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# Equivalencesinpredicatelogic

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Unformatted text preview: true.   2.  But if P(x) denotes  “x &gt; 0,” then ∃!x P(x) is false.    The uniqueness quantiﬁer is not really needed as the restriction  that there is a unique x such that P(x) can be expressed as:                                   ∃x (P(x) ∧∀y (P(y) → y =x))  Thinking about Quan*ﬁers    When the  domain of discourse is ﬁnite, we can think of  quantiﬁcation as looping through the elements of the domain.    To evaluate ∀x P(x) loop through all x in the domain.     If at every step P(x) is true, then ∀x P(x) is true.     If at a step P(x) is false, then ∀x P(x) is false and the loop  terminates.     To evaluate ∃x P(x) loop through all x in the domain.     If  at some step, P(x) is true, then ∃x P(x) is true and the loop  terminates.     If the loop ends without ﬁnding an x for which P(x) is true, then ∃x  P(x) is false.    Even if the domains are inﬁnite, we can still think of the  quantiﬁers this fashion, but the loops will not terminate in  some cases.  Proper*es of Quan*ﬁers    The truth value of ∃x P(x)  and ∀ x P(x) depend on both the propositional function P(x) and on the domain U.   Examples: If U is the  positive integers and P(x) is the statement            “x &lt; 2”, then ∃x P(x)   is true, but ∀ x P(x) is false. 2.  If Uis the negative integers and P(x) is the statement            “x &lt; 2”, then both ∃x P(x)  and  ∀ x P(x) are true. 3.  If U consists of 3, 4, and 5,  and P(x) is the statement            “x &gt; 2”, then  both ∃x P(x)   and ∀ x P(x) are true....
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## This document was uploaded on 03/06/2014 for the course MATH 320 at CSU Northridge.

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