Discrete Mathematics and Its Applicationsdase.ecnu.edu.cn/mgao/teaching/DM_2020_Spring/slides/1... · 2020-02-24 · Outline 1 Logical Equivalences 2 Propositional Satis ability 3

Discrete Mathematics and Its ApplicationsLecture 1: The Foundations: Logic and Proofs (1.3-1.5)

MING GAO

DASE @ ECNU(for course related communications)

[email protected]

Mar. 10, 2020

Outline

1 Logical Equivalences

2 Propositional Satisfiability

3 Predicates

4 Quantifiers

5 Applications of Quantifiers

6 Nested Quantifiers

7 Take-aways

MING GAO (DaSE@ECNU) Discrete Mathematics and Its Applications Mar. 10, 2020 2 / 28

Logical Equivalences

Motivation

Example

There are two kinds of inhabitants in an island, knights, who alwaystell the truth, knaves, who always lie. You encounter two people Aand B. What are A and B if A says “B is a knight” and B says “Thetwo of us are opposite types”?

Solution:

p : “A is a knight;”

q : “B is a knight;”

If A is a knight, we have p ∧ q ∧ ((¬p ∧ q) ∨ (p ∧ ¬q)).

If A is a knave, we have ¬p ∧ ¬q ∧ ((p ∧ q) ∨ (¬p ∧ ¬q)).

The problem is how to determine the truth value of the propositions.



Motivation

Example

There are two kinds of inhabitants in an island, knights, who alwaystell the truth, knaves, who always lie. You encounter two people Aand B. What are A and B if A says “B is a knight” and B says “Thetwo of us are opposite types”?Solution:








Motivation

Example









Motivation

Example









Motivation

Example









Logical equivalences

Definition

Compound propositions that have the same truth values in all possiblecases are called logically equivalent.

Compound propositions p and q are called logically equivalent ifp ↔ q is a tautology, denoted as p ≡ q or p ⇔ q.

Remark: Symbol ≡ is not a logical connectives, and p ≡ q is not aproposition.

One way to determine whether two compound propositions areequivalent is to use a truth table.




Definition








Definition








Definition







De Morgan’s laws

Laws

¬(p ∧ q) ≡ ¬p ∨ ¬q

¬(p ∨ q) ≡ ¬p ∧ ¬q

¬(p1 ∧ p2 ∧ · · · ∧ pn) ≡ ¬p1 ∨ ¬p2 ∨ ¬ · · · ∨ ¬pn, i.e.,¬∧n

i=1 pi ≡∨n

i=1 ¬pi .

¬(p1 ∨ p2 ∨ · · · ∨ pn) ≡ ¬p1 ∧ ¬p2 ∧ ¬ · · · ∧ ¬pn, i.e.,¬∨n

i=1 pi ≡∧n

i=1 ¬pi .

The truth table can be used to determine whether two compoundpropositions are equivalent.

p q p ∧ q ¬(p ∧ q) ¬p ¬q ¬p ∨ ¬q

T T T F F F FT F F T F T TF T F T T F TF F F T T T T



De Morgan’s laws

Laws

¬(p ∧ q) ≡ ¬p ∨ ¬q

¬(p ∨ q) ≡ ¬p ∧ ¬q

¬(p1 ∧ p2 ∧ · · · ∧ pn) ≡ ¬p1 ∨ ¬p2 ∨ ¬ · · · ∨ ¬pn, i.e.,¬∧n

i=1 pi ≡∨n

i=1 ¬pi .

¬(p1 ∨ p2 ∨ · · · ∨ pn) ≡ ¬p1 ∧ ¬p2 ∧ ¬ · · · ∧ ¬pn, i.e.,¬∨n

i=1 pi ≡∧n

i=1 ¬pi .


p q p ∧ q ¬(p ∧ q) ¬p ¬q ¬p ∨ ¬q




De Morgan’s laws

Laws

¬(p ∧ q) ≡ ¬p ∨ ¬q

¬(p ∨ q) ≡ ¬p ∧ ¬q

¬(p1 ∧ p2 ∧ · · · ∧ pn) ≡ ¬p1 ∨ ¬p2 ∨ ¬ · · · ∨ ¬pn, i.e.,¬∧n

i=1 pi ≡∨n

i=1 ¬pi .

