Given a graph $G=(V,E)$ with positive weights $w(e)>0$ for any edge $e\in E$. The starting node is $s$. For every node $v$ there is a path $P_{s,v}$ from $s$ to $v$. An edge $e=(u,v)\in E$ will be called useful if it's the last edge in some minimal path $P_{s,v}$.
Prove that if all the edges of $P_{s,v}$ are useful then $P_{s,v}$ is a minimal distance path.
It seems that the proof is trivial but I'm not sure:
Let $l(P_{s,a})$ be the length of a path.
For $l(P_{s,a})=1$ the proof is trivial. Suppose that $l(P_{s,b})=k$ is a minimal distance path then we'll prove that $l(P_{s,c})=k+1$ is still a minimal distance path.
We know that $P_{s,b}$ is a minimal distance path from $s$ to $b$. $P_{s,c}=P_{s,b}+e_c$ where $e=(b,c)$ is a new edge. It is given that $e_c$ is a useful edge that is it is the last edge in the minimal distance path from $s$ to $c\quad(*)$ .
$P_{s,b}$ is a minimal distance path by induction assumption and because of $(*)\implies$ $P_{s,c}$ is also a minimal distance path. If we repeat the process for all edges $e_{k+2},...,e_v$ in $P_{s,v}$ where $e_v$ is the last edge in $P_{s,v}$ then $P_{s,v}$ is a minimal distance path.