hence, since $1-\xi>0$ in (\[proof:10\]) by the definition of $\xi$, $\dot{\mathcal{V}}^*(\cdot)\le0$. Therefore, the solution $\bigl(e(t),\tilde{W}(t),e_\rom{L}(t),\tilde{x}(t)\bigl)$ given by (\[err:sys:mod\]), (\[filter\]), (\[update:law:error\]), and (\[deviation:error\]) is Lyapunov stable for all $\bigl(e_0,\tilde{W}_0,0,0\bigl) \in \IR^n \times \IR^{(s+n)\times m}\times\IR^n\times\IR^n$ and $t \in \overline{\IR}_+$.

To show $\lim_{t\rightarrow\infty}\bigl(x(t)-x_\rom{r_i}(t)\bigl)=0$, note that $\sigma\bigl(x(t)\bigl)$ is bounded for all $t \in \overline{\IR}_+$, and hence, $\dot{e}(t)$ is bounded. Furthermore, since $\dot{e}_\rom{L}(t)$ and $\dot{\tilde{x}}(t)$ are also bounded, then $\ddot{\mathcal{V}}^*\bigl(e(t),\tilde{W}(t),e_\rom{L}(t),$ $\tilde{x}(t)\bigl)$ is bounded for all $t \in \overline{\IR}_+$. Now, it follows from Barbalat’s lemma \[, Lemma 8.2\] that \_[t ]{}\^\*(e(t),(t),e\_(t),(t))=0, which consequently shows that \_[t]{}e(t)&=&0,\
\_[t]{}e\_(t)&=&0,\
\_[t]{}(t)&=&0. Therefore, it follows from x(t)-x\_(t)&=&x(t)-x\_(t)+x\_(t)-x\_(t)=e(t)+(t),\[in:ineq\] that \_[t]{}(x(t)-x\_(t))&=&0.

Finally, since $\dot{\mathcal{V}}^*\bigl(e(t),\tilde{W}(t),e_\rom{L}(t),\tilde{x}(t)\bigl)\le0$ for all $t \in \overline{\IR}_+$, this implies that \^\*(e(t),(t),e\_(t),(t))&&\^\*(e\_0,\_0,0,0).    \[proof:20\] Using ${\mathcal{V}}^*\bigl(e_0,\tilde{W}_0,0,0\bigl)\le\epsilon_\mathcal{V}$ and \^\*(e(t),(t),e\_(t),(t))&&\_(P)\^2\_2,     in (\[proof:20\]) yields $\norm{e(t)}_2\le \sqrt{\lambda_\rom{min}^{-1}(P)\epsilon_\mathcal{V}}$. Since $\norm{\cdot}_\infty \le \norm{\cdot}_2$, and this bound is uniform, then $\norm{e_\tau(t)}_{\mathcal{L}_\infty}\le \sqrt{\lambda_\rom{min}^{-1}(P)\epsilon_\mathcal{V}}$, and hence, \_[\_]{}, \[temp:05\] is a direct consequence due to the fact that the former expression holds uniformly in $\tau$. Similarly, using ${\mathcal{V}}^*\bigl(e_0,\tilde{W}_0,0,0\bigl)\le\epsilon_\mathcal{V}$ and \^\*(e(t),(t),e\_(t),(t))&&2\^[-1]{} \_\^[-1]{}(P) \_(P) \_(R)\^2\_2, yields \_[\_]{}. \[temp:06\] Now, it follows from (\[in:ineq\]) that \_[\_]{}\_[\_]{}+\_[\_]{}, \[temp:07\] and hence, (\[bound:1\]) is a direct consequence of using (\[temp:05\]) and (\[temp:06\]) in (\[temp:07\]). This completes the proof. $\square$

*Remark 5.1*. Even though the proposed architecture is predicated 