Digital Circuits

Boolean Algebra and Logic Gate

Chapter 2

Boolean Algebra and Logic Gate

Dr. Marwa Fayez Fahmy

Lecture 4

Definition of Boolean Algebra

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We need to define algebra for binary values

Developed by George Boole in 1854

Huntington postulates for Boolean algebra (1904):

B = {0, 1} and two binary operations, + and．

Closure with respect to operator + and operator ·

Identity element 0 for operator + and 1 for operator ·

Commutativity with respect to + and · x+y = y+x, x·y = y·x

Distributivity of · over +, and + over ·

x·(y+z) = (x·y)+(x·z) and x+(y·z) = (x+y)·(x+z)

Complement for every element x is x’ with x+x’=1, x·x’=0

There are at least two elements x,yB such that xy

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Boolean Algebra

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Terminology:

Literal: A variable or its complement

Product term: literals connected by •

Sum term: literals connected by +

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Postulates (فرضية)مسلمه of Two-

Valued Boolean Algebra B = {0, 1} and two binary operations, + and．

The rules of operations: AND、OR and NOT.

1. Closure (+ and‧)

2. The identity elements (1) +: 0

(2)．: 1

x y x．y

0 0 0

0 1 0

1 0 0

1 1 1

x y x+y

0 0 0

0 1 1

1 0 1

1 1 1

x x'

0 1

1 0

AND OR NOT

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Postulates of Two-Valued Boolean Algebra

3. The commutative laws تبادلية

4. The distributive laws

x·(y+z) = (x·y)+(x·z) and

x+(y·z) = (x+y)·(x+z)

x y z y+z x．(y+z) x．y x．z (x．y)+(x．z)

0 0 0 0 0 0 0 0

0 0 1 1 0 0 0 0

0 1 0 1 0 0 0 0

0 1 1 1 0 0 0 0

1 0 0 0 0 0 0 0

1 0 1 1 1 0 1 1

1 1 0 1 1 1 0 1

1 1 1 1 1 1 1 1 5

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Postulates of Two-Valued Boolean Algebra

5. Complement

x+x'=1 → 0+0'=0+1=1; 1+1'=1+0=1

x．x'=0 → 0．0'=0．1=0; 1．1'=1．0=0

Note

A set of two elements

+ : OR operation; ．: AND operation

A complement operator: NOT operation

Binary logic is a two-valued Boolean algebra

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Dualityازدواجية

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The principle of duality is an important concept. This says that if an expression is valid in Boolean algebra, the dual of that expression is also valid.

To form the dual of an expression, replace all + operators with . operators, all . operators with + operators, all ones with zeros, and all zeros with ones.

Form the dual of the expression

a + (bc) = (a + b)(a + c)

Following the replacement rules…

a(b + c) = ab + ac

Take care not to alter the location of the parentheses if they are present.

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Basic Theorems

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Boolean Theorems

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Huntington’s postulates define some rules

Need more rules to modify algebraic expressions

Theorems that are derived from postulates

What is a theorem?

A formula or statement that is derived from postulates

(or other proven theorems)

Post. 1: closure Post. 2: (a) x+0=x, (b) x·1=x Post. 3: (a) x+y=y+x, (b) x·y=y·x Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x’=1, (b) x·x’=0

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Basic theorems of Boolean algebra

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Theorem 1: Idempotent التتاثر

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denoting an element of a set which is unchanged in value when multiplied or otherwise operated on by itself.

Theorem 1(a): (x + x) = x

Theorem 1(b): (x.x)= x

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Proof of x+x=x

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We can only use Huntington postulates:

Show that x+x=x.

x+x = (x+x)·1 by 2(b)

= (x+x)(x+x’) by 5(a)

= x+xx’ by 4(b)

= x+0 by 5(b)

= x by 2(a)

Q.E.D.

We can now use Theorem 1(a) in future proofs

Huntington postulates: Post. 2: (a) x+0=x, (b) x·1=x Post. 3: (a) x+y=y+x, (b) x·y=y·x Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x’=1, (b) x·x’=0

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Proof of x·x=x

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Similar to previous proof

Show that x·x = x. x·x = xx+0 by 2(a) = xx+xx’ by 5(b) = x(x+x’) by 4(a) = x·1 by 5(a) = x by 2(b) Q.E.D.

