all unnecessary obstructions in the set $\bigcup_{1\leq i\leq s}{\rm NTObs}(i,s)$ of newly constructed obstructions.

So far we have detected unnecessary obstructions in the set $\bigcup_{1\leq i\leq s}{\rm O}(i,s)$ of newly constructed obstructions. Intuitively, we are also able to detect unnecessary obstructions in the set $\bigcup_{1\leq i\leq j\leq s-1}{\rm Obs}(i,j)$ of previously constructed obstructions. Thus, in the last step, we detect unnecessary obstructions in this set by using the new generator $g_{s}$.

\[sec3lem10\] Let ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})\in\bigcup_{1\leq i\leq j\leq s-1}{\rm NTObs}(i,j)$ be a non-trivial obstruction. If there are two words $w,w'\in\langle X\rangle$ satisfying $w_{j}\lw_{\sigma}(g_{j})w'_{j}=w\lw_{\sigma}(g_{s})w'$, then we can represent ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ as $${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})={\rm o}_{i,s}(w_{i},w'_{i};w,w')-{\rm o}_{j,s}(w_{j},w'_{j};w,w').$$ Moreover, if $S_{i,s}(w_{i},w'_{i};w,w')$ and $S_{j,s}(w_{j},w'_{j};w,w')$ have Gröbner representations in terms of $G$, then so does $S_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$.

The claimed equality follows from Definition \[sec2def4\].a and the condition $w_{j}\lw_{\sigma}(g_{j})w'_{j}$ $=w\lw_{\sigma}(g_{s})w'$. The proof of the second part is analogous to the proof of the second part of Lemma \[sec3lem3\].a.

The following example shows that the obstruction ${\rm o}_{i,s}(w_{i},w'_{i};w,w')$ in the equation of Lemma \[sec3lem10\] can be either an obstruction without overlap or a multiple of a non-trivial obstruction. In the case that ${\rm o}_{i,s}(w_{i},w'_{i};w,w')$ is a multiple of a non-trivial obstruction, say ${\rm o}_{i,s}(\tilde{w}_{i},\tilde{w}'_{i};\tilde{w},\tilde{w}')$, the example shows that it is 