\ T &= (T \setminus S) \cup (S \cap T) \\ \overline{T} &= (\overline{T} \setminus \overline{S}) \cup (\overline{S} \cap \overline{T}) \\ S \cup T &= (S \setminus T) \cup (S \cap T) \cup (T \setminus S) \\ \overline{S \cap T} &= (\overline{S} \setminus \overline{T}) \cup (\overline{S} \cap \overline{T}) \cup (\overline{T} \setminus \overline{S}) \end{align} Then \begin{align} & \operatorname{cap}(S\cup T,\overline{S\cup T}) + \operatorname{cap}(S\cap T,\overline{S\cap T}) \\ &= \operatorname{cap}(S\cup T,\overline{S}\cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{S\cap T}) \\ &= \operatorname{cap}(S\setminus T,\overline{S}\cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{S}\cap \overline{T}) + \operatorname{cap}(T\setminus S,\overline{S}\cap \overline{T}) \\ &+ \operatorname{cap}(S\cap T,\overline{S} \setminus \overline{T}) + \operatorname{cap}(S\cap T,\overline{S} \cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{T} \setminus \overline{S}) \\ &\le \operatorname{cap}(S\setminus T,\overline{S}\cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{S}\cap \overline{T}) + \operatorname{cap}(T\setminus S,\overline{S}\cap \overline{T}) \\ &+ \operatorname{cap}(S\cap T,\overline{S} \setminus \overline{T}) + \operatorname{cap}(S\cap T,\overline{S} \cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{T} \setminus \overline{S}) \\ &+ \color{red}{\operatorname{cap}(S\setminus T,\overline{S}\setminus \overline{T}) + \operatorname{cap}(T\setminus S,\overline{T}\setminus \overline{S})} \\ &= \operatorname{cap}(S\setminus T,\overline{S}\setminus \overline{T}) + \operatorname{cap}(S\setminus T,\overline{S}\cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{S} \setminus \overline{T}) + \operatorname{cap}(S\cap T,\overline{S}\cap \overline{T}) \\ &+ \operatorname{cap}(T\setminus S,\overline{T}\setminus \overline{S}) + \operatorname{cap}(T\setminus S,\overline{S}\cap \overline{T}) + \operatorname{cap}(S\cap T,\overline{T} \setminus \overline{S}) + \operatorname{cap}(S\cap T,\overline{S} \cap \overline{T}) \\ &= \operatorname{cap}(S,\overline S) + \operatorname{cap}(T,\overline {T}) \end{align}