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sav08:deriving_propositional_resolution [2008/03/12 11:04]
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 ====