[proofplan]
The proof is the standard tube-MPC invariance argument on the closed-loop feasibility prefix $I$. Define the tracking error $e_k=x_k-z_k$ and use the feedback law $u_k=v_k+Ke_k$ to show that the error evolves according to the robustly invariant error system. Robust positive invariance and the initial condition $e_0\in E$ then imply by induction that $e_k\in E$ for every $k\in I$. The Pontryagin-difference constraints convert $z_k\in X\ominus E$ and $v_k\in U\ominus KE$ into the original constraints $x_k\in X$ and $u_k\in U$.
[/proofplan]
[step:Derive the closed-loop error recursion]
For each $k\in I$, define the tracking error $e_k\in \mathbb{R}^n$ by
\begin{align*}
e_k:=x_k-z_k.
\end{align*}
Using the uncertain dynamics, the nominal dynamics, and the applied input law, we compute
\begin{align*}
e_{k+1}=x_{k+1}-z_{k+1}.
\end{align*}
Substituting $x_{k+1}=Ax_k+Bu_k+w_k$, $z_{k+1}=Az_k+Bv_k$, and $u_k=v_k+Ke_k$ gives
\begin{align*}
e_{k+1}=A(x_k-z_k)+B(u_k-v_k)+w_k.
\end{align*}
Since $x_k-z_k=e_k$ and $u_k-v_k=Ke_k$, this becomes
\begin{align*}
e_{k+1}=(A+BK)e_k+w_k.
\end{align*}
Thus the closed-loop tracking error follows exactly the robustly invariant error dynamics.
[/step]
[step:Propagate membership of the error in the invariant tube]
We prove by induction along the feasibility prefix $I$ that $e_k\in E$ for every $k\in I$. The base case is the hypothesis $e_0=x_0-z_0\in E$. For the inductive step, assume $k,k+1\in I$ and assume $e_k\in E$. Since the disturbance satisfies $w_k\in W$ and $E$ is robust positively invariant for $e_{k+1}=(A+BK)e_k+w_k$, we have
\begin{align*}
(A+BK)e_k+w_k\in E.
\end{align*}
By the error recursion derived above, this is exactly $e_{k+1}\in E$. Induction on the prefix $I$ gives $e_k\in E$ for every $k\in I$.
[guided]
We want to show that the true state never leaves the tube around the nominal state. The tube cross-section is the set $E$, so the exact statement to prove is $e_k\in E$ for every relevant time $k$, where the error variable is
\begin{align*}
e_k:=x_k-z_k.
\end{align*}
First we verify the recursion satisfied by this error. The uncertain state sequence satisfies
\begin{align*}
x_{k+1}=Ax_k+Bu_k+w_k,
\end{align*}
the nominal state sequence satisfies
\begin{align*}
z_{k+1}=Az_k+Bv_k,
\end{align*}
and the applied input is
\begin{align*}
u_k=v_k+Ke_k.
\end{align*}
Therefore
\begin{align*}
e_{k+1}=x_{k+1}-z_{k+1}.
\end{align*}
Substituting the two dynamics and the input law gives
\begin{align*}
e_{k+1}=A(x_k-z_k)+B(u_k-v_k)+w_k.
\end{align*}
Since $x_k-z_k=e_k$ and $u_k-v_k=Ke_k$, the error recursion is
\begin{align*}
e_{k+1}=(A+BK)e_k+w_k.
\end{align*}
The base case is part of the theorem's hypotheses:
\begin{align*}
e_0=x_0-z_0\in E.
\end{align*}
Now suppose that $k,k+1\in I$ and that the error already satisfies $e_k\in E$. The disturbance at that time satisfies $w_k\in W$. The robust positive invariance hypothesis says precisely that every point of $E$ remains in $E$ after applying the error update and adding any disturbance from $W$:
\begin{align*}
(A+BK)e+w\in E
\end{align*}
for every $e\in E$ and every $w\in W$. Applying this statement with $e=e_k$ and $w=w_k$ gives
\begin{align*}
(A+BK)e_k+w_k\in E.
\end{align*}
By the error recursion just derived, this is exactly the assertion that $e_{k+1}\in E$. This completes the induction step along the prefix $I$, and hence $e_k\in E$ for every $k\in I$.
[/guided]
[/step]
[step:Convert the tightened state constraint into the original state constraint]
Fix $k\in I$. From the previous step, $e_k\in E$. The nominal MPC constraint gives $z_k\in X\ominus E$. By the definition of the Pontryagin difference, $z_k+d\in X$ for every $d\in E$. Taking $d=e_k$ yields
\begin{align*}
z_k+e_k\in X.
\end{align*}
Since $e_k=x_k-z_k$, we have $x_k=z_k+e_k$. Hence
\begin{align*}
x_k\in X.
\end{align*}
[/step]
[step:Convert the tightened input constraint into the original input constraint]
Fix $k\in I$. Since $e_k\in E$, the definition of $KE$ gives
\begin{align*}
Ke_k\in KE.
\end{align*}
The nominal MPC input constraint gives $v_k\in U\ominus KE$. By the definition of the Pontryagin difference, $v_k+q\in U$ for every $q\in KE$. Taking $q=Ke_k$ yields
\begin{align*}
v_k+Ke_k\in U.
\end{align*}
The applied input is $u_k=v_k+K(x_k-z_k)=v_k+Ke_k$, so
\begin{align*}
u_k\in U.
\end{align*}
[/step]
[step:Conclude robust constraint satisfaction over all feasible closed-loop times]
The preceding arguments used only the assumptions $w_k\in W$, $x_0-z_0\in E$, robust positive invariance of $E$, and the tightened nominal constraints imposed for $k\in I$. Therefore the conclusion holds for every admissible disturbance sequence $(w_k)_{k\geq 0}$ with $w_k\in W$. For every $k\in I$, the corresponding true state and applied input satisfy
\begin{align*}
x_k\in X
\end{align*}
and
\begin{align*}
u_k\in U.
\end{align*}
This proves robust satisfaction of the original state and input constraints.
[/step]
