QUANTIFIERS TYPES OF QUANTIFIERS QUANTIFIERS WITH RESTRICTED DOMAIN NEGATION AND QUANTIFERS EXPRESS QUANTIFIER IN ENGLISH Nested QUANTIFIER

1. Group # 1
Adnan Yousaf 1001
Humayun Naseer 5054
Aiza Mukhtar 1003
Ummema Ali Sher 5064
DS Presentation
Discrete Structures

2. OUTLINE:
 QUANTIFIERS
 TYPES OF QUANTIFIERS
 QUANTIFIERS WITH RESTRICTED DOMAIN
 NEGATION AND QUANTIFERS
 EXPRESS QUANTIFIER IN ENGLISH
 Nested QUANTIFIER

3. QUANTIFIERS:
In natural languages, a quantifier turns a sentence about something having some property into a
sentence about the number (quantity) of things having the property. Examples of quantifiers in
English are "all", "some", "many", "few", "most", and "no";
examples of quantified sentences are "all people are mortal", "some people are mortal", and
"no people are mortal", they are considered to be true, true, and
Quantifiers are expressions that indicate the scope of the term to which they are attached, here
predicates. A predicate is a property the subject of the statement can have.
For example, in the statement “the sum of x and y is greater than 5”, the predicate ‘Q’ is- sum
is greater than 5,
and the statement can be represented as Q(x, y) where x and y are variables

4. TYPES:
1) UNIVERSAL QUANTIFIER:
The universal quantification of a predicate P(x) is the proposition
“P(x) is true for all values of x in the universe of discourse [the universe of discourse is the set of all things
we wish to talk about; that is, the set of all objects that we can sensibly assign to a variable in a propositional function] ”
We use the notation
∀xP(x)
which can be read “for all x”
2) EXISTENTIAL QUANTIFIER:
The existential quantification of P(x) is the proposition
“There exists an element x in the universe of discourse such that P(x) is true.”

5. Notation:
“There exists x such that P(x)” or “There is at least one x such that P(x)” is written
∃xP(x).
NOTE:
Universal(∀) – The predicate is true for all values of x in the domain.
Existential(∃) – The predicate is true for at least one x in the domain
EXAMPLE 1:
Suppose P(x) is the predicate x + 2 = 2x, and the universe of
discourse for x is the set {1, 2, 3}. Then...
• ∀xP(x) is the proposition “For every x in {1, 2, 3} x + 2 = 2x.” This proposition is false.
• ∃xP(x) is the proposition “There exists x in {1, 2, 3} such that x + 2 = 2x.”
This proposition is true.

6. EXAMPLE 2:
Let P(x) be the predicate “x must take a discrete
mathematics course” and let Q(x) be the predicate “x is a
computer science student”.
The universe of discourse for both P(x) and Q(x) is all UNL
students.
Express the statement “Every computer science student must
take a discrete mathematics course”.
∀x(Q(x) → P(x))
Express the statement “Everybody must take a discrete
mathematics course or be a computer science student”.
∀x(Q(x) ∨ P(x))

7. EXAMPLE 3:
Express the statement “for every x and for every y, x + y > 10”
Let P(x, y) be the statement x + y > 10 where the universe of
discourse for x, y is the set of integers.
Answer:
∀x∀yP(x, y)
Note that we can also use the shorthand
∀x, yP(x, y)
 QUANTIFERS WITH RESTRICTED DOMAIN:
As we know that quantifiers are meaningless if the variables they bind do not have a
domain. The following abbreviated notation is used to restrict the domain of the
variables-

8. What is the truth value of ∀ x <0 -- (x^2 =1)
Domain is all real numbers.
Solution:
Find just 1 counter example to make ∀ x <0 -- (x^2 =1) FALSE
 It can be written as
∀ ( x <0 -- x^2 =1)
( -2 <0 ___> (-2)^ 2 =1)
( -2<0 ____> 4=1)
T_____> F =F
So , ∀ x <0 -- (x^2 =1) is FALSE
Q 2:
What is truth value of ∃ x p(x) “x^2 >0 “ domain “ positive integer not exceeding 4? {1,2,3,4}
Answer:
P(x) : x ^2 > 10
P(1) = (1)^2 > 10 FALSE
P(2) =2^2 > 10 FALSE
P(3) = 3^2 > 10 FALSE
P(4) =4^2 > 10 TRUE
Here ∃ x p(x) “x^2 >0 is true

9.  NEGATION AND QUANTIFIERS:
EXAMPLE :
All Dogs Barks ______
∀ dogs , d barks _____>
¬ (∀ dogs , d barks) ____>
∃ dogs , ¬ d barks ______>
∃ dogs , d does not bark ______>
“Some Dogs does not bark”
Example :
Problem:
• Express the statement “Not everybody can ride a bike” as a
logical expression.
Solution:
• Let P(x)=“x can ride a bike.”
• The statement “everybody can ride a bike,” can be
expressed as ∀x P(x).
• We want the negation of this, which is ¬∀x P(x).
• Another way to say this is “There is somebody that cannot
ride a bike,” which can be expressed as ∃x ¬P(x).

