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Discrete Mathematics CS 2610 August 21, 2008

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Agenda Nested quantifiers Rules of inference Proofs

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Review
A **predicate **P, or propositional function, is a

function that maps objects in the universe of discourse to propositions Predicates can be quantified using the universal quantifier (“for all”) ∀ or the existential quantifier (“there exists”) ∃ Quantified predicates can be negated as follows ¬∀x P(x) ≡ ∃x ¬P(x) ¬∃x P(x) ≡ ∀x ¬P(x)

Quantified variables are called “bound” Variables that are not quantified are called “free”

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Predicate Logic and Propositions An expression with zero free variables is an actual proposition

Ex. Q(x) : x > 0, R(y): y < 10 ∃ x Q(x) ∧ ∃y R(y)

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Nested Quantifiers When dealing with polyadic predicates, each argument may be quantified with its own quantifier. Each nested quantifier occurs in the scope of another quantifier.

Examples: (L=likes, UoD(x)=kids, UoD(y)=cars) ∀x∀y L(x,y) reads ∀x(∀y L(x,y)) ∀x∃y L(x,y) reads ∀x(∃y L(x,y)) ∃x∀y L(x,y) reads ∃x(∀y L(x,y)) ∃x∃y L(x,y) reads ∃x(∃y L(x,y))

Another example ∀x (P(x) ∨ ∃y R(x,y))

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If L(x,y) means x likes y, how do you read the following quantified predicates?

∃y L(Alice,y) ∃y∀x L(x,y) ∀x∃y L(x,y) ∀x L(x, Prius)

Examples

Alice likes some car There is a car that is liked by everyone

Everyone likes some car Everyone likes the Prius

Order matters!!!

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Negation of Nested Quantifiers To negate a quantifier, move negation to the right, changing quantifiers as you go.

Example: ¬∀x∃y∀z P(x,y,z) ≡

∃x ∀y ∃z ¬P(x,y,z).

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Proofs and inference Assume that the following statements are true:

I have a total score over 96. If I have a total score over 96, then I get an A in the class.

What can we claim? I get an A in the class. How do we know the claim is true? Logical Deduction.

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Proofs • A theorem is a statement that can be proved to be

true.

• A proof is a sequence of statements that form an argument.

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Proofs: Inference Rules An Inference Rule:

“∴” means “therefore”

premise 1 premise 2 … ∴ conclusion

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Proofs: Modus Ponens

I have a total score over 96.

If I have a total score over 96, then I get an A for the class.

∴ I get an A for this class

p

p → q

∴ q

Tautology:

(p ∧ (p → q)) → q

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Proofs: Modus Tollens

•If the power supply fails then the lights go out. •The lights are on. ∴ The power supply has not failed.

Tautology:

(¬q ∧ (p → q)) → ¬p

¬q

p → q

∴ ¬p

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Proofs: Addition

•I am a student. ∴ I am a student or I am a visitor.

p

∴ p ∨ q

Tautology:

p → (p ∨ q)

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Proofs: Simplification

•I am a student and I am a soccer player. ∴ I am a student.

p ∧ q

∴ p

Tautology:

(p ∧ q) → p

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Proofs: Conjunction

•I am a student. •I am a soccer player. ∴ I am a student and I am a soccer player.

p

q

∴ p ∧ q

Tautology:

((p) ∧ (q)) → p ∧ q

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Proofs: Disjunctive Syllogism I am a student or I am a soccer player. I am a not soccer player. ∴ I am a student.

p ∨ q

¬q

∴ p

Tautology:

((p ∨ q) ∧ ¬q) → p

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Proofs: Hypothetical Syllogism

If I get a total score over 96, I will get an A in the course.

If I get an A in the course, I will have a 4.0 semester average.

∴ If I get a total score over 96 then
∴ I will have a 4.0 semester average**.**

p → q

q → r

∴ p → r

Tautology:

((p → q) ∧ (q → r)) → (p → r)

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Proofs: Resolution

I am taking CS1301 or I am taking CS2610. I am not taking CS1301 or I am taking CS 1302. ∴ I am taking CS2610 or I am taking CS 1302.

p ∨ q

¬ p ∨ r

∴ q ∨ r

Tautology:

((p ∨ q ) ∧ (¬ p ∨ r)) → (q ∨ r)

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Proofs: Proof by Cases I have taken CS2610 or I have taken CS1301. If I have taken CS2610 then I can register for CS2720 If I have taken CS1301 then I can register for CS2720 ∴ I can register for CS2720 p ∨ q

p → r

q → r

∴ r

Tautology:

((p ∨ q ) ∧ (p → r) ∧ (q → r)) → r

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Fallacy of Affirming the Conclusion

•If you have the flu then you’ll have a sore throat. You have a sore throat. ∴ You must have the flu.

Fallacy:

(q ∧ (p → q)) → p

q

p → q

∴ p

Abductive, rather than deductive reasoning! Docsity.com

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Fallacy of Denying the Hypothesis

•If you have the flu then you’ll have a sore throat. •You do not have the flu. ∴ You do not have a sore throat.

Fallacy:

(¬p ∧ (p → q)) → ¬q

¬p

p → q

∴ ¬q

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Inference Rules for Quantified Statements

∀x P(x) ∴ P(c)

∃x P(x) ∴ P(c)

P(c)___ ∴ ∀x P(x)

**Universal Instantiation
**

(for an arbitrary object c from UoD)

**Universal Generalization
**

(for any arbitrary element c from UoD)

**Existential Instantiation
**

(for some specific object c from UoD, that has not yet occurred!)

P(c)__ ∴ ∃x P(x)

**Existential Generalization
**

(for some specific object c from UoD)

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Vacuous & Trivial Proofs
p → q is vacuously true if p is false
In this case, p → q is a **vacuous proof
**
p → q is trivially true if q is true
In this case, we have a **trivial proof
**

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Proofs Direct Proof: To prove p → q, we assume p and show/derive q Indirect Proof by Contraposition: To prove p → q, we prove its contrapositive, ¬ q → ¬ p So assume ¬ q and show/derive ¬ p

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Proofs Indirect Proof by Contradiction: To prove p, we

assume ¬ p and derive a contradiction. Based on the tautology

( ¬ p → F ) → p “if the negation of p implies a contradiction then p must be true” (aka: Reductio ad Absurdum)

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More Proofs Equivalence: To prove p ↔ q, we prove

p → q and q → p

Cases: To prove (p1 v p2 v … v pn) → q, we prove (p1 → q) ∧ (p2 → q) ∧ … ∧ (pn → q)

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More Proofs Quantifiers: ∀x P(x) : provide a proof, or counterexample. ∃x P(x): Existence Constructive Proof: Find an a in the UoD

such that P(a) holds. Existence Non-Constructive Proof: Prove that ∃x

P(x) is true without finding an a in the UoD such that P(a) holds

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