$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
$\begin{array}{ll}
-& = \min_{x \in span\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
-& = \min_{x \in span\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
-& \leqslant \min_{x \in span\left( x_{k-1} \right)} ||b-Ax ||_2\\
-& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k-1} ||_2\\
-& \leqslant ||b-Ax_{k-1}||_2 .
+& = \min_{x \in span\left(S_{k-s+1}, S_{k-s+2}, \hdots, S_{k} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in span\left(x_{k-s+1}, x_{k-s}+2, \hdots, x_{k} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in span\left( x_{k} \right)} ||b-Ax ||_2\\
+& \leqslant \min_{\lambda \in \mathbb{R}} ||b-\lambda Ax_{k} ||_2\\
+& \leqslant ||b-Ax_{k}||_2\\
+& = ||r_k||_2\\
+& \leqslant \left(1-\dfrac{\alpha}{\beta}\right)^{\frac{km}{2}} ||r_0||,
\end{array}$
\end{itemize}
+which concludes the induction and the proof.
\end{proof}
We can remark that, at each iterate, the residue of the TSIRM algorithm is lower
