LARA

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sav08:non-ground_instantiation_and_resolution [2008/04/01 16:25]
vkuncak
sav08:non-ground_instantiation_and_resolution [2008/04/01 17:39]
vkuncak
Line 42: Line 42:
 \[ \[
 \frac{C \cup \{\lnot A_1\}\ \ \ D \cup \{A_2\}} \frac{C \cup \{\lnot A_1\}\ \ \ D \cup \{A_2\}}
-     ​{subst(\sigma,C \cup D)}+     ​{subst(\sigma_1)(C\cup subst(\sigma_2)(D)}
 \] \]
-such that $subst(\sigma)(A_1) = subst(\sigma)(A_2)$.+such that $subst(\sigma_1)(A_1) = subst(\sigma_2)(A_2)$.
  
-Note: $\sigma$ such that $subst(\sigma)(A_1) = subst(\sigma)(A_2)$ is called a **unifier** for $\{A_1,​A_2\}$.+Resolution with instantiation generalizes resolution and ground resolution.
  
-**Factoring ​with instantiation** +One complete proof system contains: 
-\[ +  ​instantiation 
-\frac{C ​\cup \{A_1A_2\}} +  ​resolution ​with instantiation 
-     {subst(\sigma)(C)} + 
-\+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 
-where $subst(\sigma)(A_1)=subst(\sigma)(A_2)$.+ 
 +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? Further step: do we need to consider all unifiers?