l14soln - CS 245 Winter 2009 Solution Set for Lecture 14...

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CS 245 Winter 2009 Solution Set for Lecture 14 Kelly Ch. 8 In many of these examples, the text has forgotten to put in some brackets that reduce the scope of the quantiFers. We’ve added these brackets in the solution sets. However, if you didn’t add the brackets, then you could end up with a di±erent answer. Exercise 8.1 (p. 180) 1. ( x A ( x )) ( y A ( y )) This is not a tautology. To see this consider the following environment and interpre- tation: D = { Alice, Bob } A ( Alice ) = T A ( Bob ) = ² We then have I [[( x A ( x )) IMP ( y A ( y ))]] = ( A ( Alice ) OR A ( Bob )) IMP ( A ( Alice ) AND A ( Bob )) = (T OR ²) IMP (T AND ²) = T IMP ² = ² 2. x A ( x ) ⇒ ∃ y A ( y ) This is a tautology. A proof is as follows: 1
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1. ¬ ( x A ( x ) ⇒ ∃ y A ( y )) 2. x • ¬ ( A ( x ) ⇒ ∃ y A ( y )) NOT-EXISTS 1 3. ¬ ( A ( t ) ⇒ ∃ y A ( y )) FORALL 2 4. A ( t ) 5. ¬ ( y A ( y )) NOT-IMPLIES 3 6. y • ¬ A ( y ) NOT-EXISTS 5 7. ¬ A ( t ) CLOSED 4,7 FORALL 6 3. ( x A ( x ) B ( x )) ⇒ ∃ x B ( x ) This is a tautology. A proof is as follows: 2
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1. ¬ (( x A ( x ) B ( x )) ⇒ ∃ x B ( x )) 2. x A ( x ) B ( x ) 3. ¬∃ x B ( x ) NOT-IMPLIES 1 4. x • ¬ B ( x ) NOT-EXISTS 3 5. A ( t ) B ( t ) FORALL 2 6. A ( t ) 7. B ( t ) AND 5 8. ¬ B ( t ) CLOSED 7,8 FORALL 4 4. x (( A ( x ) B ( x )) A ( x )) This is a tautology. A proof is as follows: 1. ¬ ( x (( A ( x ) B ( x )) A ( x ))) 2. x • ¬ (( A ( x ) B ( x )) A ( x )) NOT-FORALL 1 3. ¬ (( A ( t ) B ( t )) A ( t )) EXISTS 2 4. A ( t ) B ( t ) 5. ¬ A ( t ) NOT-IMPLIES 3 6. A ( t ) 7. B ( t ) CLOSED 5,6 AND 4 3
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5. x (( A ( x ) B ( x )) A ( x )) This is not a tautology. To see this consider the following environment and interpre- tation: D = { Alice, Bob } A ( Alice ) = T B ( Alice ) = F A ( Bob ) = F B ( Bob ) = T We then have I [[ x A ( x ) B ( x ) A ( x )]] = ( A ( Alice ) OR B ( Alice ) IMP A ( Alice )) AND ( A ( Bob ) OR B ( Bob ) IMP A ( Bob )) = (T OR F IMP T) AND (F OR T IMP F) = (T IMP T) AND (T IMP F) = T AND F = F 6. x ( A ( x ) ( A ( x ) B ( x ))) This is a tautology. A proof is as follows: 1. ¬ ( x A ( x ) ( A ( x ) B ( x ))) 2. x • ¬ ( A ( x ) ( A ( x ) B ( x ))) NOT-FORALL 1 3. ¬ ( A ( t ) ( A ( t ) B ( t ))) EXISTS 2 4. A ( t ) 5. ¬ ( A ( t ) B ( t )) NOT-IMPLIES 3 6. ¬ A ( t ) 7. ¬ B ( t ) CLOSED 4,6 NOT-OR 5 7. x ( A ( x ) ( ¬ A ( x ) B ( x ))) This is a tautology. A proof is as follows: 4
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¬ ( x A ( x ) ( ¬ A ( x ) B ( x ))) 2. x • ¬ ( A ( x ) ( ¬ A ( x ) B ( x ))) NOT-FORALL 1 3. ¬ ( A ( t ) ( ¬ A ( t ) B ( t ))) EXISTS 2 4. A ( t ) 5. ¬ ( ¬ A ( t ) B ( t )) NOT-IMPLIES 3 6. ¬ A ( t ) 7. ¬ B ( t ) CLOSED 4,6 NOT-IMPLIES 5
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This note was uploaded on 10/23/2009 for the course CS 245 taught by Professor A during the Spring '08 term at Waterloo.

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l14soln - CS 245 Winter 2009 Solution Set for Lecture 14...

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