¬(p1 ∨ p2 ∨ · · · ∨ pn) ≡ ¬p1 ∧ ¬p2 ∧ ¬ · · · ∧ ¬pn, i.e.,¬∨n

i=1 pi ≡∧n

i=1 ¬pi .


p q p ∧ q ¬(p ∧ q) ¬p ¬q ¬p ∨ ¬q




Logical equivalence

Table of logical equivalence

equivalence name

p ∧ T ≡ p Identity lawsp ∨ F ≡ p

p ∧ p ≡ p Idempotent lawsp ∨ p ≡ p

(p ∧ q) ∧ r ≡ p ∧ (q ∧ r) Associative laws(p ∨ q) ∨ r ≡ p ∨ (q ∨ r)

p ∨ (p ∧ q) ≡ p Absorption lawsp ∧ (p ∨ q) ≡ p

p ∧ ¬p ≡ F Negation lawsp ∨ ¬p ≡ T



Logical equivalence Cont’d


equivalence name

p ∨ T ≡ T Domination lawsp ∧ F ≡ F

p ∧ q ≡ q ∧ p Commutative lawsp ∨ q ≡ q ∨ p

(p ∧ q) ∨ r ≡ (p ∨ r) ∧ (q ∨ r) Distributive laws(p ∨ q) ∧ r ≡ (p ∧ r) ∨ (q ∧ r)

¬(p ∧ q) ≡ ¬p ∨ ¬q De Morgan’s laws¬(p ∨ q) ≡ ¬p ∧ ¬q

¬(¬p) ≡ p Double negation law



Equivalence of implication

Equivalence law

p → q ≡ ¬p ∨ q


p q p → q ¬p ¬p ∨ q

T T T F TT F F F FF T T T TF F T T T



Equivalence of implication

Equivalence law

p → q ≡ ¬p ∨ qThe truth table can be used to determine whether two compoundpropositions are equivalent.

p q p → q ¬p ¬p ∨ q

T T T F TT F F F FF T T T TF F T T T



Logical equivalences involving conditional statements

Table of logical equivalences involving conditional statements

p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r





p → q ≡ ¬p ∨ q

p → q ≡ ¬q → ¬p

p ∨ q ≡ ¬p → q

p ∧ q ≡ ¬(p → ¬q)

¬(p → q) ≡ p ∧ ¬q

(p → q) ∧ (p → r) ≡ p → (q ∧ r)

(p → q) ∨ (p → r) ≡ p → (q ∨ r)

(p → r) ∧ (q → r) ≡ (p ∨ q)→ r

(p → r) ∨ (q → r) ≡ (p ∧ q)→ r



Logical equivalences involving biconditional statements

Table of logical equivalences involving biconditional statements

p ↔ q ≡ (p → q) ∧ (q → p)

p ↔ q ≡ ¬p ↔ ¬q

p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)

¬(p ↔ q) ≡ p ↔ ¬q

Show that ¬(p ↔ q) and p ↔ ¬q are logically equivalent.

¬(p ↔ q) ≡ ¬((¬p ∨ q) ∧ (¬q ∨ p))

≡ (p ∧ ¬q) ∨ (q ∧ ¬p)

≡ (p ∨ (q ∧ ¬p)) ∧ (¬q ∨ (q ∧ ¬p))

≡ ((p ∨ q) ∧ (p ∨ ¬p)) ∧ ((¬q ∨ q) ∧ (¬q ∨ ¬p))

≡ (¬(¬q) ∨ p) ∧ (¬p ∨ ¬q))

