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Q48Sol - Section 1.4 Question 48 Show that xP(x xQ(x and...

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Section 1.4 Question 48 : Show that xP ( x ) ∨ ∀ xQ ( x ) and x y [ P ( x ) Q ( y )], where all quantifiers have the same nonempty domain, are logically equivalent. Solution: Since the scope of the second in xP ( x ) ∨ ∀ xQ ( x ) involves only Q ( x ), we can rewrite xP ( x ) ∨ ∀ xQ ( x ) as P ( x ) ∨ ∀ yQ ( y ) First, we assume that xP ( x ) ∨ ∀ yQ ( y ) is true. (We want to show that x y [ P ( x ) Q ( y )] is also true.) Now either xP ( x ) is true or yQ ( y ) is true. If xP ( x ) is true, then P ( x ) is true for every x in the domain which implies that P ( x ) Q ( y ) is true for every x and y . So x y [ P ( x ) Q ( y )] is true. If yQ ( y ) is true, then Q ( y ) is true for every y in the domain which implies that P ( x ) Q ( y ) is true for every x and y . So x y [ P ( x ) Q ( y )] is true.
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