=== Necessary Condition === Let $F$ be a Gröbner basis. Let $f_1, f_2 \in F$. Then: :$SP \left({f_1, f_2}\right) \in Ideal \left({F}\right)$ That is: :$SP \left({f_1, f_2}\right) \equiv_F 0$ {{explain|Link to a result explaining the above.}} By the relation between reduction and congruence this implies that: :$SP \left({f_1, f_2}\right) \leftrightarrow_F^* 0$ {{explain|Define via links to the appropriate pages the meaning of the above.}} Hence: :$RF \left({F, SP \left({f_1, f_2}\right)}\right) = RF \left({F, 0}\right) = 0$  because $F$ is a Gröbner basis (and because of the equivalence between the Church-Rosser property and the normal form Church-Rosser property). {{explain|Establish the above equivalence and the conclusion drawn from it in a separate page.}} {{qed|lemma}} === Sufficient Condition === Let: :$\forall f_1, f_2 \in F \left({RF \left({F, SP \left({f_1, f_2}\right)}\right) = 0}\right)$ By the generalized Newman lemma and: : $\gets_F \mathop \subseteq \succ$ it suffices to prove local connectibility. That is, it suffices to prove that under the assumption: :$g_1 \to_F h \gets_F g_2$ we have: :$g_1 \stackrel{\prec h} {\longleftrightarrow^*_F} g_2$ {{explain|All the above notation needs to be explained.}} Let $LPP \left({f}\right)$ be the leading power product of $f$. Then by the assumption, there exist $f_1, f_2 \in F$ and $t_1, t_2 \in S \left({h}\right)$ with: : $LPP \left({f_1}\right) | t_1$ and: : $LPP \left({f_2}\right) | t_2$ 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-eqn}} {{eqn | l = g_1       | r = H \left({h, t_1}\right) + 0 \cdot t_1 + B \left({h, t_1, t_2}\right) + C \left({h, t_2}\right) \cdot t_2       | c =  }} {{eqn | o =       | ro= +       | r = L \left({h, t_2}\right) - C \left({h, t_1}\right) \cdot u_1 \cdot R \left({f_1}\right)       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = g_2       | r = H \left({h, t_1}\right) + C \left({h, t_1}\right) \cdot t_1 + B \left({h, t_1, t_2}\right) + 0 \cdot t_2       | c =  }} {{eqn | o =       | ro= +       | r = L \left({h, t_2}\right) - C \left({h, t_2}\right) \cdot u_2 \cdot R \left({f_2}\right)       | c =  }} {{end-eqn}} where: : $u_1 := \dfrac {t_1} {LPP \left({f_1}\right)}$ and: : $u_2 := \dfrac {t_2} {LPP \left({f_1}\right)}$ Furthermore: {{begin-eqn}} {{eqn | l = g_2 \to_{f_1} g_{1, 2}       | r = H \left({h, t_1}\right) + 0 \cdot t_1 + B \left({h, t_1, t_2}\right) + 0 \cdot t_2 + L \left({h, t_2}\right)       | c =  }} {{eqn | o =       | ro= -       | r = C \left({h, t_1}\right) \cdot u_1 \cdot R \left({f_1}\right) - C \left({h, t_2}\right) \cdot u_2 \cdot R \left({f_2}\right)       | c =  }} {{end-eqn}} Now: : $g_1 = h - C \left({h, t_1}\right) \cdot u_1 \cdot f_1$ and: : $g_{1, 2} = g_2 - C \left({h, t_1}\right) \cdot u_1 \cdot f_1$ and, by assumption: : $h \to_F g_2$ Hence, by sum semi-compatibility: :$g_1 \downarrow_{\stackrel{*} {F} } g_{1, 2}$ and hence: :$g_1 \stackrel{\prec h}{\longleftrightarrow^{*}_F} g_2$ (Note that, in general, $g_1 \to_{f_1} g_{1, 2}$ need not be the case. Why not?) {{explain|Not {{ProofWiki}} style. Needs to be linked to a page demonstrating this fact.}} ''Case $t_1 \prec t_2$'': Analogous. ''Case $t := t_1 = t_2$'': In this case: :$g_1 = H \left({h, t}\right) + 0 \cdot t + L \left({h, t}\right) - C \left({h, t}\right) \cdot u_1 \cdot R \left({f_1}\right)$ and: :$g_2 = H \left({h, t}\right) + 0 \cdot t + L \left({h, t}\right) - C \left({h, t}\right) \cdot u_2 \cdot R \left({f_2}\right)$ Hence: {{begin-eqn}} {{eqn | l = g_1 - g_2       | r = - C \left({h, t}\right) \cdot \left({u_1 \cdot R \left({f_1}\right) - u_2 \cdot R \left({f_2}\right)}\right)       | c =  }} {{eqn | r = - C \left({h, t}\right) \cdot \left({u_1 \cdot f_1 - u_2 \cdot f_2}\right)       | c =  }} {{eqn | r = - C \left({h, t}\right) \cdot v \cdot SP \left({f_1, f_2}\right)       | c =  }} {{end-eqn}} {{explain|Explain why the above steps hold, by links to explanatory pages if necessary}} where: :$v := t / LCM \left({LPP \left({f_1}\right), LPP \left({f_2}\right)}\right)$ {{explain|Presumably here LCM means lowest common multiple?}}                                       We have assumed that: : $RF \left({F, SP \left({f_1, f_2}\right)}\right) = 0$ That is: : $SP \left({f_1, f_2}\right) \to_{\stackrel{\ast}{F}} 0$ Hence,by product compatibility: :$g_1 - g_2 = - C \left({h, t}\right) \cdot v \cdot SP \left({f_1, f_2}\right) \to_{\stackrel{*}{F}} 0$  This means that there exists a sequence $p \in P^{\ast}$ such that: : {{begin-eqn}} {{eqn | l = p_1       | r = g_1 - g_2       | c =  }} {{eqn | n = 1       | l = \forall 1 \le i < \left\vert{p}\right\vert:       | o =        | r = \left({p_i \to_F p_{i + 1} }\right)       | c =  }} {{end-eqn}} {{explain|What is $P^*$?}} and: :$p_{\left\vert{p}\right\vert} = 0 $ Furthermore note that, because of $\to_F \subseteq \succ$: :$\forall 1 \le i < \left\vert{p}\right\vert: p_i\preceq g_1 -g_2 \prec h$ Thus, by sum semi-compatibility applied to $(1)$: :$g_1 = p_1 + g_2$ :$\forall 1 \le i < \left\vert{p}\right\vert: p_i + g_2 \downarrow_{\stackrel{*}{F} } p_{i+1} + g_2$ :$g_2 = p_{\left\vert{p}\right\vert} + g + 2$ Also, we have: :$\forall 1 \le i < \left\vert{p}\right\vert: p_i + g_2 \prec h$ because: :$\forall 1 \le i < \left\vert{p}\right\vert: H \left({p_i + g_2, t}\right) = H \left({h, t}\right) \wedge C \left({p_i + g_2, t}\right) = 0$ Thus, summarizing: : $g_1 \stackrel{\prec h} {\longleftrightarrow^{*}_F} g_2$ also in this case.