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\begin{proof}[Proof of \nameref{mengers_thm}]
  Let $k$ be the minimum size of an $a-b$
  separator. We want $k$ independent $a-b$ paths.
  If not possible, pick a minimal counterexample
  (say minimum $k$, then minimum $e(G)$). Note $k
  \ge 2$ (the theorem is trivial for $k = 1$). Let
  $S$ be a separator $|S| = k$.

  \textbf{Case 1:} $S \neq \Gamma(a)$, $S \neq
  \Gamma(b)$. Form a graph $G'$ from $G$, by
  replacing the component of $b$ in $G - S$ with
  one point $b'$, joined to each point of $S$.
  Then $e(G') < e(G)$, because otherwise
  $\Gamma(b) \subset S$ whence $\Gamma(b) = S$ (by
  minimality). Also, no $a-b'$ separator in $G'$
  has size $< k$ (as it would also be an $a-b$
  separator in $G$). Hence there exist $k$
  independent $a-b'$ paths in $G'$, i.e. have
  $a-S$ paths $P_1, \ldots, P_k$ in $G$, disjoint
  except at $a$. Similarly, have $b-S$ paths $Q_1,
  \ldots, Q_k$ in $G$, disjoint except at $b$.
  Note that no $P_i$ meets a $Q_j$ (else $S$ is
  not a separator). So can pair up the $P_i$ and
  $Q_j$ to obtain $k$ independent $a-b$ paths.

  \textbf{Case 2:} $S$ a separator, $|S| = k$
  $\implies$ $S = \Gamma(a)$ or $S = \Gamma(b)$.
  Must have $\Gamma(a)$, $\Gamma(b)$ disjoint.
  Indeed suppose $x \in \Gamma(a) \cap \Gamma(b)$.
  Then $G - x$ has no $a-b$ separator of size $< k
  - 1$ (as, with $x$, it forms an $a-b$ separator
  in $G$). So by minimality, $G - x$ has $k - 1$
  independent $a-b$ paths. But now add the path
  $axb$ to obtain $k$ independent $a-b$ paths in
  $G$. Let $a x_1 x_2 \ldots x_r b$ ($r \ge 2$) be
  a shortest path from $a$ to $b$. In $G -
  x_1 x_2$, must have a separator of size $k - 1$,
  $S$ say (minimality). Have $S \neq \emptyset$ as
  $k \ge 2$. So $S \cup \{x_1\}, S \cup \{x_2\}$
  are separators in $G$ of size $k$. Now, $S \cup
  \{x_2\} \neq \Gamma(a)$, since $x_2 \not\in
  \Gamma(a)$ -- so $S \cup \{x_2\} = \Gamma(b)$.
  Also, $S \cup \{x_1\} \neq \Gamma(b)$, since $x_1
  \not\in \Gamma(b)$ -- so $S \cup \{x_1\} =
  \Gamma(a)$. Hence $\Gamma(a)$ and $\Gamma(b)$
  are not disjoint, contradiction.
\end{proof}

\begin{remark*}
  Can also prove \nameref{mengers_thm} with vertex-capacity
  from max-flow min-cut. Indeed, form a network by
  directing each edge $xy \in G$ as $\vec{xy}$ and
  $\vec{yx}$, and give each vertex capacity $1$.
  Then a family of $k$ independent $a-b$ paths is
  exactly an integer-valued flow of size $k$ from
  $a$ to $b$. So done by integrality form of
  max-flow min-cut (as each vertex cut has
  capacity $\ge k$).
\end{remark*}

\begin{flashcard}[k-connected-iff-thm]
\begin{theorem}
  \label{menger_thm_5}
  Let $G$ be a graph, $|G| > 1$. Then $G$ is
  $k$-connected if and only if \cloze{for all distinct
  $a, b \in G$, $G$ has $k$ independent $a-b$
  paths.}
\end{theorem}

\fcscrap{\vspace{-1em}

This is ``the right way to think about
$k$-connectedness''.

\cref{menger_thm_5} is also often called
``Menger's Theorem''.
}

\begin{proof}
  \phantom{}
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Leftarrow$] \cloze{Certainly $G$ connected
      and $|G| > k$. Also, for $|S| \le k - 1$,
      $G - S$ is connected (else pick $a, b$ in
      different components of $G - S$,
      contradiction).}
    \item[$\Rightarrow$] \cloze{If $a, b$ non-adjacent
      then finish by \nameref{mengers_thm}.

      Otherwise, if $a, b$ are adjacent, then have
      $G - ab$, for which \nameref{mengers_thm} gives $k
      - 1$ independent $a-b$ paths (as no $a-b$
      separator in $G - ab$ has size $< k - 1$).
      Add in $ab$ to obtain $k$ independent $a-b$
      paths in $G$.} \qedhere
  \end{enumerate}
\end{proof}
\end{flashcard}

\begin{flashcard}[edge-connectedness-defn]
\begin{definition*}[edge-connectivity]
  \cloze{
  For $G$ connected, $|G| > 1$, the
  \emph{edge-connectivity} of $G$, written
  $\lambda(G)$, is the least $|W|$, where $W
  \subset E(G)$ has $G - W$ disconnected.

  Say $G$ is \emph{$l$-edge-connected} if
  $\lambda(G) \ge l$.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  $G$ being $1$-edge-connected is equivalent to
  being connected ($|G| > 1$).

  $G$ being $2$-edge connected is equivalent to
  $G$ being connected with no bridge ($|G| > 1$).
\end{example*}

\begin{note*}
  \begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d497b73a6e5c11ee.png}
  \end{center}
  Has $\kappa(G) = 1$, $\lambda(G) = 2$.
\end{note*}

\begin{flashcard}[vertex-menger-thm]
\begin{theorem}[Edge Menger]
  \label{edge_menger}
  \cloze{
  Let $G$ be a graph, $a, b$ distinct vertices of
  $G$. Suppose that $a, b$ in different components
  of $G - W \implies |W| \ge k$ (for any $W
  \subset E$). In other words, every $a-b$
  edge-separator has size $\ge k$. Then $G$ has
  $k$ edge-disjoint $a-b$ paths.
  }
\end{theorem}
\end{flashcard}

\textbf{Idea:} ``Replace edges of $G$ by vertices
and apply \cref{menger_thm_5}.''

\begin{flashcard}[line-graph-defn]
\begin{definition*}[line graph]
  \glssymboldefn{line_graph}{$L(G)$}{$L(G)$}
  \cloze{
  For a graph $G$, the \emph{line graph} $L(G)$ has
  vertex-set $E(G)$, with $e, f$ joined if they
  share a vertex.
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  \phantom{}
  \begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/8a09d0026e5e11ee.png}
  \end{center}
\end{example*}