≡ (¬q → p) ∧ (p → ¬q)) ≡ p ↔ ¬q (1)





p ↔ q ≡ (p → q) ∧ (q → p)

p ↔ q ≡ ¬p ↔ ¬q

p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)

¬(p ↔ q) ≡ p ↔ ¬q


¬(p ↔ q) ≡ ¬((¬p ∨ q) ∧ (¬q ∨ p))

≡ (p ∧ ¬q) ∨ (q ∧ ¬p)

≡ (p ∨ (q ∧ ¬p)) ∧ (¬q ∨ (q ∧ ¬p))

≡ ((p ∨ q) ∧ (p ∨ ¬p)) ∧ ((¬q ∨ q) ∧ (¬q ∨ ¬p))

≡ (¬(¬q) ∨ p) ∧ (¬p ∨ ¬q))

≡ (¬q → p) ∧ (p → ¬q)) ≡ p ↔ ¬q (1)





p ↔ q ≡ (p → q) ∧ (q → p)

p ↔ q ≡ ¬p ↔ ¬q

p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)

¬(p ↔ q) ≡ p ↔ ¬q


¬(p ↔ q) ≡ ¬((¬p ∨ q) ∧ (¬q ∨ p))

≡ (p ∧ ¬q) ∨ (q ∧ ¬p)

≡ (p ∨ (q ∧ ¬p)) ∧ (¬q ∨ (q ∧ ¬p))

≡ ((p ∨ q) ∧ (p ∨ ¬p)) ∧ ((¬q ∨ q) ∧ (¬q ∨ ¬p))

≡ (¬(¬q) ∨ p) ∧ (¬p ∨ ¬q))

≡ (¬q → p) ∧ (p → ¬q)) ≡ p ↔ ¬q (1)





p ↔ q ≡ (p → q) ∧ (q → p)

p ↔ q ≡ ¬p ↔ ¬q

p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)

¬(p ↔ q) ≡ p ↔ ¬q


¬(p ↔ q) ≡ ¬((¬p ∨ q) ∧ (¬q ∨ p))

≡ (p ∧ ¬q) ∨ (q ∧ ¬p)

≡ (p ∨ (q ∧ ¬p)) ∧ (¬q ∨ (q ∧ ¬p))

≡ ((p ∨ q) ∧ (p ∨ ¬p)) ∧ ((¬q ∨ q) ∧ (¬q ∨ ¬p))

≡ (¬(¬q) ∨ p) ∧ (¬p ∨ ¬q))

≡ (¬q → p) ∧ (p → ¬q)) ≡ p ↔ ¬q (1)





p ↔ q ≡ (p → q) ∧ (q → p)

p ↔ q ≡ ¬p ↔ ¬q

p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q)

¬(p ↔ q) ≡ p ↔ ¬q


¬(p ↔ q) ≡ ¬((¬p ∨ q) ∧ (¬q ∨ p))

≡ (p ∧ ¬q) ∨ (q ∧ ¬p)

≡ (p ∨ (q ∧ ¬p)) ∧ (¬q ∨ (q ∧ ¬p))

≡ ((p ∨ q) ∧ (p ∨ ¬p)) ∧ ((¬q ∨ q) ∧ (¬q ∨ ¬p))

≡ (¬(¬q) ∨ p) ∧ (¬p ∨ ¬q))

≡ (¬q → p) ∧ (p → ¬q)) ≡ p ↔ ¬q (1)


Propositional Satisfiability

Propositional satisfiability

Definition

A compound proposition is satisfiable if there is an assignment oftruth values to its variables that makes it true.

Determine whether each of the compound propositions(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p).

(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ≡ (q → p) ∧ (r → q) ∧ (p → r)

Note that (p ∨¬q)∧ (q ∨¬r)∧ (r ∨¬p) is true when the three variable p,q, and r have the same truth value.Hence, it is satisfiable as there is at least one assignment of truth valuesfor p, q, and r that makes it true.




