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DKT 122/3
DIGITAL SYSTEM 1
WEEK #5
BOOLEAN ALGEBRA (PART 1)
Boolean Algebra
Contents
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Boolean Operations & Expression
Laws & Rules of Boolean algebra
DeMorgan’s Theorems
Boolean analysis of logic circuits
Simplification using Boolean Algebra
Standard forms of Boolean Expressions
Boolean Expressions & truth tables
Boolean Algebra (Cont.)
 The Karnaugh Map
Karnaugh Map SOP minimization
Karnaugh Map POS minimization
 Programmable Logic
Boolean Operations & Expression
Expression
Variable
 a symbol used to represent logical quantities (1
or 0)
 E.g : A, B,..used as variable
Complement
 inverse of variable and is indicated by bar over
variable
 E.g : Ā
Boolean Operations & Expression
Operation
Boolean Addition
– equivalent to the OR operation
A
B
X
X=A+B
Boolean Multiplication
– equivalent to the AND operation
A
X = A∙B
B
X
Laws of Boolean Algebra
 Commutative
 Addition & multiplication
 Associative
 Addition & multiplication
 Distributive
 Same as ordinary algebra
Commutative Law
Addition
Commutative law of addition:
A+B = B+A
the order of OR-ing does not matter.
Commutative Law
Multiplication
Commutative law of Multiplication
AB = BA
the order of ANDing does not matter.
Associative Law
Addition
Associative law of addition
A + (B + C) = (A + B) + C
The grouping of ORed variables does not
matter
Associative Law
Multiplication
Associative law of multiplication
A(BC) = (AB)C
The grouping of ANDed variables does not
matter
Distributive Law
A(B + C) = AB + AC
Question: (A+B)(C+D) ?
Rules of Boolean Algebra
Rule 1: A + 0 = A
In math if you add 0 you have changed nothing.
In Boolean Algebra ORing with 0 changes nothing.
Rules of Boolean Algebra
Rule 2: A + 1 = 1
ORing with 1 must give a 1 since if any input
is 1 an OR gate will give a 1.
Rules of Boolean Algebra
Rule 3: A . 0 = 0
In math if 0 is multiplied with anything you
get 0. If you AND anything with 0 you get 0.
Rules of Boolean Algebra
Rule 4: A . 1 = A
ANDing anything with 1 will yield the anything.
Rules of Boolean Algebra
Rule 5: A + A = A
ORing with itself will give the same result
Rules of Boolean Algebra
Rule 6: A + A = 1
Either A or A must be 1 so A + A =1
Rules of Boolean Algebra
Rule 7: A . A = A
ANDing with itself will give the same result
Rules of Boolean Algebra
Rule 8: A . A = 0
In digital Logic 1 =0 and 0 =1, so AA=0 since
one of the inputs must be 0.
Rules of Boolean Algebra
Rule 9: A = A
If you NOT something twice, you are back to
the beginning
Rules of Boolean Algebra
Rule 10: A + AB = A
Proof:
A + AB = A (1 + B)
DISTRIBUTIVE LAW
= A∙1
RULE 2: (1+B) = 1
=A
RULE 4: A∙1 = A
Rules of Boolean Algebra
Rule 11: A + AB = A + B
If A is 1 the output is 1 , If A is 0 the output is B
Proof:
A + AB = (A + AB) + AB
RULE 10
= (AA + AB) + AB
RULE 7
= AA + AB + AA +AB
RULE 8
= (A + A)(A + B)
FACTORING
= 1∙(A + B)
RULE 6
=A+B
RULE 4
Rules of Boolean Algebra
Rule 12: (A + B) (A + C)= A + BC
PROOF
(A + B)(A +C) = AA + AC +AB +BC
DISTRIBUTIVE LAW
= A + AC + AB + BC
RULE 7
= A(1 + C) +AB + BC
FACTORING
= A.1 + AB + BC
RULE 2
= A(1 + B) + BC
FACTORING
= A.1 + BC
RULE 2
= A + BC
RULE 4
Rules of Boolean Algebra
SUMMARY - LAWS OF BOOLEAN ALGEBRA
[1] COMMUTATIVE:
A+B=B+A
AB = BA
[2] ASSOCIATIVE:
A + (B + C) = (A + B) + C
A(BC) = (AB)C
[3] DISTRIBUTIVE:
A(B + C) = AB + AC
(A + B)(C + D) = AC + AD + BC + BD
Rules of Boolean Algebra
SUMMARY - RULES OF BOOLEAN ALGEBRA
[1] A + 0 = A
[7] A.A = A
[2] A + 1 = 1
[8] A.A = 0
[3] A.0 = 0
[9] A = A
[4] A.1 = A
[10] A + AB = A
[5] A + A = A
[11] A + AB = A + B
[6] A + A = 1
[12] (A + B)(A + C) = A + BC
Rules of Boolean Algebra
SUMMARY - RULES OF BOOLEAN ALGEBRA
[1] A + 0 = A
[7] A.A = A
[2] A + 1 = 1
[8] A.A = 0
[3] A.0 = 0
[9] A = A
[4] A.1 = A
[10] A + AB = A
[5] A + A = A
[11] A + AB = A + B
[6] A + A = 1
[12] (A + B)(A + C) = A + BC
DeMorgan’s Theorems
 Two most important theorems of Boolean
Algebra were contributed by De Morgan
 Extremely useful in simplifying expression
in which product or sum (POS) of
variables is inverted
 The TWO theorems are:
X.Y = X + Y
& X+Y = X . Y
DeMorgan’s Theorems
X.Y = X + Y
(a) & (b) Equivalent circuit implied by the
theorem
(c) Alternative symbol for the NAND function
(d) Truth table that illustrates DeMorgan’s
Theorem
Input
(d)
Output
X
Y
XY X+Y
0
0
1
1
0
1
1
1
1
0
1
1
1
1
0
0
DeMorgan’s Theorems
X+Y = X . Y
(a) & (b) Equivalent circuit implied by the
theorem
(c) Alternative symbol for the NOR function
(d) Truth table that illustrates DeMorgan’s
Theorem
Input
(d)
Output
X
Y
X+Y XY
0
0
1
1
0
1
0
0
1
0
0
0
1
1
0
0