SUBTOPIC 3 : QUANTIFIERS

The statement

P : n is odd integer.

A proposition is a statement that is either true or false. The statement p is not proposition

because whether p is true or false depends on the value of n. For example, p is true if n = 104 and

false if n = 8. Since, most of the statements in mathematics and computer a science use variable,

we must extend the system of logic to include such statements.

Example 1: Let P(n) be the statement

n is an odd integer

Then P is a propositional function with the domain of discourse Z+¿¿ since for each n ∈

Z+¿¿, P(n) is a proposition [for each n ∈ Z+¿¿, P(n) is true or false but not both]. For example, if n

= 1, we obtain the proposition.

P (1): 1 is an odd integer

(Which is true) If n = 2, we obtain the proposition/

P (2): 2 is an odd integer

(Which is false)

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1.Quantifiers

Definition:

Let P (x) be a statement involving the variable x and let D be a set. We call P a proportional function or predicate (with respect to D ) , if for each x ∈ D , P (x) is a proposition. We call D the domain of discourse of P.

A propositional function P, by itself, is neither true nor false. However, for each x is

domain of discourse, P (x) is a proposition and is, therefore, either true or false. We can think of

propositional function as defining a class of propositions, one for each element in the domain of

discourse. For example, if P is a propositional function with domain of discourse Z+¿¿, we obtain

the class of propositions.

P (1), P (2), …..

Each of P (1), P (2), …. Is either true or false.

∀x, P (x)

Similar expressions:

- For each…- For every…- For any…

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2. Universal Quantification

Definition:

Let P be a propositional function with the domain of discourse D. The universal quantification of P (x) is the statement. “For all values of x, P is true.”

3. Counterexample

A counterexample is an example chosen to show that a universal statement is FALSE.

To verify:

- ∀x, P (x) is true- ∀x, P (x) is false

Example 1:

Consider the universally quantified statement.

∀x (x2 ≥ 0)

The domain of discourse is R. The statement is true because for every real number x, it is true

that the square of x is positive or zero.

According the definition, the universally quantified statement.

∀x, P (x)

Is false for at least one x in the domain of discourse that makes P (x) is false. A value x in the

domain of discourse that makes P (x) false is called a counterexample to the statement.

Example 2:

Consider the universally quantified statement.

∀x (x2-1 ≥ 0)

The domain of discourse is R. The statement is false since, if x = 1, the proposition

12-1 > 0

Is false. The value1 is counterexample of the statement.

∀x (x2-1 ≥ 0)

Although there are values of x that make the propositional function true, the counterexample

provide show that the universally quantified statement is false.

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4. Existential Quantification

Let P be a proportional function with the domain of discourse D. The existential quantification of P (x) is the statement. “There exists a value of x for which P (x) is true.

∃x, P(x)

Similar expressions:

Example 1:

Consider the existentially quantified statement.

∃x (x

x2+1=2

5¿

The domain of discourse is R. the statement is true because it is possible to find at least one real

number x for which the proposition

x

x2+1=2

5

Is true. For example, if x = 2, we obtain the true proposition.

2

22+1=2

5

Is not the case that every value of x results in a true proposition. For example, if x = 1, the

proposition

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4. Existential Quantification

Let P be a proportional function with the domain of discourse D. The existential quantification of P (x) is the statement. “There exists a value of x for which P (x) is true.

∃x, P(x)

Similar expressions:

1

12+1=2

5

Is false.

According to definition, the existentially quantified statement

∃x, P(x)

Is false for every x in the domain of discourse, the proposition P (x) is false.

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5. De Morgan’s Law for Logic

Theorem:

(∀x, P (x)) ≡ (∃x, (P(x))

(∃x, (P(x)) ≡ (∀x, P (x))

The statement

“The sum of any two positive real numbers is positive”.

Example 1: Let P(x) be the statement

1

x2+1>1

We show that

∃x, P(x)

Is false by verifying that

∀x, ⌐ P (x)

Is true.

The technique can be justified by appealing to theorem. After we prove that proposition

is true, we may negate and conclude that is false. By theorem,

∃x, ⌐⌐P(x)

Or equivalently

∃x, P(x)

Is also false.

1. EXERCISE.

a) let P (x) be the propositional function “x ≥ x2.” Tell whether each proposition is

true or false. The domain of discourse is R

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5. De Morgan’s Law for Logic

Theorem:

(∀x, P (x)) ≡ (∃x, (P(x))

(∃x, (P(x)) ≡ (∀x, P (x))

The statement

“The sum of any two positive real numbers is positive”.

i. P (1)

ii. ⌐∃x P(x)

b) Suppose that the domain of discourse of the propositional function P is {1, 2, 3,

4}. Rewrite each propositional function using only negation, disjunction and

conjunction.

i. ∀x P (x)

c) Determine the truth value, the domain discourse R x R, justify the answer.

i. ∀x ∀y (x2< y + 1)

d) Assume that ∀x ∃y P(x, y) is false and that the domain of discourse is nonempty.

It must also be false? Prove the answer.

i. ∀x ∀y P (x,y)

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