Definition



(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ≡ (q → p) ∧ (r → q) ∧ (p → r)





Definition



(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ≡ (q → p) ∧ (r → q) ∧ (p → r)





Definition



(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ≡ (q → p) ∧ (r → q) ∧ (p → r)

Note that (p ∨¬q)∧ (q ∨¬r)∧ (r ∨¬p) is true when the three variable p,q, and r have the same truth value.

Hence, it is satisfiable as there is at least one assignment of truth valuesfor p, q, and r that makes it true.




Definition



(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ≡ (q → p) ∧ (r → q) ∧ (p → r)




Applications of satisfiability

Sudoku puzzle

For each cell with a given value, we as-sert p(i , j , n) when the cell in row i andcolumn j has the given value n.

For every row, we assert:∧9

i=1

∧9n=1

∨9j=1 p(i , j , n);

For every column, we assert:∧9

j=1

∧9n=1

∨9i=1 p(i , j , n);

For every block, we assert it contains every number:∧2r=0

∧2s=0

∧9n=1

∨3i=1

∨3j=1 p(3r + i , cs + j , n);

To assert that no cell contains more than one number, we takethe conjunction over all values of n, n

′, i , and j where each

variable ranges from 1 to 9 and n 6= n′

ofp(i , j , n)→ ¬p(i , j , n

′).




Sudoku puzzle



i=1

∧9n=1

∨9j=1 p(i , j , n);


j=1

∧9n=1

∨9i=1 p(i , j , n);


∧2s=0

∧9n=1

∨3i=1

∨3j=1 p(3r + i , cs + j , n);




ofp(i , j , n)→ ¬p(i , j , n

′).




Sudoku puzzle



i=1

∧9n=1

∨9j=1 p(i , j , n);


j=1

∧9n=1

∨9i=1 p(i , j , n);


∧2s=0

∧9n=1

∨3i=1

∨3j=1 p(3r + i , cs + j , n);




ofp(i , j , n)→ ¬p(i , j , n

′).




Sudoku puzzle



i=1

∧9n=1

∨9j=1 p(i , j , n);


j=1

∧9n=1

∨9i=1 p(i , j , n);


∧2s=0

∧9n=1

∨3i=1

∨3j=1 p(3r + i , cs + j , n);




ofp(i , j , n)→ ¬p(i , j , n

′).




Sudoku puzzle



i=1

∧9n=1

∨9j=1 p(i , j , n);


j=1

∧9n=1

∨9i=1 p(i , j , n);


∧2s=0

∧9n=1

∨3i=1

∨3j=1 p(3r + i , cs + j , n);




ofp(i , j , n)→ ¬p(i , j , n

′).


Predicates

Motivation I

In many cases, the statement we are interested in contains variables.

Example

“e is even”, “p is prime”, or “s is a student”.

As we previously did with propositions, we can use variables to rep-resent these statements.

E (x) : “x is even”;

P(y) : “y is prime”;

S(w) : “w is a student”.

You can think of E (x), P(y) and S(w) as statements that may betrue of false depending on the values of its variables.


Predicates

Motivation I


Example

“e is even”, “p is prime”, or “s is a student”.As we previously did with propositions, we can use variables to rep-resent these statements.






Predicates

Motivation I


Example







Predicates

Motivation I


Example







Predicates

Motivation I


Example







Predicates

Motivation II

Example

“Every computer connected to the university network isfunctioning properly.”No rules of propositional logic allow us to conclude the truth ofthe statement“MATH3 is functioning properly, if MATH3 is one of thecomputers connected to the university network.”

Likewise, we cannot use the rules of propositional logic toconclude from the statement“CS2 is under attack by an intruder, and CS2 is a computer onthe university network.”We can conclude the truth of “There is a computer on theuniversity network that is under attack by an intruder.”

Predicate logic is a more powerful type of logic theory.


Predicates

Motivation II

Example





Predicates

Motivation II

Example





Predicates

Predicates

Definition

“x is greater than 3” has two parts: variable x (the subject of the state-ment), and predicate P “is greater than 3”, denoted as P(x) : x > 3.

