Please check if my thoughts is correct.
Let we have the following linear programing problem $$max\, c^Tx$$
$$Ax\leq b,$$
$$x\geq 0,$$
where $A$ is a matrix with $m$ rows and $n$ columns, $x\in\mathbb{R}^n.$
To solve this problem by the simplex method, we use the canonical form
$$max\, c^Tx$$
$$Ax+Is = b,$$
$$(x,s)\geq 0,$$
where we have additional $m$ columns and $(x,s)\in\mathbb{R}^{n+m}.$
Let we have an optimal solution for the canonical form $(x^*,s^*)$. It corresponds to some basic feasible solution (bfs) with (no more than) $m$ non-zero basic coordinates. Each bfs corresponds to a certain vertex (in the Polyhedron $Ax+Is = b, (x,s)\geq 0$) which in turn is determined by active inequalities (which become equalities for that optimal solution $(x^*,s^*)$). The number of active inequalities is equal to the dimension of the space i.e. $n+m$ for canonical LP. We have $m$ equalities in the system $Ax+Is = b$. We obtain the remaining $n$ equalities from these simple inequalities $(x,s)\geq 0$ by substituting $n$ zero values of non-basic variables.
So we got $n+m$ inequalities that turned into equalities (for the solution) that determine the vertex of the solution to the problem in canonical form.
But what about the vertex of the original problem in standard form?
Let $k\leq m$ of basic coordinates correspond to vector $x^*$, and others $m-k$ to vector of slack variables $s^*$. Then we get (from $Ax\leq b$) two subsystems of inequalities $A_1x\leq b$ with $k$ inequalities and $A_2x\leq b$ with $m-k$ inequalities, where for $x^*$ $A_1x^* = b$ and $A_2x^* < b$. Therefore, we need $n-k$ more active inequalities to obtain the vertex of the solution $x^*$ to the problem in standard form (in the Polyhedron $Ax \leq b, x\geq 0$). We obtain the remaining $n-k$ equalities from these simple inequalities x≥0 by substituting $n-k$ zero values of non-basic variables in $x^*$.
Indeed, there are $n$ non-basic variables in the vector $(x^*,s^*),$ $k$ belong to the vector $x^*$ and $n-k$ corresponds to the vector $s^*.$ Therefore, among $n$ inequalities $x\geq0$ there are $n - k$ active inequalities for $x^*$. So we have $n$ active inequalities for $x^*$ in standard form LP.
Is this true approach?