we have $\lw_{\sigma}(ww_{j}g_{j}w'_{j}w')>_{\sigma}\lw_{\sigma}(ww_{k}g_{i_{k}}w'_{k}w')$ for all $k\in\{1,\dots,\mu\}$ and $S_{i,j}(ww_{i},w'_{i}w';ww_{j},w'_{j}w')=\sum^{\mu}_{k=1}c_{k}ww_{k}g_{i_{k}}w'_{k}w'$ is a Gröbner representation in terms of $G$.

To deal with trivial obstructions of type (T2), we introduce some terminology as follows.

\[sec2def7\] Let $G=\{g_1,\dots,g_s\}\subseteq K\langle X\rangle\setminus\{0\}$ with $s\geq 1$.

-   Let $w_{1},w_{2}\in\langle X\rangle$ be two words. If there exist some words $w,w',w''\in\langle X\rangle$ and $w\neq 1$ such that $w_{1}=w'w$ and $w_{2}=ww''$, or $w_{1}=ww'$ and $w_{2}=w''w$, or $w_{1}=w$ and $w_{2}=w'ww''$, or $w_{1}=w'ww''$ and $w_{2}=w$, then we say $w_{1}$ and $w_{2}$ have an *overlap* at $w$. Otherwise, we say that $w_{1}$ and $w_{2}$ have *no overlap*.

-   Let ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})\in {\rm Obs}(i,j)$ be an obstruction. If $\lw_{\sigma}(g_{i})$ and $\lw_{\sigma}(g_{j})$ have an overlap at $w\in\langle X\rangle\setminus\{1\}$ and if $w$ is a subword of $w_{i}\lw_{\sigma}(g_{i})w'_{i}$, then we say that ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ has an *overlap* at $w$. Otherwise, we say that ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ has *no overlap*.

Thus, as shown in (T2), there are infinitely many obstructions without overlaps in each ${\rm Obs}(i,j)$. The following lemma gets rid of these trivial obstructions.

\[sec2lem8\] If ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})\in {\rm Obs}(i,j)$ has no overlap, then $S_{i,j}(w_{i},w'_{i};$ $w_{j},w'_{j})$ has a Gröbner representation in terms of $G$.

See [@Mo94], Lemma 5.4.

Observe that Lemma \[sec2lem8\] is indeed a non-commutative version of the 