Statement P(x) is the value of the propositional function P at x ;

Once variable x is fixed, statement P(x) becomes a proposition andhas a truth value. e.g., P(4) (true) and P(2) (false)

Let A(x) denote “Computer x is under attack by an intruder”.Assume that CS2 in the campus is currently under attack byintruders. What are truth values of A(CS1), and A(CS2)?

Let Q(x , y) denote the statement “x = y + 3”. What are the truthvalues of the propositions Q(1, 2) and Q(3, 0)?

In general, a statement involving n variables x1, x2, · · · , xn can be denotedby P(x1, x2, · · · , xn), where P is also called an n-place predicate or a n-arypredicate, and x1, x2, · · · , xn is a n-tuple.


Predicates

Predicates

Definition








Predicates

Predicates

Definition








Predicates

Predicates

Definition








Predicates

Predicates

Definition








Predicates

Predicates

Definition








Quantifiers

Quantifiers

Quantification

Quantification expresses the extent to which a predicate is true over arange of elements. In general, all values of a variable is called the domainof discourse (or universe of discourse), just referred to as domain.

1 The universal quantification of P(x) is the statement “P(x) for allvalues of x in the domain”. Notation ∀xP(x) denotes the universalquantification of P(x), where ∀ is called the universal quantifier. Anelement for which P(x) is false is called a counterexample of ∀xP(x).

2 The existential quantification of P(x) is the proposition “Thereexists an element x in the domain such that P(x)”. Notation∃xP(x) denotes the existential quantification of P(x), where ∃ iscalled the existential quantifier.

Statement When True? When False?∀xP(x) P(x) is true for every x ∃x for which P(x) is false∃xP(x) ∃x for which P(x) is true P(x) is false for every x


Quantifiers

Quantifiers

Quantification






Quantifiers

Quantifiers

Quantification






Quantifiers

Quantifiers

Quantification






Quantifiers

Examples

Universal quantification

1 Let P(x) be “x + 1 > x”. What is the truth value of ∀xP(x) for∀x ∈ R?

Note that if the domain is empty, then ∀xP(x) is true for anypropositional function P(x) because there are no elements x in thedomain for which P(x) is false.

2 Let P(x) be “x2 > 0”. What is the truth value of ∀xP(x) for∀x ∈ Z ? (Note that x = 0 is a counterexample because x2 = 0.)

Existential quantification

1 Let P(x) denote “x > 3”. What is the truth value of ∃xP(x) for∀x ∈ R?

2 Let P(x) be “x2 > 0”. What is the truth value of ∃xP(x) for∀x ∈ Z ?


Quantifiers

Examples

Universal quantification

1 Let P(x) be “x + 1 > x”. What is the truth value of ∀xP(x) for∀x ∈ R?

Note that if the domain is empty, then ∀xP(x) is true for anypropositional function P(x) because there are no elements x in thedomain for which P(x) is false.

2 Let P(x) be “x2 > 0”. What is the truth value of ∀xP(x) for∀x ∈ Z ? (Note that x = 0 is a counterexample because x2 = 0.)


1 Let P(x) denote “x > 3”. What is the truth value of ∃xP(x) for∀x ∈ R?

2 Let P(x) be “x2 > 0”. What is the truth value of ∃xP(x) for∀x ∈ Z ?


Quantifiers

Remarks

When all the elements in the domain can be listed as x1, x2, · · · , xnUniversal quantification

∀xP(x) is the same as conjunction

P(x1) ∧ P(x2) ∧ · · · ∧ P(xn),

because this conjunction is true if and only if P(x1), P(x2), · · · ,P(xn) areall true.


∃xP(x) is the same as disjunction

P(x1) ∨ P(x2) ∨ · · · ∨ P(xn),

since the disjunction is true if and only if at least one of P(x1),P(x2), · · · ,P(xn) is true.


