Discrete Mathematics CS 2610 August 21, 2008
Agenda Nested quantifiers Rules of inference Proofs
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”
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)
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))
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)
Alice likes some car There is a car that is liked by everyone
Everyone likes some car Everyone likes the Prius
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).
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.
Proofs • A theorem is a statement that can be proved to be
• A proof is a sequence of statements that form an argument.
Proofs: Inference Rules An Inference Rule:
“∴” means “therefore”
premise 1 premise 2 … ∴ conclusion
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 → q
(p ∧ (p → q)) → q
Proofs: Modus Tollens
•If the power supply fails then the lights go out. •The lights are on. ∴ The power supply has not failed.
(¬q ∧ (p → q)) → ¬p
p → q
•I am a student. ∴ I am a student or I am a visitor.
∴ p ∨ q
p → (p ∨ q)
•I am a student and I am a soccer player. ∴ I am a student.
p ∧ q
(p ∧ q) → p
•I am a student. •I am a soccer player. ∴ I am a student and I am a soccer player.
∴ p ∧ q
((p) ∧ (q)) → p ∧ q
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
((p ∨ q) ∧ ¬q) → p
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
((p → q) ∧ (q → r)) → (p → r)
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
((p ∨ q ) ∧ (¬ p ∨ r)) → (q ∨ r)
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
((p ∨ q ) ∧ (p → r) ∧ (q → r)) → r
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.
(q ∧ (p → q)) → p
p → q
Abductive, rather than deductive reasoning! Docsity.com
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.
(¬p ∧ (p → q)) → ¬q
p → q
Inference Rules for Quantified Statements
∀x P(x) ∴ P(c)
∃x P(x) ∴ P(c)
P(c)___ ∴ ∀x P(x)
(for an arbitrary object c from UoD)
(for any arbitrary element c from UoD)
(for some specific object c from UoD, that has not yet occurred!)
P(c)__ ∴ ∃x P(x)
(for some specific object c from UoD)
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
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
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)
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)
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