Predicate Logic - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics

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CS

Discrete Mathematical Structures

Predicate Logic - everybody loves somebody

Alicia eats pizza at least once a week.

Garrett eats pizza at least once a week.

Allison eats pizza at least once a week.

Gregg eats pizza at least once a week.

Ryan eats pizza at least once a week.

Meera eats pizza at least once a week.

Ariel eats pizza at least once a week.

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Predicates

Suppose Q(x,y) = “x > y”

Proposition, YES or NO?

Q(x,y)

Q(3,4)

Q(x,9)

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NO

YES

NO

Predicate, YES or NO?

Q(x,y)

Q(3,4)

Q(x,9)

YES

NO

YES

Predicates - the universal quantifier

Another way of changing a predicate into a proposition.

Suppose P(x) is a predicate on some universe of discourse. Ex. B(x) = “x is carrying a backpack,” x is set of cs173 students.

The universal quantifier of P(x) is the proposition : “P(x) is true for all x in the universe of discourse.”

We write it ∀x P(x), and say “for all x, P(x)”

∀x P(x) is TRUE if P(x) is true for every single x. ∀x P(x) is FALSE if there is an x for which P(x) is false.

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∀x B(x)?

Predicates - the existential quantifier

Another way of changing a predicate into a proposition.

Suppose P(x) is a predicate on some universe of discourse. Ex. C(x) = “x has a candy bar,” x is set of cs173 students.

The existential quantifier of P(x) is the proposition : “P(x) is true for some x in the universe of discourse.”

We write it ∃x P(x), and say “for some x, P(x)”

∃x P(x) is TRUE if there is an x for which P(x) is true. ∃x P(x) is FALSE if P(x) is false for every single x.

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∃x C(x)?

Predicates - the existential quantifier

B(x) = “x is wearing sneakers.” L(x) = “x is at least 21 years old.” Y(x)= “x is less than 24 years old.”

Are either of these propositions true?

a) ∃x B(x) b) ∃x (Y(x) ∧ L(x))

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A: only a is true

B: only b is true

C: both are true

D: neither is true

Universe of discourse is people in this room.

Predicates - more examples

B(x) = “x is a hummingbird.” L(x) = “x is a large bird.” H(x) = “x lives on honey.” R(x) = “x is richly colored.”

All hummingbirds are richly colored.

No large birds live on honey.

Birds that do not live on honey are dully colored.

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Universe of discourse is all creatures.

∀x (B(x) → R(x))

¬∃x (L(x) ∧ H(x))

∀x (¬H(x) → ¬R(x))

Predicates - quantifier negation

Not all large birds live on honey.

∀x P(x) means “P(x) is true for every x.” What about ¬∀x P(x)? Not [“P(x) is true for every x.”] “There is an x for which P(x) is not true.” ∃x ¬P(x)

So, ¬∀x P(x) is the same as ∃x ¬P(x).

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¬∀x (L(x) → H(x))

∃x ¬(L(x) → H(x))

Predicates - quantifier negation

So, ¬∀x P(x) is the same as ∃x ¬P(x). So, ¬∃x P(x) is the same as ∀x ¬P(x).

General rule: to negate a quantifier, move negation to the right, changing quantifiers as you go.

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Predicates - quantifier negation

No large birds live on honey.

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¬∃x (L(x) ∧ H(x)) ≡ ∀x ¬(L(x) ∧ H(x)) Negation rule ≡ ∀x (¬L(x) ∨ ¬H(x)) DeMorgan’s

≡ ∀x (L(x) → ¬H(x)) Subst for →

What’s wrong with this proof?