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(CSC 102). Discrete Structures. Lecture 6. Previous Lectures Summary. Different forms of arguments Modus Ponens and Modus Tollens Additional Valid Arguments Valid Argument with False Conclusion Invalid argument with a true Conclusion Converse and Inverse error

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### (CSC 102)

Discrete Structures

Lecture 6

Previous Lectures Summary
• Different forms of arguments
• Modus Ponens and Modus Tollens
• Valid Argument with False Conclusion
• Invalid argument with a true Conclusion
• Converse and Inverse error

Today’s Lecture

• Predicates
• Set Notation
• Universal and Existential Statement
• Translating between formal and informal language
• Universal conditional Statements
• Equivalent Form of Universal and Existential statements
• Implicit Qualification
• Negations of Universal and Existential statements

Predicates

A predicate is a sentence which contains finite number of variables and becomes a statement when specific values are substituted for the variables.

The domain of a predicate variable is the set of all values that may be substituted in place of the variable

Truth Set

If P(x) is a predicate and x has domain D, the truth set of P(x) is the set of all elements of D that make P(x) true when substituted for x. The truth set of P(x) is denoted by

read as “the set of all x in D such that P(x)”.

Notation

For any two predicates P(x) and Q(x), the notation

means that every element in the truth set of P(x) is in the truth set of Q(x). The notation

means that P and Q have identical truth sets.

Consider the predicate:

The truth set of the above predicate is

Cont…

• Example
• Let P(x) = x is a factor of 8, Q(x)= x is a factor of 4 and R(x)= x < 5 and . The domain of x is assumed to be . Use symbols , to indicate true relationships among P(x), Q(x) and R(x).
• The truth set of P(x) is {1,2,4,8}, Q(x) is {1,2,4}. Since every element in the truth set of Q(x) is in the truth set of P(x), So
• The truth Set of R(x) is {1,2,4}, which is identical to the truth set of Q(x). Hence .

Cont…

Let Q(x, y) be the statement

x + y = x − y

where the domain for x and y is the set of all real numbers. Determine the truth value of:

(a) Q(5,−2).

(b) Q(4.7, 0).

(c) Determine the set of all pairs of numbers, x and y, such that Q(x, y) is true.

Solution:

(a) Q(5,−2) says that 5 + (−2) = 5 − (−2), or 3= 7, which is false.

(b) Q(4.7, 0) says that 4.7+ 0 = 4.7 − 0, which is true.

(c) x + y = x − y if and only if x + 2y = x, which is true if and only if

y = 0. Therefore, x can be any real number and y must be zero.

Universal and Existential Statements

• Let Q(x) be a predicate and D the domain of x. A universal statement is of the form “ ”. It is true if and only if Q(x) is true for all x in D and it is false if and only if Q(x) is false for at least one x in D. A value for x for which Q(x) is false is called a counterexample to the universal statement.
• Example: Let D={1,2,3,4,5} and consider the statement Show that this statement is true.
• Solution: Check that is true for each individual x in D.

Cont…..

• Hence is true.
• The technique used in first statement while showing the truthness of the universal statement is called method of exhaustion.

Consider the statement Find the counter example to show that this statement is not true.

• Counter example . Take x=1/2, then x is in R and
• Hence is false.

Existential Quantifier

Let Q(x) be a predicate and D the domain of x. An existential statement is of the form.

such that

It is true if and only if Q(x) is true for at least one x in D. It is false if and only if Q(x) is false for all x in D.

The symbol denotes “there exist” and is called the existential quantifier.

Truth and falsity of Existential statements

Suppose P(x) is the predicate “x < |x|.” Determine the truth value of ∃ x s.t. P(x) where the domain for x is:

(a) the three numbers 1, 2, 3.

(b) the six numbers −2,−1, 0, 1, 2, 3.

Solution

(a) P(1), P(2), and P(3) are all false because in each case x = |x|. Therefore, ∃ x such that P(x) is false for this domain.

(b) If we begin checking the six values of x, we find P(−2) is true. It states that −2 < |−2|. We need to check no further; having one case that makes the predicate true is enough to guarantee that ∃ x s.t. P(x) is true.

Truth and falsity of Existential statements

Consider the statement . Show that this statement is true.

Sol: observe that . Thus is true for at least one integer m . Hence is true.

Let E={5,6,7,8,9,10} and consider the statement

Show that this statement is false.

Sol: the statement is not true for every value of the E. Thus is false.

Translating from formal to informal language

Rewrite the following statements in a variety of equivalent but more informal ways. Do not use the symbol

a)

b)

c)

Solution: a) we can write the statement in many ways like “ All real numbers have non negative squares”,

“No real number has a negative square”,

“ x has a non negative square, for each value of x”.

Cont….

b). Similarly we can translate the second statement in these ways.

“ All real numbers have squares not equal to -1”,

“No real number have square equal to -1”.

c). “There is an integer whose square is equal to itself”, “we can find at least one integer equal to its own square”

Cont…

Write the following statement in English, using the predicates

F(x): “x is a Freshman”

T (x, y): “x is taking y”

where x represents students and y represents courses:

∃x (F(x) ∧ T (x, Discrete Math))

Solution

The statement ∃ x (F(x)∧T (x, Discrete)) says that there is a student x with two properties: x is a freshman and x is taking Discrete. In English, “Some Freshman is taking Discrete Math.”