Huntington postulates: Post. 2: (a) x+0=x, (b) x·1=x Post. 3: (a) x+y=y+x, (b) x·y=y·x Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x’=1, (b) x·x’=0 Th. 1: (a) x+x=x

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Theorem 2: Annulment الغاء

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Theorem 2(a): (x + 1) = 1

Theorem 2(b): (x.0)= 0

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Proof of x+1=1

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Theorem 2(a): x + 1 = 1

x + 1 = 1．(x + 1) by 2(b)

=(x + x')(x + 1) 5(a)

= x + x' 1 4(b)

= x + x' 2(b)

= 1 5(a)

Theorem 2(b): x．0 = 0 by duality

Huntington postulates: Post. 2: (a) x+0=x, (b) x·1=x Post. 3: (a) x+y=y+x, (b) x·y=y·x Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x’=1, (b) x·x’=0 Th. 1: (a) x+x=x

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Involution Theorem

Theorem 3 : (x’)’ = x

Postulate 5 defines the complement of x, x + x' = 1 and x x' = 0

The complement of x' is x is also (x')’

By means of truth table

x x’ (x’)’

0 1 0

1 0 1

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Associative Theorem

Theorem 4(a): (x + y)+z = x+(y+z)

Theorem 4(b): (x.y).z = x.(y.z)

By means of truth table

x y z x+y (x+y)+z y+z x+(y+z)

0 0 0 0 0 0 0

0 0 1 0 1 1 1

0 1 0 1 1 1 1

0 1 1 1 1 1 1

1 0 0 1 1 1 1

1 0 1 1 1 1 1

1 1 0 1 1 1 1

1 1 1 1 1 1 1 17 17

DeMorgan’s Theorem

Theorem 5(a): (x + y)’ = x’y’

Theorem 5(b): (xy)’ = x’ + y’

By means of truth table

x y x’ y’ x+y (x+y)’ x’y’ xy x’+y' (xy)’

0 0 1 1 0 1 1 0 1 1

0 1 1 0 1 0 0 0 1 1

1 0 0 1 1 0 0 0 1 1

1 1 0 0 1 0 0 1 0 0

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Absorption Property (Covering)

Theorem 6(a): x + xy = x

x + xy = x．1 + xy by 2(b)

= x (1 + y) 4(a)

= x (y + 1) 3(a)

= x．1 Th 2(a)

= x 2(b)

Theorem 6(b): x (x + y) = x by duality Report

By means of truth table (another way to proof )

x y xy x+xy

0 0 0 0

0 1 0 0

1 0 0 1

1 1 1 1

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Huntington postulates: Post. 2: (a) x+0=x, (b) x·1=x Post. 3: (a) x+y=y+x, (b) x·y=y·x Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x’=1, (b) x·x’=0 Th. 1: (a) x+x=x

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Report

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Proof That :

1. xy + x’z + yz = xy + x’z

2. (x+y)•(x’+z)•(y+z) = (x+y)•(x’+z) -- (dual)

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Description of the Laws of

Boolean Algebra

• Annulment Law – A term AND´ed with a “0”

equals 0 or OR´ed with a “1” will equal 1

• A . 0 = 0 A variable AND’ed with 0 is always

equal to 0

• A + 1 = 1 A variable OR’ed with 1 is always

equal to 1

• Identity Law – A term OR´ed with a “0”

or AND´ed with a “1” will always equal that term

• A + 0 = A A variable OR’ed with 0 is always equal

to the variable

• A . 1 = A A variable AND’ed with 1 is always

equal to the variable

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• Idempotent Law – An input that is AND´ed or OR´ed

with itself is equal to that input

• A + A = A A variable OR’ed with itself is always equal

to the variable

• A . A = A A variable AND’ed with itself is always

equal to the variable

• Complement Law – A term AND´ed with its

complement equals “0” and a term OR´ed with its

complement equals “1”

•

• A . A = 0 A variable AND’ed with its complement is

always equal to 0

• A + A = 1 A variable OR’ed with its complement is

always equal to 1

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• Commutative Law – The order of application of

two separate terms is not important

•

• A . B = B . A The order in which two variables are

AND’ed makes no difference

• A + B = B + A The order in which two variables are

OR’ed makes no difference

•

• Double Negation Law – A term that is inverted

twice is equal to the original term

•

• (A’)’ = A A double complement of a variable is

always equal to the variable

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• Distributive Law – This law permits the multiplying or

factoring out of an expression.

