Gröbner Basis

Theorem
Let $F$ be a finite set of polynomials, $SP(f_{1},f_{2})$ S-polynomial of $f_{1}$ and $f_{2}$, and if $g$ is a polynomial and $RF(g)$ is reduced form of $g$. Then

F is a Gröbner basis $\iff \forall f_{1}, f_{2} \in F (RF(F,SP(f_{1},f_{2}))=0)$.

Proof
The direction "$\Rightarrow$" is easy. Namely, if $f_{1},f_{2}\in F$, then $SP(f_{1}, f_{2})\in Ideal(F )$, i.e. $SP(f_{1},f_{2})\equiv_{F} 0$. By the relation between reduction and congruence this implies that $SP(f_{1},f_{2})\leftrightarrow_{F}^{\ast} 0$. Hence, $RF(F,SP(f_{1},f_{2}))=RF(F,0)=0$ because $F$ is a Gr\"{o}bner basis (and because of the equivalence between the Church-Rosser property and the normal form Church-Rosser property).

For the direction "$\Leftarrow$", by the generalized Newman lemma and the fact that $\rightarrow_{F}\subseteq \succ$, it suffices to prove local connectibility, i.e. it suffices to prove that under the assumption \[ g_{1}\leftarrow_{F} h \rightarrow_{F} g_{2} \] we have \[ g_{1} \stackrel{\prec h}{\longleftrightarrow^{\ast}_{F}} g_{2}. \] Let $LPP(f)$ is leading power product of $f$. Then by the assumption, there exist $f_{1},f_{2} \in F$ and $t_{1},t_{2} \in S(h)$ with $LPP(f_{1})|t_{1}$ and $LPP(f_{2})|t_{2}$, such that $h \rightarrow_{f_{1},t_{1}} g_{1}$ and $h \rightarrow_{f_{2},t_{2}} g_{2}$.

Now we have three cases.

Case $t_{1}\succ t_{2}$: In this case, \begin{align*} g_{1}=& H(h,t_{1})+0 \cdot t_{1}+B(h,t_{1},t_{2})+C(h,t_{2})\cdot t_{2}\\ &+L(h,t_{2})-C(h,t_{1})\cdot u_{1}\cdot R(f_{1}) \end{align*} and \begin{align*} g_{2}=& H(h,t_{1})+C(h,t_{1})\cdot t_{1}+B(h,t_{1},t_{2})+0 \cdot t_{2}\\ &+L(h,t_{2})-C(h,t_{2})\cdot u_{2}\cdot R(f_{2}) \end{align*} where $u_{1}:=t_{1}/LPP(f_{1}), u_{2}:=t_{2}/LPP(f_{2})$.

Furthermore, \begin{align*} g_{2}\rightarrow_{f_{1}} g_{1,2}:=& H(h,t_{1})+0\cdot t_{1}+B(h,t_{1},t_{2})+0\cdot t_{2}+L(h,t_{2})\\ &-C(h,t_{1})\cdot u_{1}\cdot R(f_{1})-C(h,t_{2})\cdot u_{2}\cdot R(f_{2}) \end{align*}

Now $g_{1}=h-C(h,t_{1})\cdot u_{1}\cdot f_{1}$ and $g_{1,2}=g_{2}-C(h,t_{1})\cdot u_{1}\cdot f_{1}$ and, by assumption, $h\rightarrow_{F} g_{2}$. Hence, by sum semi-compatibility, $g_{1}\downarrow_{\stackrel{\ast}{F}} g_{1,2}$ and, hence, $g_{1} \stackrel{\prec h}{\longleftrightarrow^{\ast}_{F}} g_{2}$. (Note that, in general, $g_{1}\rightarrow_{f_{1}} g_{1,2}$ need not be the case. Why not?)

Case $t_{1}\prec t_{2}$: Analogous.

Case $t:=t_{1}=t_{2}$: In this case, \[ g_{1}=H(h,t)+0\cdot t+L(h,t)-C(h,t)\cdot u_{1}\cdot R(f_{1}) \] and \[ g_{2}=H(h,t)+0\cdot t+L(h,t)-C(h,t)\cdot u_{2}\cdot R(f_{2}). \] Hence, \begin{align*} g_{1}-g_{2} &=-C(h,t)\cdot (u_{1}\cdot R(f_{1})-u_{2}\cdot R(f_{2}))\\ &=-C(h,t)\cdot (u_{1}\cdot f_{1}-u_{2}\cdot f_{2})\\ &=-C(h,t)\cdot v\cdot SP(f_{1},f_{2}), \end{align*} where \[ v:=t/LCM(LPP(f_{1}),LPP(f_{2})). \] We have assumed that $RF(F,SP(f_{1},f_{2}))=0$, i.e. $SP(f_{1},f_{2})\rightarrow_{\stackrel{\ast}{F}} 0$. Hence,by product compatibility, $g_{1}-g_{2}=-C(h,t)\cdot v\cdot SP(f_{1},f_{2})\rightarrow_{\stackrel{\ast}{F}} 0$. This means that there exists a sequence $p\in P^{\ast}$ such that \begin{equation} \begin{array}{l} \displaystyle p_{1}=g_{1}-g_{2}\nonumber\\ \displaystyle \forall 1 \leq i < |p| \hspace{2mm}(p_{i} \rightarrow_{F} p_{i+1}),\label{eq1} \end{array} \end{equation}

and \[ p_{|p|}=0 \]

Furthermore note that, because of $\rightarrow_{F} \subseteq \succ$, \[ \forall 1\leq i <|p| \hspace{2mm}(p_{i}\preceq g_{1}-g_{2}\prec h). \] Thus, by sum semi-compatibility applied to (\ref{eq1}), \begin{eqnarray} && g_{1}=p_{1}+g_{2},\nonumber\\ && \forall 1\leq i < |p| \hspace{2mm}(p_{i}+g_{2}\downarrow_{\stackrel{\ast}{F}} p_{i+1}+g_{2}),\nonumber\\ && g_{2}=p_{|p|}+g+{2}.\nonumber \end{eqnarray} Also, we have \[ \forall 1\leq i <|p| \hspace{2mm}(p_{i}+g_{2} \prec h) \] because \[ \forall 1\leq i <|p| \hspace{2mm}(H(p_{i}+g_{2},t)=H(h,t) \wedge C(p_{i}+g_{2},t)=0). \] Thus, summarizing, $g_{1} \stackrel{\prec h}{\longleftrightarrow^{\ast}_{F}} g_{2}$ also in this case.