G$ for all $k\in\{1,\dots,\mu\}$ such that $\lw_{\sigma}(w_{j}g_{j}w'_{j})>_{\sigma}\lw_{\sigma}(w_{k}g_{i_{k}}w'_{k})$ if $c_k\neq 0$ for some $k\in\{1,\dots,\mu\}$.

See [@Xiu12], Proposition 4.1.2.

A presentation of $S_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ as in Proposition \[sec2pro5\].b is called a *(weak) Gröbner representation* of $S_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ in terms of $G$.

Observe that there are infinitely many obstructions in each set ${\rm Obs}(i,j)$, due to the following two types of *trivial* obstructions.

-   If ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})\in {\rm Obs}(i,j)$, then, for all $w,w'\in\langle X\rangle$, we have ${\rm o}_{i,j}(ww_{i},w'_{i}w';ww_{j},w'_{j}w')\in{\rm Obs}(i,j)$.

-   For all $w\in\langle X\rangle$, we have ${\rm o}_{i,j}(\lw_{\sigma}(g_{j})w,1;1,w\lw_{\sigma}(g_{i})), {\rm o}_{i,j}(1, w\lw_{\sigma}(g_{j});$ $\lw_{\sigma}(g_{i})w,1)\in{\rm Obs}(i,j)$.

Before going on, let us get rid of these two types of trivial obstructions. The following lemma handles trivial obstructions of type (T1).

\[sec2lem6\] If the S-polynomial of ${\rm o}_{i,j}(w_{i},w'_{i};w_{j},w'_{j})\in {\rm Obs}(i,j)$ has a Gröbner representation in terms of $G$, then, for all $w,w'\in\langle X\rangle$, the S-polynomial of ${\rm o}_{i,j}(ww_{i},w'_{i}w';ww_{j},w'_{j}w')$ also has a Gröbner representation in terms of $G$.

Without loss of generality, we assume that $S_{i,j}(w_{i},w'_{i};w_{j},w'_{j})$ is non-zero. We write $S_{i,j}(w_{i},w'_{i};w_{j},w'_{j})=\sum^{\mu}_{k=1}c_{k}w_{k}g_{i_{k}}w'_{k}$, where $c_{k}\in K\setminus\{0\}$, $w_{k},w'_{k}\in\langle X\rangle$, and $g_{i_{k}}\in G$ such that $\lw_{\sigma}(w_{j}g_{j}w'_{j})>_{\sigma}\lw_{\sigma}(w_{k}g_{i_{k}}w'_{k})$ for all $k\in\{1,\dots,\mu\}$. For all $w,w'\in\langle X\rangle$, it is clear that $S_{i,j}(ww_{i},w'_{i}w';ww_{j},w'_{j}w')$ $=\sum^{\mu}_{k=1}c_{k}ww_{k}g_{i_{k}}w'_{k}w'$. Since the word ordering $\sigma$ is compatible with multiplication, we have $w\lw_{\sigma}(w_{j}g_{j}w'_{j})w'>_{\sigma}w\lw_{\sigma}(w_{k}g_{i_{k}}w'_{k})w'$ for all $k\in\{1,\dots,\mu\}$. Hence 