Babai's nearest plane algorithm solves approximate-CVP (Closest Vector Problem) where the approximation factor is $2(\frac{2}{\sqrt{3}})^n$.
Let $b_1,...,b_n$ be a basis and $t$ be the target. This algorithm finds an integer $c$ such that the hyperplane $c b_n^*+span(b_1,...,b_{n-1})$ is close to $t$. Recursively, it finds a lattice point $x^\prime \in L(b_1,...,b_{n-1})$ that is closest to the projection $t^\prime$ of $t-cb_n$ onto $span(b_1,....,b_{n-1})$ and output $x=x^\prime+cb_n$.
In proof, we have $$||x-t|| \le 2(\frac{2}{\sqrt{3}})^n||y-t||$$ where y is the nearest point to $t$.
If $||t-y|| \le \frac{b_n^*}{2}$ then $y \in c b_n^*+span(b_1,...,b_{n-1})$ and $y^\prime =y-cb_n \in L(b_1,....,b_n)$ is the nearest point to $t^\prime$. I could not understand how $x^\prime$ is within the distance $2(\frac{2}{\sqrt{3}})^{n-1}||t^\prime-y^\prime||$ and how to prove in this case.
