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--- sav08:non-ground_instantiation_and_resolution [2008/04/01 16:09]
vkuncak
+++ sav08:non-ground_instantiation_and_resolution [2008/04/01 17:45]
vkuncak
@@ Line 15: / Line 15: @@
 \]
 ++++
-===== Non-Ground Factoring ====
-\[
-\frac{C \cup \{\lnot A,A\}}
-     {C}
-\]
 ===== Non-Ground Instantiation ====
@@ Line 46: / Line 39: @@
 If we try to do the proof, how do we know what to instantiate with? ++|we instantiate to enable subsequent resolution rule.++
-Resolution with instantiation:
+**Instantiation followed by resolution:**
 \[
 \frac{C \cup \{\lnot A_1\}\ \ \ D \cup \{A_2\}}
-     {subst(\sigma,C \cup D)}
+     {subst(\sigma_1)(C) \cup subst(\sigma_2)(D)}
 \]
-such that $subst(\sigma,A_1) = subst(\sigma,A_2)$.
+such that $subst(\sigma_1)(A_1) = subst(\sigma_2)(A_2)$.
+This rule generalizes resolution and ground resolution.
+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
+One complete proof system contains:
+  * instantiation followed by resolution
+  * instantiation
-Note: $\sigma$ such that $subst(\sigma,A_1) = subst(\sigma,A_2)$ is called a **unifier** for $\{A_1,A_2\}$.
+Note: $\sigma$ such that $subst(\sigma)(A_1) = subst(\sigma)(A_2)$ is called a **unifier** for $\{A_1,A_2\}$.
-Does resolution with instantiation subsume:
+Further step: do we need to consider all possible unifiers?
-  * ++non-ground resolution rule?|yes, take empty substitution++
-  * ++instantiation rule?|not by itself++
-Further step: do we need to consider all unifiers?
+Most general unifier for $\{A_1,A_2\}$, denoted $mgu(A_1,A_2)$
-Most general unifier.
+To compute it we can use the standard [[Unification]] algorithm.