1. CT455: Computer Organization Boolean Algebra

2. Lecture 3: Boolean Algebra
Digital circuits
Boolean Algebra
Two-Valued Boolean Algebra
Boolean Algebra Postulates
Precedence of Operators
Truth Table & Proofs
Duality

3. Lecture 3: Boolean Algebra
Basic Theorems of Boolean Algebra
Boolean Functions
Complement of Functions
Standard Forms
Minterm & Maxterm
Canonical Forms
Conversion of Canonical Forms
Binary Functions

4. Digital Circuits
Digital circuit can be represented by a black-box with inputs on one side, and outputs on the other.
Digital circuit
inputs
outputs :
The input/output signals are discrete/digital in nature, typically with two distinct voltages (a high voltage and a low voltage).
High
Low
In contrast, analog circuits use continuous signals.

5. Digital Circuits Advantages of Digital Circuits over Analog Circuits:
more reliable (simpler circuits, less noise-prone)
specified accuracy (determinable)
but slower response time (sampling rate)
Important advantages for two-valued Digital Circuit:
Mathematical Model – Boolean Algebra
Can help design, analyse, simplify Digital Circuits.

7. Boolean Algebra What is an Algebra? (e.g. algebra of integers)
set of elements (e.g. 0,1,2,..)
set of operations (e.g. +, -, *,..)
postulates/axioms (e.g. 0+x=x,..)
Boolean Algebra named after George Boole who used it to study human logical reasoning – calculus of proposition.
Events : true or false
Connectives : a OR b; a AND b, NOT a
Example: Either “it has rained” OR “someone splashed water”, “must be tall” AND “good vision”.

8. Boolean Algebra
Later, Shannon introduced switching algebra (two-valued Boolean algebra) to represent bi-stable switching circuit.

9. Two-valued Boolean Algebra
Set of Elements: {0,1}
Set of Operations: { ., + , ¬ }
Sometimes denoted by ’, for example a’ x y x y
x+y x x'
x.y
Signals: High = 5V = 1; Low = 0V = 0

10. Boolean Algebra Postulates
A Boolean algebra consists of a set of elements B, with two binary operations {+} and {.} and a unary operation {'}, such that the following axioms hold:
Commutative laws:
A + B = B + A
A . B = B . A

11. Boolean Algebra Postulates
Associative laws:
(A + B) + C = A + (B + C) = A + B + C
(A . B) . C = A .( B . C) = A . B . C
Distributive laws:
A . (B + C) = (A . B) + (A . C)
A + (B . C) = (A + B) . (A + C)

12. Boolean Algebra Postulates
Complement: For every x in B, there exists an element x' in B such that
x + x' = 1
x . x' = 0

13. Precedence of Operators
To lessen the brackets used in writing Boolean expressions, operator precedence can be used.
Precedence (highest to lowest): '
Examples:
a . b + c = (a . b) + c
b' + c = (b') + c
a + b' . c = a + ((b') . c)

14. Precedence of Operators
Use brackets to overwrite precedence.
Examples:
a . (b + c)
(a + b)' . c

15. Rules for Boolean Algebra
1. A + 0 = A
9. (A’)’ = A
2. A + 1 = 1
10. A + AB = A
3. A . 0 = 0
11. A + A’B = A + B
4. A . 1 = A
12. (A + B)(A + C) = A+BC
5. A + A = A
6. A + A’ = 1
7. A . A = A
8. A . A’ = 0

16. 10. A + AB = A
A + AB = A(1 + B) DISTRIBUTIVE LAW
= A . 1 RULE 2
= A RULE 4 A B AB
A+AB 1

17. Theorem 10 (Absorption)
(a) a + ab = a (b) a(a + b) = a
Examples:
(X + Y) + (X + Y)Z = X + Y [T4(a)]
AB'(AB' + B'C) = AB' [T4(b)]

18. A B A’B A+A’B A+B 1 11. A + A’B = A + B A + A’B = (A + AB) + A’B
= (AA + AB) + A’B
= AA + AB + AA’ + A’B
= (A + A’)(A + B)
= 1. (A + B)
= ( A + B) A B
A’B
A+A’B
A+B 1

