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sav08:partial_congruences [2008/04/23 07:50]
sav08:partial_congruences [2008/04/26 17:42]
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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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 \[ \[
 \begin{array}{l} \begin{array}{l}
-   ​\forall x_1,​\ldots,​x_n,​y_1,​\ldots,​y_n. \bigwedge_{i=1}^n (x_i,y_i) \in r\ \land \\  +   ​\forall x_1,​\ldots,​x_n,​y_1,​\ldots,​y_n. \bigwedge_{i=1}^n (x_i,y_i) \in r\ \land \ f(x_1,​\ldots,​x_n) \in T \land f(y_1,​\ldots,​y_n) \in T  \rightarrow \\ 
-    ​f(x_1,​\ldots,​x_n) \in T \land f(y_1,​\ldots,​y_n) \in T  \rightarrow \\ +
     (f(x_1,​\ldots,​x_n),​f(y_1,​\ldots,​y_n)) \in r     (f(x_1,​\ldots,​x_n),​f(y_1,​\ldots,​y_n)) \in r
 \] \]