03_ECE MATH 311_Validity of Arguments and Logical Equivalence

DISCRETE MATHEMATICS ECE – MATH 311

TOPIC 3: VALIDITY OF ARGUMENTS & LOGICAL EQUIVALENCE

By: Edison A. Roxas, MSECE

Validity of Arguments & Logical Equivalence 2

OBJECTIVES

At the end of the topic, the students should be able to:

1. Recall Boolean Algebra Postulates and Theorems.

2. Define and distinguish tautology from contradiction;

3. Examine the truth table and its implications to some premises;

4. Examine Logic Puzzle;

5. Determine if conclusion is valid for some premises;

6. Enumerate different logical equivalence; and

7. Establish and apply some of the logical equivalences.

earoxas @ UST 2013

Table 3.1: Boolean Algebra

Validity of Arguments & Logical Equivalence 3

IDENTITY X + 0 = X X . 1 = X

COMPLEMENT X + X’ = 1 X . X’ = 0

IDEMPOTENT X + X = X X . X = X

DOMINATION X + 1 = 1 X . 0 = 0

INVOLUTION (X’)’ = X

COMMUTATIVE X + Y = Y + X XY = YX

ASSOCIATIVE X + (Y + Z) = (X + Y) + Z X(YZ) = (XY)Z

DISTRIBUTIVE X (Y + Z) = XY + XZ X + YZ = (X + Y) (X + Z)

DE MORGAN (X + Y)’ = X’Y’ (XY)’ = X’ + Y’

ABSORPTION X + XY = X X (X + Y) = X

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EXAMPLE 3.1:

Simplify using Boolean Algebra the examples in Example 1.5 and compare results.

a. F = xy + xy’

b. F = (x+y)(x+y’)

c. F = xyz + x’y + xyz’

earoxas @ UST 2013 Validity of Arguments & Logical Equivalence 4

TAUTOLOGY & CONTRADICTION

It is an important class of compound propositions that consists of those that are always TRUE for all possible combinations of p and q (or p, q and r). This is called tautology.

A compound proposition that is always FALSE is called contradiction or absurdity.

A compound proposition p is a contradiction if and only if ~p is a tautology.

A compound proposition is a contingent if it is neither a tautology or contradiction.

Validity of Arguments & Logical Equivalence 5 earoxas @ UST 2013

EXAMPLE 3.2: Determine whether the following are tautology

or contradiction.

1. p → p 2. ~ p ˅ p 3. ~[A → (~A →B)] 4. [p ˄ (p → q)] → q 5. (p ˄ ~q) ˄ (~p ˅ q) 6. (p → q) ↔ (~q → ~p) 7. [(p → q) ˄ p] → q 8. A → [ ~ A ↔ (A → B)]


LOGIC PUZZLE

EXAMPLE 3.3:

Tony and his girlfriend Pepper was in a room together with the other members of the organization SHIELD. He was with Bruce; Natasha; and their leader Nick. Suddenly a momentary power interruptions was experienced; when the power was restored they found Nick murdered. An inquiry was held; and these are their statements:


EXAMPLE 3.3 (cont’d):

Bruce: I am innocent ; Natasha was talking to Nick when the power was interrupted.

Natasha: I am innocent; I was not talking to Nick when the power was out.

Tony: I am innocent; Pepper committed the murder. Pepper: I am innocent; one of the men committed the

murder. Four of these eight statements are TRUE and four are

FALSE. Assuming only one person committed the murder, who did it?


VALIDITY OF ARGUMENTS

Complicated statements are analyze using connectives and form a simpler statement.

This simpler statement are then tested for truth or falsity using a truth table.

If the final column of the truth table is a TAUTOLOGY, then argument is considered valid.


EXAMPLE 3.4:

1. Show that the value R = (A → (A ˄ B)) is a contingent.

2. If A → B is false, determine the truth table of (~A) ˅ (A↔B)?


EXAMPLE 3.5:

- Prove the validation of the argument given as:

The competition will start on time implies that all contestants are present; if and only if all the contestants are present or the competition will not start on time.


EXAMPLE 3.6: Prove that the following is a valid argument.

p: Claire studies.

q: Claire plays volleyball.

r: Claire passes the board examination.

===

P1: If Claire studies, then she will pass the board examination.

P2: If Claire doesn’t play volleyball, then she’ll study.

P3: Claire failed the board examination.

===

Prove that (P1^P2^P3) q is valid.


LOGICAL EQUIVALENCE

Compound values that have the same truth values in all possible cases are called logically equivalent.

The compound proposition p and q are called logically equivalent if p↔q is a tautology .

The notation p ≡ q denotes that p and q are logically equivalent.

The symbol “≡” is not a compound proposition but rather is the statement that p↔q is a tautology .


Table 3.2: LOGICAL EQUIVALENCE

1. Double Negation: ~~p ≡ p 2. Commutative Laws: a. (pνq) ≡ (qνp)

b. (pΛq) ≡ (qΛp) c. (p↔q) ≡ (q↔p)

3. Associative Laws: a. [(pνq)νr] ≡ [pν(qνr)] b. [(pΛq)Λr] ≡ [pΛ(qΛr)]

4. Distributive Laws: a. [pν(q^r)] ≡ [(pνq)^(pνr)] b. [p^(qνr)] ≡ [(p^q)ν(p^r)]

5. Idempotent Laws: a. (pνp) ≡ p b. (p^p) ≡ p

earoxas @ UST 2013 14 Validity of Arguments & Logical Equivalence


6. Identity Laws: a. (pνF) ≡ p b. (p^T) ≡ p

7. Domination Laws / Inverse Laws: a. (pνT) ≡ T b. (p^F) ≡ F

8. Complement Laws / Negation Laws: a. (pν~p) ≡ T b. (p^~p) ≡ F



9. DeMorgan Laws: a. ~(pνq) ≡ (~p^~q) b. ~(p^q) ≡ (~pν~q)

c. (pνq) ≡ ~(~p^~q) d. (p^q) ≡ ~(~pν~q)

10. Contrapositive: (p→q) ≡ (~q→~p)

11. Implication: a. (p→q) ≡ (~pνq) b. (p→q) ≡ ~(p^~q)



12. Absorption Laws:

a. pν(p^q) ≡ p b. p^(pνq) ≡ p

13. Equivalence:

a. (pνq) ≡ (~p→q) b. (p^q) ≡ ~(p→~q)


EXAMPLE 3.7:

1. Prove that (AνB)^~(~A^B) ≡ A

2. Show that ~(p+(~pxq)) and (~px~q) are logically equivalent.

3. Show that (p^q) (pνq) is a tautology. Use the truth table.


03_ECE MATH 311_Validity of Arguments and Logical Equivalence

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