19. Theorem 11
(a) a + a'b = a + b (b) a(a' + b) = ab
Examples:
B + AB'C'D = B + AC'D [T11(a)]
(X + Y)((X + Y)' + Z) = (X + Y)Z [T11(b)]

20. 11. A + A’B = A + B
เปลี่ยนเป็นรูปอื่นๆได้
A + A’B’ = A + B’
A’ + AB = A + B
A + A’BC = A + BC
(AB)’ + ABCD = (AB)’ + CD

21. 12. (A + B)(B + C) = AA + AC + AB + BC
= A.1 + AB + BC
= A(1+B) +BC
= A.1 +BC
= A . BC
A B C
A+B
A+C
(A+B)(A+C)
BC A+BC
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1 1

22. (a) ab + ab' = a (b) (a + b)(a + b') = a Examples:
Theorem
(a) ab + ab' = a (b) (a + b)(a + b') = a
Examples:
ABC + AB'C = AC [T6(a)]
(W' + X' + Y' + Z')(W' + X' + Y' + Z)(W' + X' + Y + Z')(W' + X' + Y + Z)
= (W' + X' + Y')(W' + X' + Y + Z')(W' + X' + Y + Z)
= (W' + X' + Y')(W' + X' + Y)
= (W' + X')

23. Theorem 7
(a) ab + ab'c = ab + ac (b) (a + b)(a + b' + c) = (a + b)(a + c)
Examples:
wy' + wx'y + wxyz + wxz' = wy' + wx'y + wxy + wxz' [T7(a)]
= wy' + wy + wxz' [T7(a)]
= w + wxz' [T7(a)]
= w [T7(a)]
(x'y' + z)(w + x'y' + z') = (x'y' + z)(w + x'y') [T7(b)]

24. Basic Theorems of Boolean Algebra
6. DeMorgan.
(a) (x + y)' = x'.y'
(b) (x.y)' = x' + y'
7. Consensus.
(a) x.y + x'.z + y.z = x.y + x'.z
(b) (x+y).(x'+z).(y+z) = (x+y).(x'+z)

25. Laws of Boolean Algebra
DeMorgan's Law X 1 Y 1 X 1 Y 1 X + Y 1
X•Y 1
(X + Y)' = X'  Y'
NOR is equivalent to AND
with inputs complemented X 1 Y 1 X 1 Y 1
X•Y 1 X + Y 1
(X  Y)' = X' + Y'
NAND is equivalent to OR
with inputs complemented
DeMorgan's Law can be used to convert AND/OR expressions
to OR/AND expressions or NAND/NAND expression
Example:
Z = A' B' C + A' B C + A B' C + A B C’ AND - OR
= {(A’B’C)’  (A’BC)’  (AB’C)’  (ABC’)’ } ‘ NAND - NAND
Z' = (A + B + C')  (A + B' + C')  (A' + B + C')  (A' + B' + C)

26. Theorem 8 (DeMorgan's Theorem)
(a) (a + b)' = a'b' (b) (ab)' = a' + b'
Generalized DeMorgan's Theorem
(a) (a + b + … z)' = a'b' … z' (b) (ab … z)' = a' + b' + … z'
Examples:
(a + bc)' = (a + (bc))'
= a'(bc)' [T8(a)]
= a'(b' + c') [T8(b)]
= a'b' + a'c' [P5(b)]
Note: (a + bc)' ¹ a'b' + c'

27. More Examples for DeMorgan's Theorem
(a(b + z(x + a')))' = a' + (b + z(x + a'))' [T8(b)]
= a' + b' (z(x + a'))' [T8(a)]
= a' + b' (z' + (x + a')') [T8(b)]
= a' + b' (z' + x'(a')') [T8(a)]
= a' + b' (z' + x'a) [T3]
= a' + b' (z' + x') [T5(a)]
(a(b + c) + a'b)' = (ab + ac + a'b)' [P5(b)]
= (b + ac)' [T6(a)]
= b'(ac)' [T8(a)]
= b'(a' + c') [T8(b)]

28. Truth Table
Provides a listing of every possible combination of inputs and its corresponding outputs.
Example (2 inputs, 2 outputs):

29. Switching Functions
Switching algebra: Boolean algebra with the set of elements K = {0, 1}
If there are n variables, we can define switching functions.
Sixteen functions of two variables (Table 2.3):
A switching function can be represented by a table as above, or by a switching expression as follows:
f0(A,B)= 0, f6(A,B) = AB' + A'B, f11(A,B) = AB + A'B + A'B' = A' + B, ...
Value of a function can be obtained by plugging in the values of all variables:
The value of f6 when A = 1 and B = 0 is: = = 1.

