chapter i - symbolic logic

PropositionLogical operations

Generation of operatorsPropositions with quantifiers ∀, ∃

Symbolic logics

NGUYEN CANH Nam1

1Faculty of Applied MathematicsDepartment of Applied Mathematics and Informatics

Hanoi University of [email protected]

HUT - 2010

NGUYEN CANH Nam Mathematics I - Chapter 1



References1 Nguyen Dinh Tri, Ta Van Dinh, Nguyen Ho Quynh, Toan hoc cap cap - Tap I,

Nha xuat ban giao duc.

2 Nguyen Van Nghi, Nguyen Canh Luong, Phuong phap giai bai tap toan caocap - Tap I, Nha xuat ban khoa hoc va ky thuat.

3 Kenneth H.Rosen, Discrete Mathematics and Its Applications, Mc Graw Hill,International edition 2007.

4 D.A.R. Wallace, Groups, Rings and Fields, Springer 1998.

5 Titu Andreescu and Dorin Andrica, Complex Numbers from A to. . . Z,Birkhäuser Boston 2006.

6 P.B.Bhattacharya S.K.Jain and S.R.Nagpaul, Basic Abstract Algebra,Cambridge University Press, Second edition, 1996.

7 www...




Agenda1 Proposition2 Logical operations

Negation operator NOTConjunction operator AND (∧)Disjunction operator OR (∨)Implication operator IMP (→)Equivalence operator IFF (↔)Tautologies, contradictions

3 Generation of operatorsBinary XOR operatorBinary operator NOR (↑)Binary Operator NAND (↓ )

4 Propositions with quantifiers ∀,∃NGUYEN CANH Nam Mathematics I - Chapter 1



Proposition

DefinitionA proposition is a statement which is either true or false,although we may not know which, but not both.Propositions are denoted by lower letters as p,q, r The truth orfalsity is called truth value of the proposition. The truth value ofthe proposition p is denoted by V (p).If p is true then V (p) = 1 or T . If p is false then V (p) = 0 or F .




PropositionExamples

Example 1:p = "Hanoi is the capital of Vietnam" then V (p) = 1q = "The sun rises in the east " then V (q) = 1r = "1 + 1 = 2" then V (r) = 1s = "The sun rises in the west" then V (s) = 0

Example 2:The proposition t = "There exists life outside the earth".Up to now we can not know the truth value of the statement t .




PropositionExample (continue...)

Consider the following sentences1 What time is it?2 Read this carefully3 x + 1 = 2

Sentences 1 and 2 are not proposition because they are notdeclarative sentences.





Negation operator NOT

DefinitionLet p be a proposition. The negation of p, denoted by p, is thestatement

"It is not the case that p"

The truth table of NOT operator is as follow

p p0 11 0

Table: NOT truth table





Negation operator NOTExamples

Example 1 : Find the negation of the proposition

"Today is Monday"

The negation is

"It is not the case that today is Monday"

or simply"Today is not Monday"





Negation operator NOTExamples

Example 2 : Find the negation of the proposition

"At least 100mm of rain fell today in Hanoi"

The negation is

"It is not the case that at least 100mm of rain fell today in Hanoi"

or simply

"Less than 100mm of rain fell today in Hanoi"





Conjunction operator AND (∧)

DefinitionLet p and q be propositions. The conjunction of p and q, denoted byp ∧ q, is the proposition "p and q".The truth table of AND operator is as follow

p q p ∧ q0 0 00 1 01 0 01 1 1

Table: AND truth table

Note : The conjunction p ∧ q is true when both p and q are true.NGUYEN CANH Nam Mathematics I - Chapter 1




Conjunction operator AND (∧)Example

Let p = "Pigs are mammals" and q = "Pigs fly",

then p ∧ q is interpreted as

"Pigs are flying mammals".





Disjunction operator OR (∨)

DefinitionLet p and q be propositions. The disjunction of p and q, denoted byp ∨ q, is the proposition "p or q".The truth table of OR operator is as follow

p q p ∧ q0 0 00 1 11 0 11 1 1

Table: OR truth table

Note : The disjunction p ∨ q is false when both p and q are false.





Disjunction operator OR (∨)Example

Let p = "Today is Monday" and q = "It is raining today",then p ∨ q is interpreted as

"Today is Monday or it is raining today".

This proposition is true on any day that is either a Monday or arainy day (including rainy Mondays).





