Differences

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--- sav08:non-ground_instantiation_and_resolution [2008/04/01 16:21]
vkuncak
+++ sav08:non-ground_instantiation_and_resolution [2008/04/01 17:05]
vkuncak
@@ Line 42: / Line 42: @@
 \[
 \frac{C \cup \{\lnot A_1\}\ \ \ D \cup \{A_2\}}
-     {subst(\sigma,C \cup D)}
+     {subst(\sigma_1)(C) \cup subst(\sigma_2)(D)}
 \]
-such that $subst(\sigma)(A_1) = subst(\sigma)(A_2)$.
+such that $subst(\sigma_1)(A_1) = subst(\sigma_2)(A_2)$.
+Resolution with instantiation generalizes resolution and ground resolution.
+One complete proof system contains:
+  * instantiation
+  * resolution with instantiation
+Note: if we apply instantiation that renames variables in each clause, then $\sigma_1$ and $\sigma_2$ can have disjoint domains and we let $\sigma = \sigma_1 \cup \sigma_2$, obtaining
 Note: $\sigma$ such that $subst(\sigma)(A_1) = subst(\sigma)(A_2)$ is called a **unifier** for $\{A_1,A_2\}$.
+Further step: do we need to consider all unifiers?
+Most general unifier.  To compute it we can use the standard [[Unification]] algorithm.
+----
 **Factoring with instantiation**
 \[
-\frac{C \cup \{\lnot A_1, A_2\}}
+\frac{C \cup \{A_1, A_2\}}
-     {subst(\sigma,C \cup D)}
+     {subst(\sigma)(C)}
 \]
 where $subst(\sigma)(A_1)=subst(\sigma)(A_2)$.
-Further step: do we need to consider all unifiers?
-Most general unifier.  To compute it we can use the standard [[Unification]] algorithm.