30. Truth Tables (1)
Shows the value of a function for all possible input combinations.
Truth tables for OR, AND, and NOT (Table 2.4):

31. Truth Tables (2)
Truth tables for f(A,B,C) = AB + A'C + AC' (Table 2.5)

32. Truth Table
Example (3 inputs, 2 outputs):

33. Proof using Truth Table
Can use truth table to prove by perfect induction.
Prove that: x . (y + z) = (x . y) + (x . z)
(i) Construct truth table for LHS & RHS of above equality.
(ii) Check that LHS = RHS
Postulate is SATISFIED because output column 2 & 5 (for LHS & RHS expressions) are equal for all cases.

34. Basic Theorems of Boolean Algebra
Apart from the axioms/postulates, there are other useful theorems.
1. Idempotency.
(a) x + x = x (b) x . x = x
Proof of (a):
x + x = (x + x) (identity)
= (x + x).(x + x') (complementarity)
= x + x.x' (distributivity)
= x (complementarity)
= x (identity)

35. Basic Theorems of Boolean Algebra
2. Null elements for + and . operators.
(a) x + 1 = (b) x . 0 = 0
3. Involution. (x')' = x
4. Absorption.
(a) x + x.y = x (b) x.(x + y) = x
5. Absorption (variant).
(a) x + x'.y = x+y (b) x.(x' + y) = x.y

36. Basic Theorems of Boolean Algebra
Theorem 4a (absorption) can be proved by:
x + x.y = x.1 + x.y (identity)
= x.(1 + y) (distributivity)
= x.(y + 1) (commutativity)
= x (Theorem 2a)
= x (identity)
By duality, theorem 4b:
x.(x+y) = x
Try prove this by algebraic manipulation.

37. Boolean Functions
Boolean function is an expression formed with binary variables, the two binary operators, OR and AND, and the unary operator, NOT, parenthesis and the equal sign.
Its result is also a binary value.
We usually use . for AND, + for OR, and ' or ¬ for NOT. Sometimes, we may omit the . if there is no ambiguity.

38. Boolean Functions Examples: F1= x.y.z' F2= x + y'.z
F3=(x'.y'.z)+(x'.y.z)+(x.y')
F4=x.y'+x'.z
From the truth table, F3=F4.
Can you also prove by algebraic manipulation that F3=F4?

39. Simplifying logic expressions
CS1103
Chapter 3: Boolean Algebra/Logic Gates

40. EXAMPLE
A(A+B) = AA + AB
= A + AB
= A(1 + B)
= A
A(A’ + B) = AA’ + AB
= AB
( AB + A’B’)’ = (A’ + B’)(A + B)
= A’A + A’B + AB’ + BB’
= AB’ + A’B

41. Chapter 3: Boolean Algebra/Logic Gates
X = ABC + ABC’
= AB ( C + C’)
= AB.1
= AB
A = X(X’YZ + X’YZ’)
= XX’YZ + XX’YZ’
= 0.YZ YZ’
= 0
CS1103
Chapter 3: Boolean Algebra/Logic Gates

42. A B AB’ A’B (AB)’ A+B 0 0 0 1 1 0 1 1 1 AB’+A’B (AB)’(A+B) 1
EXAMPLE AB’ + A’B = AB’+AA’+A’B+BB’
= A(B’+A’) + B(A’+B’)
= (A’+B’)(A+B)
= (AB)’(A+B)
A B
AB’
A’B
(AB)’
A+B 1
AB’+A’B
(AB)’(A+B) 1