Quantifiers

Remarks

When all the elements in the domain can be listed as x1, x2, · · · , xnUniversal quantification

∀xP(x) is the same as conjunction

P(x1) ∧ P(x2) ∧ · · · ∧ P(xn),

because this conjunction is true if and only if P(x1), P(x2), · · · ,P(xn) areall true.


∃xP(x) is the same as disjunction

P(x1) ∨ P(x2) ∨ · · · ∨ P(xn),

since the disjunction is true if and only if at least one of P(x1),P(x2), · · · ,P(xn) is true.


Quantifiers

Quantifiers with restricted domains

Example

What do the statements ∀x < 0(x2 > 0) and ∃y > 0(y 2 = 2) mean, wherethe domain in each case consists of the real numbers?

1 The statement ∀x < 0(x2 > 0) states that for every real number xwith x < 0, x2 > 0, i.e., it states “The square of a negative realnumber is positive”. This statement is the same as∀x(x < 0→ x2 > 0).

2 The statement ∃y > 0(y 2 = 2) states “There is a positive squareroot of 2”, i.e., ∃y(y > 0 ∧ y 2 = 2).

Note that the restriction of a universal quantification is the same asthe universal quantification of a conditional statement.

On the other hand, the restriction of an existential quantification isthe same as the existential quantification of a conjunction.


Quantifiers


Example







Quantifiers


Example







Quantifiers


Example







Quantifiers


Example







Quantifiers

Precedence of quantifiers and binding variables

Precedence of quantifiers

The quantifiers ∀ and ∃ have higher precedence than all logical operatorsfrom propositional calculus.

∀xP(x) ∧ Q(x) ≡ (∀xP(x)) ∧ Q(x), rather than ∀x(P(x) ∧ Q(x)).

∃xP(x) ∨ Q(x) ≡ (∃xP(x)) ∨ Q(x).

Bound and free

In statement ∃x(x + y = 1), variable x is bound by the existential quantifi-cation ∃x , but variable y is free because it is not bound by a quantifier andno value is assigned to this variable. This illustrates that in the statement,x is bound, but y is free.The part of a logical expression to which a quantifier is applied is called thescope of this quantifier. Consequently, a variable is free if it is outside thescope of all quantifiers in the formula that specify this variable.


Quantifiers

Precedence of quantifiers and binding variables

Precedence of quantifiers

The quantifiers ∀ and ∃ have higher precedence than all logical operatorsfrom propositional calculus.

∀xP(x) ∧ Q(x) ≡ (∀xP(x)) ∧ Q(x), rather than ∀x(P(x) ∧ Q(x)).

∃xP(x) ∨ Q(x) ≡ (∃xP(x)) ∨ Q(x).

Bound and free

In statement ∃x(x + y = 1), variable x is bound by the existential quantifi-cation ∃x , but variable y is free because it is not bound by a quantifier andno value is assigned to this variable. This illustrates that in the statement,x is bound, but y is free.The part of a logical expression to which a quantifier is applied is called thescope of this quantifier. Consequently, a variable is free if it is outside thescope of all quantifiers in the formula that specify this variable.


Quantifiers

Logical equivalences involving quantifiers

Definition

Statements involving predicates and quantifiers are logically equiva-lent if and only if they have the same truth value no matter whichpredicates are substituted into these statements and which domainof discourse is used for the variables in these propositional functions.We use the notation S ≡ T to indicate that two statements S andT involving predicates and quantifiers are logically equivalent.


equivalence name

∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x) Distributive law

¬∀xP(x) ≡ ∃x¬P(x) Negation law¬∃xP(x) ≡ ∀x¬P(x)

¬∀x(P(x)→ Q(x)) ≡ ∃x(P(x) ∧ ¬Q(x))


Quantifiers

Logical equivalences involving quantifiers

Definition

Statements involving predicates and quantifiers are logically equiva-lent if and only if they have the same truth value no matter whichpredicates are substituted into these statements and which domainof discourse is used for the variables in these propositional functions.We use the notation S ≡ T to indicate that two statements S andT involving predicates and quantifiers are logically equivalent.


equivalence name

∀x(P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀xQ(x) Distributive law

¬∀xP(x) ≡ ∃x¬P(x) Negation law¬∃xP(x) ≡ ∀x¬P(x)

¬∀x(P(x)→ Q(x)) ≡ ∃x(P(x) ∧ ¬Q(x))


Applications of Quantifiers

Quantifiers in system specifications

Example

Use predicates and quantifiers to express the system specifications “Everymail message larger than one megabyte will be compressed” and “If a useris active, at least one network link will be available.”

