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# quiz2_key - 5 over ∨ ⇔ ¬ 2200 x p x ∨ 5 x q...

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COT 3100 Summer 2001 Quiz #2 (25pts) 05/24/2001 Name: SSN: 1. (5pts) Find a formula involving only connectives ¬ and that is equivalent to p q p q ¬ ( ¬ p ∨¬ q )……………………….DeMorgan’s Law ¬ ( p →¬ q )…………………………by equivalence p q ⇔¬ p q 2. (10pts) Use laws of logic to determine whether or not ( p q ) ( p r ) is equivalent to p ( q r ). ( p q ) ( p r ) ( ¬ p q ) ( ¬ p r )………… by equivalence p q ⇔¬ p q ⇔¬ p ( q r )……………….by distributive law p ( q r )…………………by equivalence p q ⇔¬ p q 3. (10pts) Distribute existential quantifier over the implication: 5 x [ p ( x ) q ( x )] (i.e. find equivalent form of this proposition where quantifiers apply to p ( x ) and to q ( x )). 5 x [ p ( x ) q ( x )] 5 x [ ¬ p ( x ) q ( x )]…………… by equivalence p q ⇔¬ p q [ 5 x ¬ p ( x )] [ 5 x q ( x )]………..by distributive property of
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Unformatted text preview: 5 over ∨ ⇔ ¬ [ 2200 x p ( x )] ∨ [ 5 x q ( x )]……….by generalized DeMorgan’s law ⇔ [ 2200 x p ( x )] → [ 5 x q ( x )]………. by equivalence p → q ⇔¬ p ∨ q Example : Let p ( x ) : student x comes to the lecture today q ( x ) : student x comes to the recitation today Then 5 x [ p ( x ) → q ( x )] means, that there exists a student that if he comes to the lecture today, then he goes to the recitation today. The equivalent proposition [ 2200 x p ( x )] → [ 5 x q ( x )] reads: if all students come to the lecture today, then there will be at least one student at the recitation today (because we do not know whether or not any other student x follows the rule p ( x ) → q ( x ))....
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