, thus $S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}}= P_0$ and $B=$[$S_{0,0,0}$]{}. We know by Lemma \[lem:grundyStars\] that ${\mathcal{G}}(B)=0={\mathcal{G}}(S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}})$ so $S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}}+B$ is a ${\mathcal{P}}$-position;  -   $S$ is a single vertex, thus $S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}}= P_0$ and $B=$[$S_{0,0,0}$]{}. Considering Figure \[fig:tabgrun\], we get ${\mathcal{G}}(B)=0={\mathcal{G}}(S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}})$ so $S{\tikz[baseline=-0]{\draw (0,0) node[circle,fill=black,minimum size=0,inner sep=0]{} -- (0.0,0) node[circle,fill=black, minimum size=0, inner sep=0] {};}}+B$ is a ${\mathcal{P}}$-position;  -   $S