Let S(m, y) be “Mail message m is larger than y megabytes,” wherevariable x has the domain of all mail messages and variable y is apositive real number, and let C (m) denote “Mail message m will becompressed.” Then “Every mail message larger than one megabytewill be compressed” can be represented as ∀m(S(m, 1)→ C (m)).

Let A(u) represent “User u is active,” where variable u has thedomain of all users, let S(n, x) denote “Network link n is in state x ,”where n has the domain of all network links and x has the domain ofall possible states for a network link. Then “If a user is active, atleast one network link will be available” can be represented by∃uA(u)→ ∃nS(n, available).




Example







Example





Nested Quantifiers

Nested quantifiers

Definition

Nested quantifiers is one quantifier within the scope of another.

For example, ∀x∃y(x + y = 0). Note that everything within thescope of a quantifier can be thought of as a propositional function.∀x∃y(x + y = 0) is the same thing as ∀xQ(x), where Q(x) is∃yP(x , y), where P(x , y) is x + y = 0.Please translate following nested quantifiers into statements

∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.

∀x∀y∀z((x + y) + z = x + (y + z)) for ∀x , y , z ∈ R.

∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition

Nested quantifiers is one quantifier within the scope of another.For example, ∀x∃y(x + y = 0). Note that everything within thescope of a quantifier can be thought of as a propositional function.∀x∃y(x + y = 0) is the same thing as ∀xQ(x), where Q(x) is∃yP(x , y), where P(x , y) is x + y = 0.Please translate following nested quantifiers into statements

∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition


∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition


∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition


∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition


∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Nested quantifiers

Definition


∀x∀y(x + y = y + x) for ∀x , y ∈ R.

∀x∃y(x + y = 0) for ∀x , y ∈ R.


∀x∀y((x > 0) ∧ (y > 0)→ (xy < 0)) for ∀x , y ∈ R.

∀ε > 0∃δ > 0∀x(0 < |x − a| < δ → |f (x)− L| < ε).


Nested Quantifiers

Order of quantifiers

Order is important

It is important to note that the order of the quantifiers is important,unless all the quantifiers are universal quantifiers or all are existentialquantifiers.

Statement When True? When False?

∀x∀yP(x , y) P(x , y) is true There is a pair x , y for∀y∀xP(x , y) for every pair x , y which P(x , y) is false

∀x∃yP(x , y) For every x , there is a y There is an x such thatfor which P(x , y) is true P(x , y) is false for ∀y

∃x∀yP(x , y) There is an x for which For every x , there is a yP(x , y) is true for every y for which P(x , y) is false

∃x∃yP(x , y) There is a pair x , y P(x , y) is false∃y∃xP(x , y) for which P(x , y) is true for every pair x , y


Nested Quantifiers

Order of quantifiers

Order is important

It is important to note that the order of the quantifiers is important,unless all the quantifiers are universal quantifiers or all are existentialquantifiers.

Statement When True? When False?

∀x∀yP(x , y) P(x , y) is true There is a pair x , y for∀y∀xP(x , y) for every pair x , y which P(x , y) is false

∀x∃yP(x , y) For every x , there is a y There is an x such thatfor which P(x , y) is true P(x , y) is false for ∀y

∃x∀yP(x , y) There is an x for which For every x , there is a yP(x , y) is true for every y for which P(x , y) is false

∃x∃yP(x , y) There is a pair x , y P(x , y) is false∃y∃xP(x , y) for which P(x , y) is true for every pair x , y


Nested Quantifiers

Applications of nested quantifiers

Nested quantifiers translation

∀x(C (x) ∨ ∃y(C (y) ∧ F (x , y)))

C (x): x has a computer;

F (x , y): x and y are friends;

Domain for both x and y consists of all students in your school.

