DISCRETE
MATHEMATICS
AND
ITS APPLICATIONS
Book: Discrete Mathematics and Its Applications
Author: Kenneth H. Rosen
Sixth Edition
McGraw-Hill International Edition
Chapter 1
The Foundations:
Logic and Proofs
Objectives
Explain what makes up a correct
mathematical argument
 Introduce tools to construct arguments

Contents
1.1-Propositional Logic
1.2-Propositonal Equivalences
1.3-Predicates and Quantifiers
1.4-Nested Quantifiers
1.5-Rules of Inference
1.1- Propositional
Logic
1.1.1- Definitions and Truth Table
1.1.2- Precedence of Logical Operators
1.1.1- Definitions and Truth
Table
Proposition is a declarative sentence that is either
true or false but not both.
 Proposition is a sentence that declares a fact.
 Examples:
• Ha Noi is the capital of Vietnam.
TRUE
• Toronto is the capital of Canada.
FALSE

1.1.1- Definitions and Truth
Table

Proposition is a declarative sentence that is either
true or false but not both.
• 2 + 2 = 3.
FALSE
• What time is it?
A question.
• X +Y = Z
The sentence cannot be shown to be either true of
false.
1.1.1- Definitions and Truth
Table
Example
1.1.1- Definitions…

Truth table
2+2=4
p
True/ T / 1
False / F / 0
1.1.1- Definitions…

Let p be a proposition. The negation of p, denoted
by p (also denoted by p ) , is the statement “It is
not the case that p”.

The proposition p is read “not p”. The truth
vale of the negation of p, p , is the opposite of
the truth value of p.
p
1
0
0
1
1.1.1- Definitions…
Negation of proposition p is the statement “ It is not
case that p”.
 Notation: p (or p )

p
1
0
0
1
1.1.1- Definitions…
Find the negation of the proposition
“Sarah’s PC runs Windows.”
and express this in simple English.

Solution:
The negation is:
“It is not the case that Sarah’s PC runs Windows.”


This negation can be more simply expressed by:
“Sarah’s PC does not run Windows.”
1.1.1- Definitions…
Example
1.1.1- Definitions…
Let p and q be proposition. The conjunction of p and
q, denoted by p^q, is the proposition “ p and q”.
 The conjunction p^q is true when both p and q are
true and false otherwise.

p
q
p^q
F
F
T
T
F
T
F
T
F
F
F
T
1.1.1- Definitions…

Conjunction of propositions p and q is the
proposition “ p and q” and denoted by p^q.
p
q
p^q
0
0
1
1
0
1
0
1
0
0
0
1
1.1.1- Definitions…
1.1.1- Definitions…

Disjunction of propositions p and q is the proposition
“ p or q” and denoted by p v q
p
q
pvq
0
0
1
1
0
1
0
1
0
1
1
1
1.1.1- Definitions…

p v q (disjunction of p and q): the proposition “p
or q,” which is true if and only if at least one of p
and q is __T_ .
1.1.1- Definitions…
Exclusive-or (XOR) of propositions p and q, denoted
by p  q
 Is the proposition that is true when exactly one
of p and q is true and is false otherwise.

q
p
q
pq
0
0
1
1
0
1
0
1
0
1
1
0
1.1.1- Definitions…
Implication: p → q (p implies q)
 p: hypothesis / antecedent / premise
 q: conclusion / consequence
 p → q can be expressed as:

1.1.1- Definitions…




-
-
Implication: p → q (p implies q)
p: hypothesis / antecedent / premise
q: conclusion / consequence
p → q can be expressed as:
q if p
If p, then q
p is sufficient condition for q
q is necessary condition for p
p
0
0
1
1
q
0
1
0
1
p→q
1
1
0
1
“If 1 + 1 = 3, then dogs can fly”
 TRUE
(p  q)
p=0, q=0 ,
so (pq) is true.
1.1.1- Definitions…


