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l14soln

# 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 quantifiers. We’ve added these brackets in the solution sets. However, if you didn’t add the brackets, then you could end up with a different 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 ) = F We then have I [[( x A ( x )) IMP ( y A ( y ))]] = ( A ( Alice ) OR A ( Bob )) IMP ( A ( Alice ) AND A ( Bob )) = (T OR F) IMP (T AND F) = T IMP F = F 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
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
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-IMPLIES 5 Exercise 8.2 (p. 184) 1. In course notes. 2. | = ST x • ∀ y ( A ( x,y ) ( B ( x ) A ( x,y ))) 5

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1. ¬∀ x • ∀ y ( A ( x,y ) ( B ( x ) A ( x,y ))) 2. x • ¬∀ y ( A ( x,y ) ( B ( x ) A ( x,y ))) NOT-FORALL 1 3. ¬∀ y ( A ( s,y ) ( B ( s ) A ( s,y ))) EXISTS 2 4. y • ¬ ( A ( s,y ) ( B ( s ) A ( s,y ))) NOT-FORALL 3 5. ¬ ( A ( s,t ) ( B ( s ) A ( s,t ))) EXISTS 4 6. A ( s,t ) 7. ¬ ( B ( s ) A ( s,t )) NOT-IMPLIES 5 8. B ( s ) 9. ¬ A ( s,t ) CLOSED 6,9 NOT-IMPLIES 7 3. | = ST x • ∀ y (( A ( x ) ( B ( x,y ) C ( x,y ))) (( A ( x ) B ( x,y )) ( A ( x ) C ( x,y )))) 6
1. ¬∀ x • ∀ y (( A ( x ) ( B ( x,y ) C ( x,y ))) (( A ( x ) B ( x,y )) ( A ( x ) C ( x,y )))) 2. x • ¬∀ y (( A ( x ) ( B ( x,y ) C ( x,y ))) (( A ( x ) B ( x,y )) ( A ( x ) C ( x,y )))) NOT-FORALL 1 3.

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

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