Prove or disprove: $p$ is the shortest path from $s\in V$ to $t\in V$ with $w'=w_{1}+w_{2}$ I saw the following statement:

Let $G=(V,E)$ be a graph and two $w_{1},w_{2}\,:\,E\to\mathbb{R}$ weight functions so there are no negative cycles in graph. Let $p$ be the shortest path from $s\in V$ to $t\in V$ with $w_1$ and $w_2$. Prove or disprove: $p$ is the shortest path from $s\in V$ to $t\in V$ with $w'=w_{1}+w_{2}$.

I could not disprove it and I believe this statement it true. But I could not prove it formally. How do I translate "$p$ be the shortest path from $s\in V$ to $t\in V$ with $w_1$ and $w_2$" into math?
 A: This statement is true.
If $p=e_1e_2\ldots e_k$, $e_1,e_2,\ldots e_k\in E(G)$,
then $w_i(p)=w_i(e_1)+w_i(e_2)+\ldots+w_i(e_k)=d_i$ for $i=1,2$ and
$w'(p)=w'(e_1)+w'(e_2)+\ldots+w'(e_k)=d_1+d_2$.
Let $q=e'_1e'_2\ldots e'_m$ be another path from $s$ to $t$.
Then $w_i(q)\geq d_i$ and $w'(q)=w_1(q)+w_2(q)\geq d_1+d_2=w'(p)$.
This means that $p$ is the shortest path from $s$ to $t$ with $w'$.
Correct me if I'm wrong.
A: *

*Let G(V,E) be an undirected graph, so that any path from s to t can be also traversed from t to s in reverse and with the same cost.

*Let $p_{shortest}$ be the shortest path from s to t (and from t to s in the reverse direction); let the cost of this shortest path be $\omega_{min}$.

*Let $\pi_{1}$ and $\pi_{2}$ be two distinct paths from s-t, and consider their cost to be: $\omega_{1}$ and $\omega_{2}$ respectively.

*Then, since G is undirected, there exist a cycle c(s-s) = c(t-t), with cost: $\Omega=\omega_{1}+\omega_{2}$.

*If $\omega_{1}=\omega_{2}=\omega$; hence, from (4), $\Omega=2\omega$.

*If $\omega_{1}=\omega_{2}=\omega_{min}$, then, from (5), $\Omega=2\omega_{min}$.

*If both paths are different from $p_{shortest}$, the sum of their cost must be greater than $2\omega_{min}$; else the $p_{shortest}$ premise is false. In this case, $\Omega>2\omega_{min}$.

*If $|\pi_{1}|=|p_{shortest}|$, and $|\pi_{2}|=!|p_{shortest}|$, then, $\omega_{2}>\omega_{min}$; or again $p_{shortest}$ is not the shortest path. In this case, let $\omega_{2}=\omega_{min}+\Delta \omega$. Then, $\Omega = 2\omega_{min}+\Delta \omega$.
And similarly for $\omega_{1}>\omega_{min}$ with $\omega_{2}=\omega_{min}$.

