$\sigma z_i=z_{i+1}$ for all $i$. Then $G$ preserves the lexicographic order on ${\mathrm{Mon}}$ for which $z_{i+1}>z_i>y>x$ for all $i$. Now consider the set $B=\{y+1\}$. We have $$G(y+1) \times G(y+1)=G(y+1,y+1) \cup G(y+1,xz_1+1) \cup
G(xz_1+1,y+1),$$ so we may take $H$ from Definition \[de:EqS\] equal to $\{(1,1),(1,\sigma),(\sigma,1)\}.$ The S-polynomial $S(y+1,y+1)$ is zero, and the S-polynomials $S(y+1,xz_1+1)$ and $S(xz_1+1,y+1)$ reduce to zero modulo $y+1$ and $\sigma(y+1)=xz_1+1$. Hence we have a complete set of S-polynomials of $y+1$ with itself that all $G$-reduce to zero modulo $B$. Nevertheless, $B$ is not a $G$-Gröbner basis of the $G$-stable ideal that it generates, since that ideal also contains $S(xz_1+1,xz_2+1)=z_2-z_1$, which does not $G$-reduce to zero modulo $B$.

Here is an example where EGB4 is not fulfilled.

\[ex:EGB4\] Let $X=\{x,y\}$, let $G$ be the multiplicative monoid of the positive integers, where $m$ acts by $x \mapsto x^m$ and $y \mapsto y^m$. Now $$Gx \times Gy = \{(x^i,y^j) \mid i,j \in {{\mathbb Z}}_{>0}\},$$ while the diagonal $G$-orbit of the pair $(x^i,y^j)$ equals $\{(x^{ai},y^{aj}) \mid a \in {{\mathbb Z}}_{>0}\}$. Hence $Gx \times Gy$ is not the union of finitely many $G$-orbits.

However, if we do assume EGB4, then every pair $(b_0,b_1)$ has a finite and computable complete set of $S$-polynomials and 