Biconditional statement p q is the proposition “ p if and
only if q”
p → q (p only if q) and p q (p if q)
p q
p→q
q→p
(p→q) ^ (q→p)
p↔q
0 0
1
1
1
1
0 1
1
0
0
0
1 0
0
1
0
0
1 1
1
1
1
1
1.1.2- Precedence of Logical Operators
(1) Parentheses from inner to outer
1.1.2- Precedence of Logical Operators
1.2- Propositional
Equivalences
1.2.1- Tautology and Contradiction
1.2.2- Logical Equivalences
1.2.3- De Morgan’s Laws
1.2.1- Tautology and Contradiction
Tautology is a proposition that is always true
 Contradiction is a proposition that is always
false
 When p ↔ q is tautology, we say “p and q are
called logically equivalence”. Notation: p ≡ q

Example 3 p.23

Show that p  q and ¬p v q are logically
equivalent.
Example

Which proposition is logically equivalent to
1.2.2- Logical Equivalences…
Equivalence
Name
p^T≡p
pvT≡ T
pvF≡p
p^F ≡ F
Identity laws
Domination Laws
pvp≡ p
¬(¬p) ≡ p
p^p ≡ p
Idempotent Laws
Double Negation Laws
pvq≡qvp
p^q ≡q^p
(p v q) v r ≡ p v (q v r)
(p ^ q) ^ r ≡p^(q^r)
Commutative Laws
Associative Laws
pv (q^r) ≡ (pvq) ^ (pvr)
p^ (qvr) ≡ (p^q) v (p^r)
Distributive Laws
1.2.2- Logical Equivalences…
Example: Given
x, y 
 2 x  y  3 and 4 x
p

2
, then:
 e  1   2 x  y  3
y
q

p
1.2.2- Logical Equivalences…
Equivalence
¬ (p^q) ≡ ¬pv¬q
pv (p^q)≡ p
pv¬p ≡ T
Name
¬(pvq) ≡ ¬p^¬q De Morgan Laws
p^(pvq)≡ p Absorption Laws
p^¬p≡ F
Negation Laws
1.2.2- Logical Equivalences…
Example:
C  " 3a  8  1"
 C  "3a  8  0 and
3a  8  1"
Using De Morgan Laws for C, then:
C  " 3a  8  0 or 3a  8  1"
1.2.2- Logical Equivalences…
Example:
The negation of the statement:
“The summer in Maine is hot and sunny”
Is the statement:
“The summer in Maine is NOT hot OR NOT sunny.”
1.2.2- Logical Equivalences…
Example:
The negation of the statement:
“Tom has a cellphone AND he has a laptop computer”
Is the statement:
“Tom does NOT have a cellphone OR he does NOT hav
a laptop computer.”
1.2.2- Logical Equivalences…
Equivalences
Equivalences
p→q ≡ ¬pvq
p→q ≡ ¬q → ¬p
p↔q ≡ (p→q) ^ (q→p)
p↔q ≡ ¬p ↔ ¬q
pvq ≡ ¬ p → q
p^q ≡ ¬ (p → ¬q)
¬(p→q) ≡ p^¬q
p↔q ≡ (p ^ q) v (¬p ^ ¬q)
¬ p↔q ≡ p↔ ¬q
(p→q) ^(p→r) ≡ p → (q^r)
(p→r) ^ (q→r) ≡ (pvq) → r
(p→q) v (p→r) ≡ p→ (qvr)
(p→r) v (q→r) ≡ (p^q) → r
1.2.2- Logical Equivalences…
Example:
Show that E   p   p  q    q is a tautology.
Solution:
E   p   p  q    q
  p  p    p  q    q
 0   p  q    q
  p  q  q
 pq q
 p 1
 1.
1.3- Predicates and Quantifiers
Introduction
 Predicates
 Quantifiers

1.3.1- Introduction

A type of logic used to express the meaning of a wide
range of statements in mathematics and computer
science in ways that permit us to reason and
explore relationships between objects.
1.3.2- Predicates – vị từ
Propositional
Logic



Predicate Logic



Q =“ 5 is a prime number.”
Q ≡ True
X>0
P(X)=“X is a prime number” ,
called propositional function at X.
P(2)=”2 is a prime number” ≡True
P(4)=“4 is a prime number” ≡False
1.3.2- Predicates – vị từ
Q(X1,X2,…,Xn) , n-place/ n- predicate
 Example: “x=y+3”  Q(x,y)

Q(1,2) ≡ “1=2+3” ≡ false
Q(5,2) ≡ “5=2+3” ≡ true
1.3.2- Predicates – vị từ
1.3.2- Predicates…

