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Quantifiers and Negation Predicates. Predicate Will define function. The domain of a function is the set from which possible values may be chosen for.

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Presentation on theme: "Quantifiers and Negation Predicates. Predicate Will define function. The domain of a function is the set from which possible values may be chosen for."— Presentation transcript:

1 Quantifiers and Negation Predicates

2 Predicate Will define function. The domain of a function is the set from which possible values may be chosen for the argument of the function. A predicate is a function whose possible values are True and False.

3 Universal Quantifier  The universal quantifier. Upside down A. “For All”

4 Universal Quantifier Let P(x) be a predicate with domain D. A universal statement is a statement in the form “  x, P(x)”. It is true iff P(x) is true for all x in D. It is false iff P(x) is false for at least one x in D. A value of x which shows that P(x) is false is called a counterexample to the universal statement.

5 Universal Quantifier Examples Let D = {1, 2, 3, 4, 5}.  x, x² >= x Let D=Reals.  x  Reals, x² >= x Let D= Natural numbers. Let P(x)=x is prime, E(x)=x is even.  x E(x)  x P(x)  x (E(x) V E(x+1))

6 Existential Quantifier  Existential quantifier. Backwards E. There Exists

7 Existential Quantifier Let P(x) be a predicate with domain D. An existential statement is a statement in the form “  x, P(x)”. It is true iff P(x) is true for at least one x in D. It is false iff P(x) is false for all x in D. Examples: D=Integers,  m, m² = m D = {5, 6, 7, 8, 9},  m, m² = m

8 Universal Conditional Statement Universal conditional statement  x, (P(x)  Q(x)) Example: D=Reals.  x, (x > 2  x 2 > 4) A universal conditional statement is vacuously true or true by default iff P(x) is false for every x in D. i.e. ~  x  D, P(x) iff  x  D (P(x)  Q(x)) is vacuously true.

9 Universal Conditional Statement Examples: D=Natural numbers. E(x)=x is even. P(x)=x is prime  x (E(x)  ~E(x+1))  x ((E(x) Λ x > 2)  ~P(x))  x (P(x)  ~E(x))

10 Negation of Quantified Statements ~ (  x  D, P(x)) ≡  x  D, ~P(x) ~(  x  D, P(x)) ≡  x  D, ~P(x) ~  x  D, P(x)  Q(x) ≡  x  D, ~(P(x)  Q(x)) ≡  x  D, P(x)  ~Q(x)

11 Multiply Quantified Statements Let the D=Integers. Let P(x, y) = y<x. There exists a number x, there exists a number y, such that y<x  x,  y, P(x, y) ≡  y,  x, P(x, y) True (both True) For all numbers x, for all numbers y, y < x  x,  y, P(x, y) ≡  y,  x, P(x, y) True (both False)

12 Multiply Quantified Statements For all numbers x, there exists a number y such that x < y  x,  y, P(x, y) True There exists a number y such that for all numbers x, x < y  y,  x, P(x, y) False  x,  y, P(x, y) ≡  y,  x, P(x, y) False

13 Negation of Multiply Quantified Statements ~  x,  y, P(x, y) ≡  x,  y, ~P(x, y) ~  x,  y, P(x, y) ≡  x,  y,~P(x, y) ~  x,  y, P(x, y) ≡  x,  y, ~P(x, y) ~  x,  y, P(x, y) ≡  x,  y, ~P(x, y)

14 Example a) Symbolize: The sum of any two even integers is even. b) Negate it.

15 Example Solution Let E(x)=“x is even”. Let E(x, y)=“x+y is even”. Let the domain be the Integers. a)  x,  y, E(x)^E(y)  E(x, y) b)  x,  y, E(x) ^ E(y) ^ ~E(x, y)

16 Which of the following is true?  x  D, (P(x)  Q(x)) ≡ (  x  D, P(x))  (  x  D, Q(x))  x  D, (P(x)  Q(x)) ≡ (  x  D, P(x))  (  x  D, Q(x))  x  D, (P(x)  Q(x)) ≡ (  x  D, P(x))  (  x  D, Q(x))  x  D, (P(x)  Q(x)) ≡ (  x  D, P(x))  (  x  D, Q(x))

17 , , Λ, ν Let P(x) be a predicate. Let D={x 1, x 2, …, x n } be the domain of x.  x  D, P(x)≡P(x 1 ) Λ P(x 2 ) Λ…Λ P(x n )  x  D, P(x) ≡P(x 1 ) ν P(x 2 ) ν…ν P(x n )


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