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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. |