Translating from informal Language to Formal language

• “Every freshman at the College is taking CSC 102.”
• Solution: There are various ways to answer this question, depending on the domain.
• If we take as our domain all freshmen at the College

and use the predicate T (x) : “x is taking CSC 102”,

then the statement can be written as ∀x, T(x).

• We are making a conditional statement:

“If the student is a freshman, then the student is taking

CSC 101;”

• ∀x, (F(x) → T (x)).
• Notethat we cannot say ∀ x (F(x) ∧ T (x)), because this says that every student is a freshman, which is not
• something we can assume here.

Cont…..

• “Every freshman at the College is taking some Computer Science course.”
• Sol: If we take as our domain for people all freshmen at the College and our domain for courses, all Computer Science courses.
• Then we can use the predicate
• T (x, y): “x is taking y”
• The statement can be written as
• ∀x ∃y T(x, y).

Universal Conditional Statements

A reasonable argument can be made that the most important form of statement in mathematics is the universal conditional statement:

∀ x, if P(x) then Q(x)

Example: “Everyone who visited France stayed in Paris.”

Sol: However, if we take all people as the universe , then we need to introduce the predicate F(x) for “x visited France.” and P(x) is the predicate “x stayed in Paris.” In this case, the proposition can be written as

∀ x, (F(x) → P(x)).

We can write the following statements in a variety of informal ways.

• if then
• Sol:
• if a real number is greater then 2, then the square is greater than 4.
• Whenever a real number is greater then 2, its square is greater than 4.
• The squares of real number, greater than 2, are greater than 4.

Exercise

Rewrite the following statements in the form

∀ ,ifthen .

If a real number is an integer, then it is a rational

number.

All bytes have eight bits.

No fire trucks are green.

Sol: a).

b). ∀ x, if x is a byte, then x has eight bits.

c). ∀ x, if x is a fire truck, then x is not green.

Equivalent Forms of Universal and Existential statements

Observe that the two statements “∀ real numbers x, if x is an integer then x is rational” and “∀ integers x, x is rational” mean the same thing.

In fact, a statement of the form

if P(x) then Q(x).

Can always be rewritten in the form

Can be rewritten as

∀x, if x is in D then Q(x).

Contd.

The following statements are equivalent

∀ polygons P, if P is square, then P is a rectangle.

And

∀ squares P, P is a rectangle

The existential statements

∃ x belongs to U such that P(x) and Q(x).

And

∃ x belongs to D such that Q(x)

Are also equivalent provided D is taken to consist of all elements in U that make P(x) true.

Equivalence form for existential statement

The following statements are equivalent:

∃ a number n such that n is prime and n is even

And

∃ a prime number n such that n is even.

Implicit Quantifications

• Consider “ If a number is an integer, then it is a rational number”
• The clue to indicate its universal quantifications comes from the presence of the indefinite article “a”.
• Existential quantification can also be implicit.
• for instance, “ the number 24 can be written as a sum of sum of two integers”
• “∃ even integers m and n such that 24=m + n.”

Negations of Quantified Statement

The negation of the statement of the form

∀ x in D, Q(x)

is logically equivalent to a statement of the form

∃ x in D such that ~Q(x)

Symbolically:

Note: the negation of universal statement is logically equivalent to existential statement.

Cont….

The negation of the statement of the form

∃ x in D such that Q(x)

is logically equivalent to a statement of the form

∀ x in D, ~Q(x)

Symbolically:

Note: the negation of existential statement is logically equivalent to universal statement.

Examples

Negate “Some integer x is positive and all integers y are negative.”

Solution: Using all integers as the universe for x and y, the statement is ∃ x s.t. (x > 0) ∧ ∀ y, (y < 0). The negation is

~{∃x (x > 0) ∧ ∀ y (y < 0)}≡ ~∃x s.t. (x > 0) ∨ ~∀y, (y < 0): byDe Morgan’s law

≡ ∀ x, ~(x > 0) ∨ ∃ y s.t. ~(y < 0) properties of negation

≡ ∀ x, (x ≤ 0) ∨ ∃ y s.t. (y ≥ 0).

Therefore, the negation is “Every integer x is non positive or there is an integer y that is nonnegative.”

Cont….

Negate “There is a student who came late to class and there is a student who is absent from class.”

Solution: In symbols, if L(x) : “x came late to class” and A(x) : “x is absent from class,” this statement can be written as ∃ x st L(x) ∧ ∃ y st A(y).

Note that we must use a second variable y. By one of De Morgan’s laws the negation can be written as

~(∃ x st L(x)) ∨ ~(∃ y st A(x)) ≡ ∀x, ~L(x) ∨ ∀ y, ~A(x).

In English this is “No student came late to class or no student is absent from class.”

Lecture Summary

• Predicates
• Set Notation
• Universal and Existential Statement
• Translating between formal and informal language
• Universal conditional Statements
• Equivalent Form
• Implicit Qualification
• Negations