Solution:The statement says that for every student x in your school, x has acomputer or there is a student y such that y has a computer and xand y are friends.That is, every student in your school has a computer or has a friendwho has a computer.


Nested Quantifiers



∀x(C (x) ∨ ∃y(C (y) ∧ F (x , y)))




Solution:The statement says that for every student x in your school, x has acomputer or there is a student y such that y has a computer and xand y are friends.

That is, every student in your school has a computer or has a friendwho has a computer.


Nested Quantifiers



∀x(C (x) ∨ ∃y(C (y) ∧ F (x , y)))




Solution:The statement says that for every student x in your school, x has acomputer or there is a student y such that y has a computer and xand y are friends.That is, every student in your school has a computer or has a friendwho has a computer.


Nested Quantifiers

Applications of nested quantifiers Cont’d

Sentence translation I

“If a person is female and is a parent, then she is someones mother”.Solution:∀x((F (x) ∧ P(x))→ ∃yM(x , y))

F (x): x is female;

P(x): x is a parent;

M(x , y): x is the mother of y ;

Sentence translation II

“There is a woman who has taken a flight on every airline in the world”.Solution:∃w∀a∃f (P(w , f ) ∧ Q(f , a))

P(w , f ): w has taken f ;

Q(f , a): f is a flight on a;

Or∃w∀a∃fR(w , f , a)

R(w , f , a): w has taken f on a.


Nested Quantifiers

Applications of nested quantifiers Cont’d

Sentence translation I

“If a person is female and is a parent, then she is someones mother”.Solution:∀x((F (x) ∧ P(x))→ ∃yM(x , y))

F (x): x is female;

P(x): x is a parent;

M(x , y): x is the mother of y ;

Sentence translation II

“There is a woman who has taken a flight on every airline in the world”.Solution:∃w∀a∃f (P(w , f ) ∧ Q(f , a))



Or∃w∀a∃fR(w , f , a)

R(w , f , a): w has taken f on a.


Nested Quantifiers

Negation of nested quantifiers

Example

Use quantifiers to express the statement that “There does not exist a womanwho has taken a flight on every airline in the world”

¬∃w∀a∃f (P(w , f ) ∧ Q(f , a))



¬∃w∀a∃f (P(w , f ) ∧ Q(f , a)) ≡ ∀w¬∀a∃f (P(w , f ) ∧ Q(f , a))

≡ ∀w∃a¬∃f (P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f ¬(P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f (¬P(w , f ) ∨ ¬Q(f , a))

“For every woman there is an airline such that for all flights, this womanhas not taken that flight or that flight is not on this airline.”


Nested Quantifiers


Example


¬∃w∀a∃f (P(w , f ) ∧ Q(f , a))




≡ ∀w∃a¬∃f (P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f ¬(P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f (¬P(w , f ) ∨ ¬Q(f , a))



Nested Quantifiers


Example


¬∃w∀a∃f (P(w , f ) ∧ Q(f , a))




≡ ∀w∃a¬∃f (P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f ¬(P(w , f ) ∧ Q(f , a))

≡ ∀w∃a∀f (¬P(w , f ) ∨ ¬Q(f , a))



Take-aways

Take-aways

Conclusion

Logic equivalences


Predicates

Quantifiers

Applications of predicates and quantifiers

Nested quantifiers


Documents

Discrete Mathematics and Its Applicationsdase.ecnu.edu.cn/mgao/teaching/DM_2020_Spring/slides/1... · 2020-02-24 · Outline 1 Logical Equivalences 2 Propositional Satis ability 3