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**ICS 253: Discrete Structures I Predicates and Quantifiers**

Spring Semester 2014 (2013-2) Predicates and Quantifiers Dr. Nasir Al-Darwish Computer Science Department King Fahd University of Petroleum and Minerals

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**Propositional Predicate**

Definition: A propositional predicate P(x) is a statement that has a variable x. Examples of P(x) P(x) = “The Course x is difficult” P(x) = “x+2 < 5” Note: a propositional predicate is not a proposition because it depends on the value of x. Example: If P(x) = “x > 3”, then P(4) is true but not P(1). A propositional predicate is also called a propositional function

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**Propositional Predicate – cont.**

It also possible to have more than one variable in one predicate, e.g., Q(x,y) =“x > y-2” P(x) is a function (a mapping) that takes a value for x and produce either true or false. Example: P(x) = “x2 > 2” , P: Some Domain {T, F} Domain of x is called the domain (universe) of discourse

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Quantification A predicate (propositional function) could be made a proposition by either assigning values to the variables or by quantification. Predicate Calculus: Is the area of logic concerned with predicates and quantifiers.

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Quantifiers 1. Universal quantifier: P(x) is true for all (every) x in the domain. We write x P(x) 2. Existential quantifier: there exists at least one x in the domain such that P(x) is true. We write x P(x) 3. Others: there exists a unique x such that P(x) is true. We write !x P(x)

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**Universal Quantification**

Uses the universal quantifier (for all) x P(x) corresponds to “p(x) is true for all values of x (in some domain)” Read it as “for all x p(x)” or “for every x p(x)” Other expressions include “for each” , “all of”, “for arbitrary” , and “for any” (avoid this!) A statement x P(x) is false if and only if p(x) is not always true (i.e., P(x) is false for at least one value of x) An element for which p(x) is false is called a counterexample of x P(x); one counterexample is all we need to establish that x P(x) is false

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**Universal Quantification - Examples**

Example 1: Let P(x) be the statement “x + 1 > x” . What is the truth value of ∀x P(x), where the domain for x consists of all real numbers? Solution: Because P(x) is true for all real numbers x, the universal quantification ∀x P(x) is true. Example 2: Suppose that P(x) is “x2 > 0” . What is the truth value of ∀x P(x), where the domain consists of all integers. Solution: We show ∀x P(x) is false by a counterexample. We see that x = 0 is a counterexample because for x = 0, x2 = 0, thus there is some integer x for which P(x) is false.

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**Existential Quantification**

Uses the existential quantifier (there exists) x P(x) corresponds to “There exists an element x (in some domain) such that p(x) is true” In English, “there is”, “for at least one”, or “for some” Read as “There is an x such that p(x)”, “There is at least one x such that p(x)”, or “For some x, p(x)” A statement x P(x) is false if and only if “for all x, P(x) is false”

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**Existential Quantification - Examples**

Example 1: Let P(x) denote the statement “x > 3”. What is the truth value of ∃x P(x), where the domain for x consists of all real numbers? Solution: Because “x > 3” is true for some values of x , for example, x = 4, the existential quantification ∃x P(x) is true. Example 2: Let Q(x) denote the statement “x = x + 1” . What is the truth value of ∃x Q(x), where the domain consists of all real numbers? Solution: Because Q(x) is false for every real number x, the existential quantification ∃x Q(x) is false.

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**Predicates and Negations**

When true When false Negation x P(x) For all x, P(x) is true There is at least one x s.t. P(x) is false x P(x) x P(x) There is at least one x s.t. P(x) is true For all x, P(x) is false x P(x)

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**Domain (or Universe) of Discourse**

Cannot tell if a quantified predicate P(x) is true (or false) if the domain of x is not known. The meaning of the quantified P(x) changes when we change the domain. The domain must always be specified when universal or existential quantifiers are used; otherwise, the statement is ambiguous.

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**Quantification Examples**

P(x) = “x+1 = 2” Domain is R (set of real numbers) Proposition Truth Value x P(x) x P(x) x P(x) x P(x) !x P(x) !x P(x) F F T T T F

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**Quantification Examples**

P(x) = “x2 > 0” Domain Proposition Truth Value R x P(x) Z x P(x) Z - {0} x P(x) Z !x P(x) N={1,2, ..} x P(x) F F T T F

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**Quantification Examples**

Proposition Truth Value xR (x2 x) !xR (x2 < x) x(0,1) (x2 < x) x{0,1} (x2 = x) x P(x) F F T T T

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**Logically Equivalence**

Definition: Two statements involving predicates & quantifiers are logically equivalent if and only if they have the same truth values independent of the domains and the predicates. Examples: ( x P(x) ) x P(x) ( x P(x) ) x P(x)

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**x P(x) x P(x) Theorem P(x1) P(x2) ... P(xn)**

If the domain of discourse is finite, say Domain = {x1, x2, …, xn}, then x P(x) x P(x) P(x1) P(x2) ... P(xn) P(x1) P(x2) ... P(xn)

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**((x P(x) ) ((!x Q(x) ) (x P(x) )))**

Precedence , , , , , Example: ((x P(x) ) ((!x Q(x) ) (x P(x) )))

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**Correct Equivalences x ( P(x) Q(x) ) x P(x) x Q(x)**

This says that we can distribute a universal quantifier over a conjunction x ( P(x) Q(x) ) x P(x) x Q(x) This says that we can distribute an existential quantifier over a disjunction The preceding equivalences can be easily proven if we assume a finite domain for x = {x1, x2, …, xn} Note: we cannot distribute a universal quantifier over a disjunction, nor can we distribute an existential quantifier over a conjunction.

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**Wrong Equivalences x 1 2 P(x) T F Q(x)**

x ( P(x) Q(x) ) x P(x) x Q(x) Read as “there exists an x for which both P(x) and Q(x) are true is equivalent to there exists an x for which P(x) is true and there exists an x for which Q(x) is true”. One can construct an example that makes the above equivalence false. Consider the following x ( P(x) Q(x) ) = F but x P(x) x Q(x) = T, since P(1) is true and Q(2) is true x 1 2 P(x) T F Q(x)

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**Wrong Equivalences x 1 2 P(x) T F Q(x)**

x ( P(x) Q(x) ) x P(x) x Q(x) One can construct an example that makes the above equivalence false. Consider the following x ( P(x) Q(x) ) = T but x P(x) x Q(x) = F F = F x 1 2 P(x) T F Q(x)

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**Wrong Equivalences ( x P(x) ) Q(x) x ( P(x) Q(x) )**

Notice that the LHS = ( x P(x) ) Q(x) is not fully quantified. So it cannot be equivalent to RHS.

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**Quantifiers with restricted domains**

What do the following statements mean for the domain of real numbers? Be careful about → and ˄ in these statements

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© by Kenneth H. Rosen, Discrete Mathematics & its Applications, Sixth Edition, Mc Graw-Hill, 2007 Chapter 1: (Part 2): The Foundations: Logic and Proofs.

© by Kenneth H. Rosen, Discrete Mathematics & its Applications, Sixth Edition, Mc Graw-Hill, 2007 Chapter 1: (Part 2): The Foundations: Logic and Proofs.

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