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• A(B + C) = A.B + A.C (OR Distributive Law)

• A + (B.C) = (A + B).(A + C) (AND Distributive Law)

•

• Absorptive Law – This law enables a reduction in a

complicated expression to a simpler one by absorbing

like terms.

•

• A + (A.B) = A (OR Absorption Law)

• A(A + B) = A (AND Absorption Law)

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• Associative Law – This law allows the removal of

brackets from an expression and regrouping of the

variables.

• A + (B + C) = (A + B) + C = A + B + C (OR

Associate Law)

• A(B.C) = (A.B)C = A . B . C (AND Associate Law)

• de Morgan´s Theorem – There are two “de

Morgan´s” rules or theorems,

(1) Two separate terms NOR´ed together is the same

as the two terms inverted (Complement) and AND´ed

for example: A+B = A . B

(2) Two separate terms NAND´ed together is the

same as the two terms inverted (Complement)

and OR´ed for example: A.B = A + B

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Operator Precedence

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The operator precedence for evaluating Boolean Expression

is

Parentheses

NOT

AND

OR

Examples

x y' + z

(x y + z)'

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Truth Tables for the Laws of

Boolean

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Boolean

Expression Description

Equivalent

Switching Circuit

Boolean Algebra

Law or Rule

A + 1 = 1 A in parallel with

closed = “CLOSED” Annulment

A + 0 = A A in parallel with

open = “A” Identity

A . 1 = A A in series with

closed = “A” Identity

A . 0 = 0 A in series with

open = “OPEN” Annulment

A + A = A A in parallel with

A = “A” Idempotent

A . A = A A in series with

A = “A” Idempotent

NOT A’ = A NOT NOT A

(double negative) = “A” Double Negation

A + A = 1 A in parallel with

NOT A = “CLOSED” Complement

A . A = 0 A in series with

NOT A = “OPEN” Complement

A+B = B+A A in parallel with B =

B in parallel with A Commutative

A.B = B.A A in series with B =

B in series with A Commutative

A+B = A.B invert and replace OR with AND de Morgan’s Theorem

A.B = A+B invert and replace AND with OR de Morgan’s Theorem

Report

Reports

Report 1 Slide 19

Report 2 Slide 20

Report 3 Slide 29

Boolean Algebra Examples

Boolean Algebra Example No1

Construct a Truth Table for the logical functions at

points C, D and Q in the following circuit and identify

a single logic gate that can be used to replace the whole circuit.

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Inputs Output at

A B C D Q

0 0 1 0 0

0 1 1 1 1

1 0 1 1 1

1 1 0 0 1

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Boolean Algebra Example No 2

Find the Boolean algebra expression for the following

system.

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Inputs Intermediates Output

B A A.B A + B Q

0 0 0 1 1

0 1 0 0 0

1 0 0 0 0

1 1 1 0 1

Boolean Algebra Example No 3

Find the Boolean algebra expression for the

following system.

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Inputs Intermediates Output

C B A A.B.C B C B+C A.(B+C) Q

0 0 0 0 1 1 1 0 0

0 0 1 0 1 1 1 1 1

0 1 0 0 0 1 1 0 0

0 1 1 0 0 1 1 1 1

1 0 0 0 1 0 1 0 0

1 0 1 0 1 0 1 1 1

1 1 0 0 0 0 0 0 0

1 1 1 1 0 0 0 0 1

Report

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Proof That

1. xy + x’z + yz = xy + x’z

2. (x+y)•(x’+z)•(y+z) = (x+y)•(x’+z) -- (dual)

Proof:

xy + x’z + yz = xy + x’z + (x+x’)yz

= xy + x’z + xyz + x’yz

= (xy + xyz) + (x’z + x’zy)

= xy + x’z

QED (2 true by duality).

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