hw2-soln - CISC 404/604 Homework 2 Solutions 1a To show...

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CISC 404/604 Homework 2 Solutions 1a. To show that ± res (( A B ) ( B C )) (( ¬ C A ) B ): First, negate what we’re trying to prove (i.e., to show ± res α consider ¬ α ): ¬ ((( A B ) ( B C )) (( ¬ C A ) B )) Second, convert from statement-form into a set of clauses: { [ ¬ ((( A B ) ( B C )) (( ¬ C A ) B ))] } (starting with { [ ¬ α ] } ) { [( A B ) ( B C )] , [ ¬ (( ¬ C A ) B )] } (applying rule 3b) { [ A B ] , [ B C ] , [ ¬ (( ¬ C A ) B )] } (applying rule 3a) { [ ¬ A,B ] , [ B C ] , [ ¬ (( ¬ C A ) B )] } (applying rule 2b) { [ ¬ A,B ] , [ ¬ B,C ] , [ ¬ (( ¬ C A ) B )] } (applying rule 2b) { [ ¬ A,B ] , [ ¬ B,C ] , [ ¬ C A ] , [ ¬ B ] } (applying rule 3b) { [ ¬ A,B ] , [ ¬ B,C ] , [ ¬ C ] , [ A ] , [ ¬ B ] } (applying rule 3a) Let S = { [ ¬ A,B ] , [ ¬ B,C ] , [ ¬ C ] , [ A ] , [ ¬ B ] } . Third, attempt to derive the empty clause via resolution on S : 1. [ ¬ A,B ] - member of S 2. [ A ] - member of S 3. [ B ] - resolving 1 and 2 4. [ ¬ B ] - member of S 5. [ ] - resolving 3 and 4. ± res (( A B ) ( B C )) (( ¬ C A ) B ). 1b. To show that { (( A B ) ( C D )) } ± res (( A ( C D )) ( B ( C D ))): First, convert { α i ,...,α n } into a set of clauses (where { α i ,...,α n } = { (( A B ) ( C D )) } ): { [( A B ) ( C D )] } (starting with { [ α 1 ] } ) { [( A B ) , ( C D )] } (applying rule 2a) { [( A B ) , ¬ C,D ] } (applying rule 2b) { [ A, ¬ C,D ] , [ B, ¬ C,D ] } (applying rule 3a) Let S 1 = { [ A, ¬ C,D ] , [ B, ¬ C,D ] } . Second, consider ¬ β and convert it into a set of clauses (where β = (( A ( C D )) ( B ( C D )))): { [ ¬ (( A ( C D )) ( B ( C D )))] } (starting with { [ ¬ β ] } ) { [ ¬ ( A

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hw2-soln - CISC 404/604 Homework 2 Solutions 1a To show...

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