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sav08:interpolants_from_resolution_proofs [2008/03/12 13:31]
vkuncak
sav08:interpolants_from_resolution_proofs [2008/03/18 10:49]
tatjana
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 Let $A$, $B$ be sets of clauses such that $A \land B$ is not satisfiable. ​ An interpolant is a formula $H$ such that  Let $A$, $B$ be sets of clauses such that $A \land B$ is not satisfiable. ​ An interpolant is a formula $H$ such that 
   * $\models (A \rightarrow H)$ and $H \land B$ is not satisfiable and    * $\models (A \rightarrow H)$ and $H \land B$ is not satisfiable and 
-  * $FV(H) \subseteq FV(A) \cup FV(B)$+  * $FV(H) \subseteq FV(A) \cap FV(B)$
  
 Construct resolution tree that derives a contradiction from $A \cup B$.  The tree has elements of $A \cup B$ as leaves and results of application of resolution as nodes. ​ For each clause $C$ in resolution tree we define recursively formula $I(C)$ and show that $I(C)$ is interpolant for these two formulas: Construct resolution tree that derives a contradiction from $A \cup B$.  The tree has elements of $A \cup B$ as leaves and results of application of resolution as nodes. ​ For each clause $C$ in resolution tree we define recursively formula $I(C)$ and show that $I(C)$ is interpolant for these two formulas:
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     * if $q \in FV(A) \setminus FV(B)$ then $I(C) = I(C_1) \lor I(C_2)$     * if $q \in FV(A) \setminus FV(B)$ then $I(C) = I(C_1) \lor I(C_2)$
     * if $q \in FV(B) \setminus FV(A)$ then $I(C) = I(C_1) \land I(C_2)$     * if $q \in FV(B) \setminus FV(A)$ then $I(C) = I(C_1) \land I(C_2)$
-    * if $q \in FV(A) \cap FV(B)$ then $I(C) = ite(q,I(C_1),I(C_2))$+    * if $q \in FV(A) \cap FV(B)$ then $I(C) = ite(q,I(C_2),I(C_1))$
  
 We prove that $I(C)$ has the desired property by induction on the structure of the resolution proof tree. We prove that $I(C)$ has the desired property by induction on the structure of the resolution proof tree.