Given a polyhedron $P=\{x\in\mathbb{R}^{n}|a_{i}x\ge b_{i}, i=1,\dots,m\}$ and two adjacent vertices $v_{1},v_{2}$, that both of them satisfy without loss of generality the first n-1 relations with equality $a_{i}x= b_{i}, i=1,\dots,n-1$ and $a_{1},\dots,a_{n-1}$ are linearly independent vectors. I want to prove that the edge of the polyhedron $L=\{lv_{1}+(1-l)v_{2}|l\in [0,1]\}$ can also be expressed as $\{x\in P| a_{i}x= b_{i}, i=1,\dots,n-1\}$ .
I came up with a solution but I think that it 's not complete: Let $F=\{x\in P| a_{i}x= b_{i}, i=1,\dots,n-1\}$. $F$ is a face of $P$ and it's dimension, given the linear independence of the vectors $a_{1},\dots,a_{n-1}$, is $n-(n-1)=1$. So $F$ is an edge of $P$, and because $v_{1},v_{2}\in P$, $F$ is exactly the line connecting these two vertices. Therefore, F=L. Are there any logical gaps in my answer?