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11slides(3)

# 11slides(3) - CHAPTER 11 Automated Proof Systems(3 RS...

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Unformatted text preview: CHAPTER 11: Automated Proof Systems (3) RS: Counter Models Generated by Decomposition Trees RS: Proof of COMPLETENESS THEOREM 1 Counter-model generated by the decompo- sition tree. Example: Given a formula A : (( a ⇒ b ) ∩ ¬ c ) ∪ ( a ⇒ c )) and its decomposition tree T A . ((( a ⇒ b ) ∩¬ c ) ∪ ( a ⇒ c )) | ( ∪ ) (( a ⇒ b ) ∩¬ c ) , ( a ⇒ c ) ^ ( ∩ ) ( a ⇒ b ) , ( a ⇒ c ) | ( ⇒ ) ¬ a,b, ( a ⇒ c ) | ( ⇒ ) ¬ a,b, ¬ a,c ¬ c, ( a ⇒ c ) | ( ⇒ ) ¬ c, ¬ a,c 2 Consider a non-axiom leaf: ¬ a,b, ¬ a,c Let v be any variable assignment v : V AR-→ { T,F } such that it makes this non-axiom leaf FALSE, i.e. we put v ( a ) = T,v ( b ) = F,v ( c ) = F. Obviously, we have that v * ( ¬ a,b, ¬ a,c ) = ¬ T ∪ F ∪ ¬ T ∪ F = F. Moreover, all our rules of inference are sound (to be proven formally in the next section). Rules soundness means that if one of pre- misses of a rule is FALSE, so is the con- clusion. 3 Hence, the soundness of the rules proves (by induction on the degree of sequences Γ ∈ T A ) that v , as defined above falsifies all sequences on the branch of T A that ends with the non-axiom leaf ¬ a,b, ¬ a,c . In particular, the formula A is on this branch, hence v * ((( a ⇒ b ) ∩ ¬ c ) ∪ ( a ⇒ c )) = F and v is a counter-model for A ....
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11slides(3) - CHAPTER 11 Automated Proof Systems(3 RS...

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