Lemma 3.1

Let $G$ be a graph, and $u$ and $v$ two vertices of $G$. There is a path from $u$ to $v$ in $G$ if and only if there is a simple path from $u$ to $v$ in $G$

Proof

Let $u$ and $v$ be vertices of a graph $G$

$(⟸)$ A simple path from $u$ to $v$ in $G$ is a path from $u$ to $v$, so the result trivially holds

$(⟹)$ Let $P$ be a path from $u$ to $v$. If $P$ has no repeated vertices, then $P$ is a simple path and we are done. So assume that $P$ has a repeated vertex $w$. Let $P^{\prime }$ be the path obtained by going along $P$ from $u$ to the first appearance of $w$, and then going along $P$ from the last appearance of $w$ to $v$. So $P^{\prime }$ is a path from $u$ to $v$ whose length is smaller than the length of $P$. If $P^{\prime }$ has no repeated vertices, then we are done. Otherwise, repeat this procedure. By repeating this procedure until there are no more repeated vertices, we end up with a simple path from $u$ to $v$