Q48Sol - Section 1.4 Question 48: Show that ∀xP (x) ∨...

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Unformatted text preview: 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. Secondly, we assume that ∀xP (x) ∨ ∀yQ(y ) is false. (We want to show that ∀x∀y [P (x) ∨ Q(y )] is also false.) Now both ∀xP (x) and ∀yQ(y ) are false. Hence there is an x such that P (x) is false, and there is an y such that Q(y ) is false. For this particular x and y , P (x) ∨ Q(y ) is false. Therefore ∀x∀y [P (x) ∨ Q(y )] is false. We have shown that ∀xP (x) ∨ ∀xQ(x) and ∀x∀y [P (x) ∨ Q(y )] have the same truth values, so they are logically equivalent. 1 ...
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