De Morgan’s Law

Theorem (De Morgan’s Law)For any two propositions p and q we have

(i) V (p ∧ q) = V (p) ∨ V (q)(ii) V (p ∨ q) = V (p) ∧ V (q)

CorollaryThe disjunction operator OR may be defined by NOT and ANDoperators:

V (p ∨ q) = V (p ∧ q)





Distributive Laws

Theorem (Distributive Laws)For any three propositions p,q, r , we have

(i) V (p ∧ (q ∨ r)) = V ((p ∧ q) ∨ (p ∧ r))(ii) V (p ∨ (q ∧ r)) = V ((p ∨ q) ∧ (p ∨ r))





Commutative and associative Laws

Theorem (Commutative and associative Laws)For any propositions p,q, r , we have

(i) V (p ∧ q) = V (q ∧ p)(ii) V (p ∨ q) = V (q ∨ p)(iii) V ((p ∧ q) ∧ r) = V (p ∧ (q ∧ r))(iv) V ((p ∨ q) ∨ r) = V (p ∨ (q ∨ r))





Implication operator IMP (→)

DefinitionLet p and q be propositions. The implication p → q is the proposition"If p then q".The truth table of IMP operator is as follow

p q p → q0 0 10 1 11 0 01 1 1

Table: IMP truth table

Note : The implication p → q is false when p is true and q is false.





Implication operator IMP (→)

In implication p → q, p is called the hypothesis and q is calledthe conclusion (or consequence)Because implication play such an essential role in mathematicalreasoning, a variety of terminology is used to express p → q

"if p then q" "p implies q""p is sufficient for q" "a sufficient condition for q is p""q follows from p" "q unless p"





Implication operator IMP (→)Example

Let p = "Minh learns discrete mathematics" andq = "Minh will find a good job",then p → q

"If Minh learns discrete mathematics, then he will will a good job"

There are many other ways to express this conditionstatement, for examples

"Minh will find a good job when he learns discrete mathematics"

"For Minh to get a good job,it is sufficient for him to learns discrete mathematics"

"Minh will find a good jobunless he does not learn discrete mathematics"





AssertionIf p,q are propositions then we have

(i) The proposition p → (p ∨ q) is true,(ii) The proposition (p ∧ q)→ p is true.

TheoremThe implication operator IMP may be built from the negationoperator NOT and the conjunction operator AND:

V (p → q) = V (p ∧ q)





Equivalence operator IFF (↔)

DefinitionLet p and q be propositions. The equivalence p ↔ q is theproposition "p if and only if q".The truth table of IFF operator is as follow

p q p ↔ q0 0 10 1 01 0 01 1 1

Table: IFF truth table

Note : The equivalence p ↔ q is true when p and q have the sametruth value.





Equivalence operator IFF (↔)

The equivalence p ↔ q is true when both implication p → qand q → p are true and is false otherwise. That is why we usethe word "if and only if".

TheoremFor propositions p,q we have

V (p ↔ q) = V ((p → q) ∧ (q → p))

There are some other common ways to express p ↔ q

"p is necessary and sufficient for q""if p then q, and conversely""p iff q"





Equivalence operator IFF (↔)Example

Let p be the statement "You can take the flight" and let q bethe statement "You buy a ticket".Then p ↔ q

" You can take the flight if and only if you buy a ticket"





TheoremThe equivalence operator may be built from the negationoperator NOT and the conjunction operator AND:

V (p ↔ q) = V ((p ∧ q) ∧ (q ∧ p))





Tautologies

DefinitionA compound proposition is called a tautology if it is alwaystrue regardless truth values of atomic components.





TautologiesExamples

a) The proposition (p ∧ q)→ p is a tautology.p q p ∧ q (p ∧ q)→ p0 0 0 10 1 0 11 0 0 11 1 1 1

b) ((p ∨ q) ∧ p)→ q is a tautology. The truth table isp q p ∨ q (p ∨ q) ∧ p ((p ∨ q) ∧ p)→ q0 0 0 0 10 1 1 1 11 0 1 0 11 1 1 0 1





Contradictions

DefinitionA compound proposition is called a contradiction if it’s valuesare always false regardless truth values of atomic components.