Predicates are pre-conditions and postconditions of a program.
Pre-condition (P(…)) : condition describes
valid input.
 If x>0 then x:=x+1
– Predicate: “x>0”  P(x)
– Precondition: P(x)
– Postcondition: Q(x)
Post-condition (Q(…)) : condition
describes valid output of the codes.
Show the verification that a program
always produces the desired output:
P(…) is true
Executing Step 1.
Executing Step 2.
…..
Q(…) is true
1.3.2- Predicates…
State the value of x after the statement:
if x >1 then x:=1 is executed, where P(x) is the
statement “x >1”, if the value of x when this
statement is reached is
a) x = 5.
b)x = 0.
c) x = 1.
d)x = 3.
1.3.3- Quantifiers – Lượng từ






The words in natural language: all, some, many, none,
few, ….are used in quantifications.
Predicate Calculus : area of logic that deals with
predicates and quantifiers.
The universal quantification of P(x) is the statement
“P(x) for all values of x in the domain”. Notation : xP(x)
The existential quantification of P(x) is the statement
“There exists an element x in the domain such that P(x)”.
Notation : xP(x)
Uniqueness quantifier: !x P(x) or 1xP(x)
xP(x) v Q(y) :
 x is a bound variable
 y is a free variable
1.3.5- Precedence of Quantifiers
Quantifiers have higher precedence than
all logical operators from propositional
calculus.
 xP(x) v Q(x)  (xP(x)) v Q(x)
  has higher precedence. So,  affects on
P(x) only.

1.3.6- Logical Equivalences
Involving Quantifiers
Statements involving predicates and quantifiers are
logically equivalent if and only if they have the same
truth value no matter which predicates are substituted
into the statements and which domain of discourse is
used for the variables in these propositional functions.

x (P(x) ^ Q(x)) ≡ xP(x) ^ xQ(x)
– Proof: page 39
Expression
Equivalence
Expression
Negation
¬xP(x)
x ¬P(x)
xP(x)
x ¬P(x)
¬ xP(x)
x ¬P(x)
xP(x)
x ¬P(x)
1.3.7- Translating
For every student in the class has studied
calculus
 For every student in the class, that student
has studied calculus
 For every student x in the class, x has
studied calculus
 x (S(x) → C(x))

1.3.7- Translating
1.3.7- Translating
Negating nested quantifiers
¬ xy(xy=1) ≡ x ¬y (xy=1) // De Morgan laws
≡ (x) (y) ¬(xy=1)
≡ (x) (y) (xy  1)
Negating nested quantifiers
Negating nested quantifiers
Negating nested quantifiers
Negating nested quantifiers
1.5- Rules of Inference
Definitions
 Rules of Inferences

1.5.1- Definitions
Proposition 1 Hypothesis / antecedent / premise.
 Proposition 2
Arguments 2,3,4 are
 Proposition 3
premises of argument 5
 Proposition 4
 Proposition 5
Arguments
Propositional Equivalences
 ………
 Conclusion

1.5.2- Rules Inferences
Rule
pq
p
q
pq
q
p
Tautology
[(p→q) ^p] → q
(If you work hard then you will
pass the examination)
AND (You work hard)
you will pass the examination
[ (p → q) ^¬q ] → ¬p
If she is good at learning she will
get a prize
AND she did not get a prize
She is not good at learning
Name
Modus ponen
Modus tollen
1.5.2- Rules Inferences
Rule
Tautology
p  q [(p →q) ^(q →r)] →(p→r)
If the prime interest rate goes up then
q  r the stock prices go down.
 p  r AND If the stock prices go down then
most people are unhappy.
If the prime interest rate goes up then
most people are unhappy.
Name
Hypothetical
syllogism
Rules Inferences…
Rule
Tautology
Name
p  q [(pvq) ^¬p] → q
p
q
Power puts off or the lamp is
malfunctional
AND Power does not put off
Disjunctive
syllogism
the lamp is malfunctional
p
pq
p →(pvq)
It is below freezing now
Addition
It is below freezing now or raining now
pq
p
(p^q) →p
It is below freezing now and raining now
It is below freezing now
Simplication
Rules Inferences…
Rule
p
q
p^q
pvq
¬pvr
qvr
Tautology
Name
[(p) ^(q)) → (p^q)
Conjunction
[(pvq) ^(¬pvr)] →(qvr)
Jasmin is skiing OR it is not snowing
AND It is not snowing OR Bart is
playing hockey
Jasmin is skiing OR Bart is playing
hockey
Resolution
Rules Inferences…
Rule
p
q
p^q
pvq
¬pvr
qvr
Tautology
Name
[(p) ^(q)) → (p^q)
Conjunction
[(pvq) ^(¬pvr)] →(qvr)
Jasmin is skiing OR it is not snowing
AND It is not snowing OR Bart is
playing hockey
Jasmin is skiing OR Bart is playing
hockey
Resolution
1.5.3- Fallacies