10. EXAMPLE :
• Express the statement “Nobody can fly.” as a logical
expression.
Solution:
• Let P(x)=“x can fly.”
• The statement “somebody can fly,” can be expressed as
∃x P(x).
• We want the negation of this, which is ¬∃x P(x).
• Another way to say this is “Everybody can not fly,” which
can be expressed as ∀x ¬P(x).

11.  EXPRESS QUANTIFIER IN ENGLISH STATEMENTS:
QUESTION 1:
∀x P(x). mean if p(x) “ x is perfect “
Domain is “of people in your locality”
ANSWER:
AS we know that ∀x P(x).
For all x P(x).
For every x P(x).
For each x P(x).
All of x P(x).
For any x P(x).
It can be written as :
1. All people in your locality are perfect .
2. Every people in your locality are perfect
3. Each people in your locality are perfect

12. QUESTION 2:
There is a student who has taken more than 21 credit hour in a semester and received all A grade
ANSWER:
Domain: “All student”
P(x) = x student who has taken more than 21 credit hour in a semester
And
S(x) = received all A grade
So;
∃x[ P(x) ^ s(x)]
QUESTION 3:
Let N(x) be the statement “ x has visited north korea “ where the domain is :
consist of the students in your school “ .. Express ∀x N(x) in English ?
ANSWER:
1. For every x in the domain of the students in your school , x has visited North korea.
2. All students in your school has visited north korea .

13.  NESTED QUANTIFIER:
Two quantifiers are nested if one is within the
scope of the other.
Example:
∀x ∃y (x + y = 0)
∀x Q(x)
Q(x) is ∃y P(x,y)
P(x,y) is (x + y = 0)
EXAMPLE:
Translate the following statement into English.
x y (x + y = y + x)
Domain: real numbers

14. Domain: real numbers
Solution:
For all real numbers x and y, x + y = y + x
Example-1:
∀x ∃y (x+y=5)
Here ‘∃’ (read as-there exists) and ‘∀’ (read as-for all) are quantifiers
for variables x and y.
The statement can be represented as-
∀x Q(x)
Q(x) is ∃y P(x, y) Q(x)-the predicate is a function of only x because the
quantifier applies only to variable x.
P(x, y) is (x + y = 5)

15. EXAMPLE :
Let x and y be the real numbers and p(x,y) denotes “x+y=0”
Find the truth value of :
a) ∀x ∀y p(x,y)
b) ∀x ∃y p(x,y)
c) ∃y ∀x p(x,y)
d) ∃x∃y p(x,y)
SOLUTION:
Domain : All real numbers
a) ∀x ∀y p(x,y) ≡ ∀x ∀y (x=y=0)
“for all real numbers x and y , x+y=0”
It is not TRUE
e.g 2+1 not equal to 0
so it is FALSE

16. b) ∀x ∃y p(x,y)
“for ever real numbers x ,there exist a real number y such that x+y=0”
IT IS TRUE:
For example:
X=1 ,-1 , 1/2
Y =-1 ,1 ,-1/2
STEPS:
Consider all different value of x.
Find just 1 value of y for each of x such that p(x,y) becomes true.
Value of y depend on the value of x

17. c) ∃y ∀x p(x,y)
“there exist some real number y such that for every real number x ,x+y=0”
IT IS not TRUE
Prove:
It is asking us to find a real number y for which p(x,y) becomes true by plugging in every real
number x.
As we know that :
P(x,y) =x+y=0
First take sum real number y=1
P(x,1) =x+1 =0
Now , plug in all real numbers of x
P(1/2,1) is false , p(1,1) is false and soo on…
No matter what y you choose , p(x,y) is always false for all real numbers
SO IT IS FALSE

18. d) ∃x∃y p(x,y)
“there exist some real numbes x and y such that x+y=0:”
Surely : There exist some combination of real numbers x and y exist for which
x +y=0
For example :
Take x=1 and y=-1
1+(-1) =0 is True
So this is true