ContradictionsExamples

a) The proposition p ∧ p is a contradiction.

p p p ∧ p0 1 01 0 0

b) The proposition (p ∧ q)→ p is a contradiction because thetruth values of this are always false:

p q p ∧ q (p ∧ q)→ p0 0 0 00 1 0 01 0 0 01 1 1 0




Binary XOR operatorBinary operator NOR (↑)Binary Operator NAND (↓ )








Binary XOR operator (∨)

DefinitionLet p and q be propositions. The proposition p ∨ q is theproposition that is true when exactly one of p and q is true.The truth table of XOR operator is as follow

p q p∨q0 0 00 1 11 0 11 1 0

Table: XOR truth table





Binary XOR operator (∨)

AssertionThe XOR operator is the negation of the IFF operator:

V (p∨q) = V (p ↔ q)





Binary operator NOR (↑)

DefinitionGiven two proposition p,q. The proposition p ↑ q is theproposition defined by the truth table

p q p ↑ q0 0 10 1 01 0 01 1 0

Table: NOR truth table

This means neither p nor q





Binary operator NOR (↑)

AssertionThe operator NOR is negation of OR:

V (p ↑ q) = V (p ∨ q)





Binary Operator NAND (↓ )

DefinitionGiven two proposition p,q. The proposition p ↓ q is theproposition defined by the truth table

p q p ↓ q0 0 10 1 11 0 11 1 0

Table: NAND truth table





Binary Operator NAND (↓ )

AssertionThe operator NAND is negation of AND:

V (p ↓ q) = V (p ∧ q)




The universal quantifier ∀

DefinitionLet p(x) is a statement for x ∈ X . The universal quantificationof P(x) is the statement

"P(x) for all values of x in X”

The notation ∀x P(x) denotes the universal quantification ofP(x). Here ∀ is called the universal quantifier.We read ∀x P(x) as "for all x P(x)" or "for every x P(x)".




The existential quatifier

DefinitionLet p(x) is a statement for x ∈ X . The existential quantificationof P(x) is the proposition

"There is an element x in the domain such that P(x)"

We use the notation ∃x P(x) for existential quantification ofP(x). Here ∃ is called existential quantifier.




Analogously, we have propositions as

∀x∃y ,P(x , y);∃x∃y ,P(x , y) or ∀x∀y ,P(x , y).

In general, we have propositions containing ∀, ∃ and astatement P(x1, ..., xn).




Propositions with quantifiersExamples

Example 1 :

a) The proposition "∀x ∈ IR,x2 + 1 ≥ 0", is true.b) The proposition "∀x ∈ IR,x2 − 1 ≥ 0", is false.c) The proposition "∃x ∈ IR,x2 − 1 ≥ 0", is true.d) The proposition "∀x ∈ IR, ∃y ∈ IR,x + y ≥ 0", is true.e) The proposition "∃y ∈ IR, ∀x ∈ IR,x + y ≥ 0", is false.




Propositions with quantifiersExamples (continue...)

Example 2 :The function f (x) is continuous at the point x0 if

”∀ε > 0,∃δ > 0,∀x, (|x− x0| < δ)→(|f(x)− f(x0)| < ε).”




Propositions with quantifiers

RemarkTo receive the negation of a proposition containingqualifiers ∀, ∃ and statement P(x1, ..., xn), we change ∀ by ∃,change ∃ by ∀ and change P(x1, ..., xn) by P(x1, ..., xn).

In Example 2, using the above note we have that a functionf (x) is not continuous at the point x0 if

”∃ε > 0,∀δ > 0,∃x, (|x− x0| < δ)∧(|f(x)− f(x0)| > ε).”




Exercises

Exercise 1:Determine the truth table of the following compositepropositions and state whether they are tautologies,contradictions or indeterminates?

a) (p ∨ q)→ (p ∧ q)b) (p ∧ q) ∨ (p → q)




Exercises

Exercise 2:Write a logically equivalent statement using NOT, AND and OR.

a) (p → q)b) (p → q)→ r




Exercises

Exercise 3:To express the following proposition by symbolic logics: Forsubset A of IR , "m is called infimum of A, denoted bym = inf(A), if for all x in A, we have x isn’t smaller as m andfor all real number n, if x isn’t smaller as n for all x in Athen m isn’t smaller as n".




Exercises

Exercise 4:Determine the truth table of the following compositepropositions and state whether they are tautologies,contradictions or indeterminate?

a) (p → q)→ p b) p → (q → p)c) (p → q)→ q d) (p → q)→ re) ((p → q) ∧ (q → r)))→ (p → r) f) (p ∧ q)↔ (p∨q)g) (p ↓ q)→ (p ∨ q)




Exercises

Exercise 5:Prove the following assertions

a) V (p∨q) = V (p ↔ q)b) V (p ↑ q) = V (p ∨ q)




Exercises

Exercise 6:Show that the logical operators NOT and OR are sufficient togenerate AND, IFF, IMP, XOR, NOR and NAND.


chapter i - symbolic logic

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