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sav08:partial_congruences [2008/04/23 07:50]
vkuncak
sav08:partial_congruences [2015/04/21 17:30] (current)
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**Proof:​** ​ **Proof:​** ​

-Show $C^i(s) \cap T^2 = s$ for all $i \ge 0$.+Show $C^i(s) \cap T^2 = s$ for all $i \ge 0$, by induction.
+
+  * $i = 0$: s = s
+
+  * $i \rightarrow i+1$:
+$C^{i+1}(s) \cap T^2 = C(C^i(s) \cap T^2) \cap T^2$, by induction hypothesis $C^i(s) \cap T^2 =s$.
+Thus $C(C^i(s) \cap T^2) \cap T^2 = C(s) \cap T^2$.
+As $C$ adds only needed term for congruence, the added term are either not in $T^2$ or $s$ in not a congruence.
+By hypothesis $s$ is a congruence, so $C(s) \cap T^2 = s$.
+Therefore $C^{i+1}(s) \cap T^2 = s$.

**Proof End.** **Proof End.**
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We apply the congruence condition only to terms that already exist in the set, using congruence condition: We apply the congruence condition only to terms that already exist in the set, using congruence condition:
-$+\begin{equation*} \begin{array}{l} \begin{array}{l} ​\forall x_1,​\ldots,​x_n,​y_1,​\ldots,​y_n. \bigwedge_{i=1}^n (x_i,y_i) \in r\ \land \ f(x_1,​\ldots,​x_n) \in T \land f(y_1,​\ldots,​y_n) \in T \rightarrow \\ ​\forall x_1,​\ldots,​x_n,​y_1,​\ldots,​y_n. \bigwedge_{i=1}^n (x_i,y_i) \in r\ \land \ f(x_1,​\ldots,​x_n) \in T \land f(y_1,​\ldots,​y_n) \in T \rightarrow \\ (f(x_1,​\ldots,​x_n),​f(y_1,​\ldots,​y_n)) \in r (f(x_1,​\ldots,​x_n),​f(y_1,​\ldots,​y_n)) \in r -$+\end{equation*}