Distance between vertices of minimum cycle coincides in the entire graph [Proof Verification] In order to further improve my intuition about graphs, and motivated by this exercise, I want to prove or disprove the following

Let G be a simple, connected graph that is not a tree, and $v_0v_1...v_kv_0$ a cycle on $G$ of minimal length. If $1 \leq i < j \leq k$, the distance between $v_i$ and $v_j$ in the graph is achieved in the cycle, that is 

$$
d_G(v_i,v_j) = \min\{j-i,(k +1) - (j-i)\}
$$
If what I have written is correct, the former should be affirmative. What follows is my attempt to prove this, I would really appreciate if you could first and foremost check my work, and secondly, suggest a shorter argument.
Let's proceed by induction. If $n \leq 2$ there are no graphs that satisfy the hypothesis, so we start the base case at $n  = 3$, in which case the only such graph is $K_3$, which is itself a cycle, so there is nothing to prove. 
Now, let $G$ be a graph of size $n$ and suppose that the result is true for any graph of size $m < n$, and let $C = v_0v_1...v_kv_0$ be a minimal cycle, that is, $l(C) = g(G)$.
If $G = C$, there is nothing to prove. Otherwise, there exists $v \in G \setminus C$. Since the induced subgraph $H := G[G \setminus \{v\}]$ has size less than $n$, it verifies the inductive hypotheses. Moreover, a cycle in $H$ induces a cycle in $G$ of the same length, and $C \subset H$, so $C$ is a minimal cycle in $H$. Therefore,
$$
d_H(v_i,v_j) = \min\{j-i,(k +1) - (j-i)\}
$$
We want to now show, then, that we cannot shorten the distance between these vertices in $G$. That is, $d_G(v_i,v_j) = d_H(v_i,v_j)$. Let $1 \leq i < j \leq k$ and  $\xi = v_iw_1...w_lv_j$ a minimal path from $v_i$ to $v_j$. If $\xi$ consists of all vertices in $H$, we are done. If not, it is of the form 
$$
v_iw_1 \ ... \ w_svw_{s+1} \ ... \ w_lv_j
$$
To conclude, we can prove by induction on the size of $I_{\xi} := \{w_r\}_{1\leq r \leq l} \cap C$ that no such path is minimal. To be more precise, the proposition we will do induction on is

$p(a) = $ "No path of the form $\xi = v_iw_1 \ ... \ w_svw_{s+1} \ ... \ w_lv_j$ with $v_i,v_j \in C$ can be of length less than $d_H(v_i,v_j)$, if $|I_{\xi}| = a$".

If $a = 0$, this is clear, because
$$
v_0 ... v_{i-1} \ \xi \ v_{j+1} ... v_k
$$
would contradict the minimality of $C$. Let $1 \leq a \leq n$ and suppose $p(b)$ holds for $b < a$. Now, let $w_r = v_d$ be the first appearance of an element in $C$. There are two cases, since either $r \leq s$ or $r \geq s+1$.
In the first scenario, if $r \leq s$ we know by the (nested) inductive hypothesis that $w_r...v_j$ has greater length than the path $v_d...v_j$  of vertices in $C$. Since $\xi$ is minimal, $v_i$ must lie in $v_d ... v_j$, because if not we can replace the last part of $\xi$ to obtain a shorter path,
$$
v_i...w_{r-1}v_d...v_j
$$
But then again, we reach a condradiction: if $v_i$ lies on $v_d...v_j$,
$$
l(v_i v_{i+1} ...v_{j}) \leq l(v_d v_{d+1}...v_j) \leq l(w_r...v_j) <  l(\xi)
$$
but once again, $\xi$ is minimal.
If instead $r \geq s+1$, the last appearance of an element of $C$ in $\xi$ occurs after $v$, in which case we can copy the prior argument interchanging roles.
We have then proved that no path containing $v$ between vertices on $C$ can be minimal, which concludes the proof.
Thoughts?
 A: EXE : $v_0,w_1,\cdots ,w_s,v_k$ is shortest path. Then $$
d_G(w_i,w_{i+t})=t$$
Proof : Note that $d_G$ is a metric. If $d_G(w_i,w_{i+t})<t$, then
$$
s+1=d_G(v_0,v_k)\leq d_G(v_0,w_i) +
 d_G(w_i,w_{i+t}) + d_G(w_{i+t},v_k) <
  s+1 $$
EXE : For convenience, we assume that cycle
$(v_0,\cdots,v_k,\cdots,v_{k+l},v_0)$ of length $k+l+1$ is
minimal.
Then prove that $d_G(v_0,v_k)=k$ if $k<l+1$. 
Proof : Assume that $$v_0,w_1,\cdots ,w_s,v_k\ \ast$$ is shortest path of
length $s+1<k$ $(A)$. 
Here we assume that $s$ is smallest in number satisfying $(A)$.
If $$\{ w_1,\cdots, w_s\} \bigcap \{ v_1,\cdots, v_{k-1}\} =
\emptyset $$ then there is a cycle of length $k+s+1$ so that it is a
contradiction.
If $$\{ w_1,\cdots, w_s\} \bigcap \{ v_{k+1},\cdots, v_{k+l} \} =
\emptyset $$ then there is a cycle of length $s+l+2$ so that it is
a contradiction. 
Hence $ v_a,\ v_b\in \{w_1,\cdots ,w_s\} $ with $0<a<k<b\leq k+l$.
Here $d_{G}(v_a,v_b)\leq s-1$. That is, $$v_a=w_{\alpha},\cdots,w_{\alpha+\beta}=v_b\ \ast\ast$$ contradicts to a definition of $s$. Even though we assume that path in $\ast$ of type $\ast$ is smallest, we have a path of type $\ast$, i.e. $\ast\ast$. 
