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sav08:definition_of_propositional_resolution [2008/03/12 10:55] vkuncak |
sav08:definition_of_propositional_resolution [2008/03/12 17:09] vkuncak |
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\[ | \[ | ||
- | \frac{C \cup \{p\}\ \ \ D \cup \{\lnot p\}} | + | \frac{C \cup \{\lnot p\}\ \ \ D \cup \{p\}} |
{C \cup D} | {C \cup D} | ||
\] | \] | ||
Here $C,D$ are clauses and $p \in V$ is a propositional variable. | Here $C,D$ are clauses and $p \in V$ is a propositional variable. | ||
+ | |||
+ | Intuition: consider equivalent formulas | ||
+ | \[ | ||
+ | \frac{((\lnot C) \rightarrow (\lnot p))\ \ \ ((\lnot p) \rightarrow D)} | ||
+ | {(\lnot C) \rightarrow D} | ||
+ | \] | ||
===== Applying Resolution Rule to Check Satisfiability ===== | ===== Applying Resolution Rule to Check Satisfiability ===== | ||
Line 38: | Line 44: | ||
Therefore, if $C_1,C_2 \in S$ and $C_3$ is the result of applying resolution rule to $C_1,C_2$, then $I \models S$ if and only if $I \models (S \cup \{ C_3 \})$. Consequently, if we obtain an empty clause (false) by applying resolution, then the original set is not satisfiable either. | Therefore, if $C_1,C_2 \in S$ and $C_3$ is the result of applying resolution rule to $C_1,C_2$, then $I \models S$ if and only if $I \models (S \cup \{ C_3 \})$. Consequently, if we obtain an empty clause (false) by applying resolution, then the original set is not satisfiable either. | ||
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