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sav08:deriving_propositional_resolution [2008/03/12 11:03]
vkuncak 		sav08:deriving_propositional_resolution [2008/03/19 17:12]
tatjana
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 We next consider proof rules for checking [[Satisfiability of Sets of Formulas]]. We next consider proof rules for checking [[Satisfiability of Sets of Formulas]].
  
-We extending the notion of [[Substitution Theorems for Propositional Logic|substitution on formulas]] to sets of formulas by+We are extending the notion of [[Substitution Theorems for Propositional Logic|substitution on formulas]] to sets of formulas by
 \[ \[
     subst(\sigma,​S) = \{ subst(\sigma,​F) \mid F \in S \}     subst(\sigma,​S) = \{ subst(\sigma,​F) \mid F \in S \}
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 We first derive a more abstract proof system and that show that resolution is a special case of it. We first derive a more abstract proof system and that show that resolution is a special case of it.
 +
  
 ==== Key Idea ==== ==== Key Idea ====
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 Then we conclude that $\exists p. S$ is equivalent to $ProjectSet(S,​p)$ defined by Then we conclude that $\exists p. S$ is equivalent to $ProjectSet(S,​p)$ defined by
 \[ \[
-   Proj(S,p) = \{ ProjectForm(F_1,​F_2,​p) \mid F_1,F_2 \in S \}+   ProjectSet(S,p) = \{ ProjectForm(F_1,​F_2,​p) \mid F_1,F_2 \in S \}
 \] \]
  
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     A = \bigcup_{i=1}^M P_i(S)     A = \bigcup_{i=1}^M P_i(S)
 \] \]
-By definition of $P_i$, we can show that the set $A$ contains the expansion of+By definition of $P_i$, we can show that the set $A$ contains ​the conjunctive normal form of the expansion of
 \[ \[
     \exists p_1,​\ldots,​p_M. (F_1 \land \ldots \land F_n)     \exists p_1,​\ldots,​p_M. (F_1 \land \ldots \land F_n)
 \] \]
-This expansion ​is a ground formula, so it evaluates to either //true// or //​false//​. ​ By assumption, $P^*(S)$ and therefore $A$ do not contain ground contradiction. ​ Therefore, $\exists p_1,​\ldots,​p_M. (F_1 \land \ldots \land F_n)$ is true and $T$ is satisfiable. +Each of these conjuncts ​is a ground formula ​(all variables $p_1,​\ldots,​p_M$ have been instantiated), so the formula ​evaluates to either //true// or //​false//​. ​ By assumption, $P^*(S)$ and therefore $A$ do not contain ​a ground contradiction. ​ Therefore, ​each conjunct of $\exists p_1,​\ldots,​p_M. (F_1 \land \ldots \land F_n)$ is true and $T$ is satisfiable.
- +
-==== Improvement:​ Simplification Rules ==== +
- +
-Of course, we do not need to wait until we reach a ground contradiction. ​ Whenever we substitute variable with //true// or //false//, we can immediately simplify the formula using sound simplification rules. +
- +
-When we introduce simplifications we still manipulate equivalent formulas, so soundness and completeness remain the same.+
  
 ==== Improvement:​ Subsumption Rules ==== ==== Improvement:​ Subsumption Rules ====
  
-Note also that if $\models (F \rightarrow G)$, $F$ has been derived before, and $FV(G) \subseteq FV(F)$ ​ then deriving $G$ does not help derive a ground contradiction,​ because the contradiction would also be derived using $F$.  If we derive such formula, we can immediately delete it so that it does not slow us down.  ​+Note also that if $\models (F \rightarrow G)$, where $F$ has been derived before, and $FV(G) \subseteq FV(F)$, then deriving $G$ does not help derive a ground contradiction,​ because the contradiction would also be derived using $F$.  If we derive such formula, we can immediately delete it so that it does not slow us down.  ​
  
 In particular, a ground true formula can be deleted. In particular, a ground true formula can be deleted.
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 Therefore, for clauses, projection (with some elimination of redundant conclusions) is exactly the resolution proof rule. Therefore, for clauses, projection (with some elimination of redundant conclusions) is exactly the resolution proof rule.
-