s09_t08 - A C does not logically follow from A ⊆ B,B C A...

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UNIVERSITY OF WATERLOO School of Computer Science CS245 Logic and Computation Spring 2009 TUTORIAL 8 1. Formalize the following sentences in set theory. Do not use types, quantifiers, or set comprehension. Use only the following sets and relations in your formulas: People – the set of all people Courses – the set of all courses Cars – the set of all cars BlueCars – the set of all blue cars, where BlueCars Cars owns : People Cars – the relationship between people and the cars they own drives : People Cars – the relationship between people and the cars they drive enrolled : People Courses – the relationship between people and the courses they are enrolled in Sentences to formalize: (a) Not everyone who owns a car is enrolled in a course. dom( Owns ) 6⊆ dom( Enrolled ) Alternative: dom( Owns ) ( People - dom( Enrolled )) 6 = (b) All the cars owned by people who are enrolled in courses are blue cars. ran(dom( Enrolled ) / Owns ) BlueCars (c) All the blue cars that are driven are owned by their drivers. ( Drives . BlueCars ) Owns 1
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2. Provide a counterexample to show that
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Unformatted text preview: A * C does not logically follow from A ⊆ B,B * C . A counter-example: A = { a } B = { a,b } C = { a } The premises are true, but the conclusion is false. A smaller counterexample is: A = ∅ ,B = { a } ,C = ∅ 3. Given B ⊆ A , show A ∩ B = ∅ using natural deduction. 1. B ⊆ A premise 2. ∀ x • x ∈ B ⇒ x ∈ A 1 , subset 3. ∀ x • x ∈ B ⇒ ( x ∈ U ∧ x 6∈ A ) 2 , complement 4. z g 5. z g ∈ ( A ∩ B ) assumption 6. z g ∈ A ∧ z g ∈ B 5 , set intersection 7. z g ∈ A 6 , ∧ E 8. z g ∈ B 6 , ∧ E 9. z g ∈ B ⇒ ( z g ∈ U ∧ z g 6∈ A ) 3 , ∀ E 10. z g ∈ U ∧ z g 6∈ A 8 , 9 , ⇒ E 11. z g 6∈ A 10 , ∧ E 12. false 7 , 11 , ¬ E 13. ¬ ( z g ∈ ( A ∩ B )) 5-12 , ¬ I 14. ∀ z • ¬ ( z ∈ ( A ∩ B )) 4-13 , ∀ I 15. A ∩ B = ∅ 14 , empty set 2...
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This note was uploaded on 01/19/2010 for the course CS 246 taught by Professor Wormer during the Spring '08 term at Waterloo.

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s09_t08 - A C does not logically follow from A ⊆ B,B C A...

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