If you do every problem in this book then you will
learn discrete mathematic
You learned mathematic
(p → q) ^q
=(¬ p v q) ^ q
(absorption law)
=q
 No information for p
p can be true or false  You may learn discrete
mathematic but you might do some problems only.
Fallacies…


(p → q)^q  p is not a tautology
( it is false when p = 0, q = 1)
(p  q)^¬p  ¬q is not a tautology
(it is false when p = 0, q = 1)
1.5.4- Rules of Inference for
Quantified Statements
Rule
Name
xP(x)
P(c)
Universal Instantiation
P(c) for arbitrary c
xP(x)
Universal generalization
xP(x)
P(c) for some element c
Existential instantiation
P(c) for some element c
xP(x)
Existential generalization
Rules of Inference for Quantified Statements…
“All student are in this class had taken the
course PFC”
 “Harry is in this class”
 “Had Harry taken PFC?”
 x(P(x) → Q(x))
 P(Harry) → Q(Harry)
 P(Harry)
 Q(Harry) // conclusion

Rules of Inference for Quantified Statements…
“All student are in this class had taken the
course PFC”
 “Harry is in this class”
 “Harry had taken PFC”. CORRECT
Premise
 x(P(x) → Q(x))
Universal Instantiation
 P(Harry) → Q(Harry)
Modus ponens
 P(Harry)
 Q(Harry) // conclusion

Rules Inferences…
a) Everyone who eats granola every day is healthy.
Linda is not healthy.
Therefore, Linda does not eat granola every day.
P(x) = “x eats granola every day ”.
Q(x) = “x is healthy”.
Step
Reason
x ( P  x   Q  x ) (1)
Premise
( P  Linda   Q  Linda )
Universal inst
Linda does not eat granola every
Q  Linda 
day.
P  Linda 
Premise
Modus tollens
Rules Inferences…
a) Everyone who eats granola every day is healthy.
Linda is not healthy. Therefore, Linda does not eat
granola every day. Correct.
Step
P(x) = “x eats granola every day ”.
x ( P  x   Q  x ) (1)
Q(x) = “x is healthy”.
Q  Linda 
Linda does not eat granola every day
( P  Linda   Q  Linda )
Reason
Premise
Premise
Universal
instantiation
from (1)
Rules Inferences…
b) All parrots like fruit.
My pet bird is not a parrot.
Therefore, my pet bird does not like fruit. Incorrect.
P(x) = “x is a parrot”.
Q(x) = “x like fruit”.
Step
x ( P  x   Q  x )
( P  bird   Q  bird )
P  bird 
Reason
Premise
Universal instantiation
Premise
Rules Inferences…
b) All parrots like fruit. My pet bird is not a parrot.
Therefore, my pet bird does not like fruit.
P(x) = “x is a parrot”.
Q(x) = “x like fruit”.
Step
x ( P  x   Q  x )
P  bird 
( P  bird   Q  bird )
Reason
Premise
Premise
Rules Inferences…
c) If Tom knows French, Tom is smart. But Tom does
not know French. So, he is not smart. NO VALID
Rules Inferences…
d) Ceci can not go fishing if she does not have a bike.
Last week, Ceci went fishing with her friends.
Therefore, she has got a bike.
Summary
Propositional Logic
 Propositional Equivalences
 Predicates and Quantifiers
 Nested Quantifiers
 Rules and Inference

Review


F
T
T

T

F
F
Review
Review
Review
Review
F
F
Review
p
q
Not p
Review
P(x)
Q(x)
P( Mrs.Bean)  Q( M
P( Mrs.Bean)
THANK YOU