Suppose that the polytope $P=\{x\mid Ax\leq b\}$ is totally dual integral (TDI), and that the inequality $\tilde{a}^Tx\leq \tilde{b}$ is satisfied at every point of $P$.
Does this imply that the system $\{Ax\leq b, \ \ \tilde{a}^Tx\leq \tilde{b}\}$ is TDI ?
I am not sure how to (dis)prove that. I tried to apply the definition of TDI to the new system and I don't quite see why it should be TDI. On the other hand it would seem strange that adding a redundant inequality breaks the TDI property.
Total dual integrality is a property of a system and not of a polyhedron. For every rational polyhedron $P$ there exists a TDI system $Ax \leq b$ such that $P = \{ x\in \mathbb{R}^n\ :\ Ax \leq b\}$. You need $b$ to be integral in order to be able to deduce that $P$ is integral.
Adding a redundant inequality to a TDI system results to a TDI system. Consider the primal problem with an integral $c$ \begin{align} \max&.\ cx\\ s.t.\ &Ax \leq b \end{align}
Since the Primal system before adding the redundant inequality is TDI there exists an integral optimal solution $u'$ for the corresponding Dual Problem.
Now consider the dual problem \begin{align} \max. ub &+ v \bar{b}\\ s.t. uA &= c,\ \\ v\bar{a} &=c,\\ u \geq 0,\ &v \geq 0 \end{align} To prove that the Primal system is TDI we have to construct an optimal integer solution $u^*$ for the Dual problem. Notice that adding a redudant inequality does not change the optimal value of the Primal problem. Then the solution $u^* =(u', 0)$ is a feasible optimal integral solution for the Dual.
Nevertheless adding redundant inequalities to systems which are not TDI may result to a TDI system. The following system is clearly non TDI. \begin{align} \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \leq \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{align} Still we can produce a TDI system for the same polyhedron by adding a redudant inequality to obtain the TDI system \begin{align} \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \leq \begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix} \end{align}
This paper may be useful if you want to see which operations preserve Total Dual Integrality.