43. EXAMPLE
Y = AB + A’B +A’B’
Y = B(A + A’) +A’B’
Y = B.1 +A’B’
Y = B +A’B’
Y = B +A’

44. Chapter 3: Boolean Algebra/Logic Gates
W = AB(A’C + B)
= ABA’C + ABB
= 0.BC + AB
= 0 + AB
= AB
CS1103
Chapter 3: Boolean Algebra/Logic Gates

45. Z = ( X.Y) + [( X’ . ( Y + X )]
Z = ( X.Y) + [( X’.Y) + (X.X’)]
Z = ( X.Y) + [( X’.Y) + (0)]
Z = ( X.Y) + ( X’.Y)
Z = Y. ( X + X’)
Z = Y. 1
Z = Y

46. Z = ( X.Y’) + [( X . Y ) + Y’ )]
Z = [( X . Y’ ).(X . Y )] + [( X . Y’ ) + Y’ )]
Z = [( X . X ).(Y . Y’ )] + [ X . (Y’ + Y’ )]
Z = [( X . X ).( 0 )] + [ X . (Y’ + Y’ )]
Z = 0 + [ X . Y’]
Z = X . Y’

47. Z = X + X’.Y + Y’.X’
Z = X + X’. (Y + Y’)
Z = X + X’. (1)
Z = 1

48. Z = X( X’ + Y )(Y’ + X’)
Z = (X.X’ + XY )( Y’ + X’ )
Z = ( 0 + XY)( Y’ + X’
Z = XY( Y’ + X’)
Z = XYY’ + XYX’
Z =

49. Chapter 3: Boolean Algebra/Logic Gates
Z = w( y + w )( x + y + w )
= (wy + ww)( x + y + w )
= (wy + w)( x + y + w )
= (w + wy)( x + y + w )
= w( x + y + w )
= wx + wy + ww
= wx + ( wy + w )
= wx + ( w + wy)
= wx + w
= w + wx
= w
CS1103
Chapter 3: Boolean Algebra/Logic Gates

50. Simplify
AB + BC(B + C)
AB + BBC + BCC
AB + BC + BC
AB + BC
B( A + C)
AB(A + B’C +C)
ABA + ABB’C + ABC
AB + ABB’C + ABC
AB + ABC
AB(1 + C) AB

51. EXAMPLE SIMPLIFY THE EXPRESSION
AB + A(B +C) + B(B + C) USING BOOLEAN ALGEBRA
AB + AB + AC +BB + BC
AB + AB + AC + B + BC
AB + AC + B + BC
AB + AC + B(1 + C)
AB + AC + B.1
AB + AC + B
B(A + 1) +AC
B.1 + AC
B + AC

52. SIMPLIFY THE EXPRESSION
[AB’(C + BD)+ A’B’]C USING BOOLEAN ALGEBRA
[AB’(C + BD)+ A’B’]C
(AB’C + AB’BD + A’B’)C
(AB’C + A.0.D + A’B’)C
(AB’C A’B’)C
(AB’C + A’B’)C
AB’CC + A’B’C
AB’C + A’B’C
B’C(A + A’)
B’C.1
B’C

53. SIMPLIFY THE EXPRESSION
A’BC + AB’C’ + A’B’C’ + AB’C + ABC USING BOOLEAN ALGEBRA
A’BC + AB’C’ + A’B’C’ + AB’C + ABC
BC(A’ + A) + AB’C’ + A’B’C’ + AB’C
BC.1 + AB’(C’ + C) + A’B’C’
BC + AB’.1 + A’B’C’
BC + B’(A + A’C’)
BC + B’(A + C’)
BC + AB’ + B’C’

54. Gate Logic: Laws of Boolean Algebra
Apply the laws and theorems to simplify Boolean equations
Example: full adder's carry out function
identity
Cout = A' B Cin + A B' Cin + A B Cin' + A B Cin
= A' B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin
= A' B Cin + A B Cin + A B' Cin + A B Cin' + A B Cin
= (A' + A) B Cin + A B' Cin + A B Cin' + A B Cin
= (1) B Cin + A B' Cin + A B Cin' + A B Cin
= B Cin + A B' Cin + A B Cin' + A B Cin + A B Cin
= B Cin + A B' Cin + A B Cin + A B Cin' + A B Cin
= B Cin + A (B' + B) Cin + A B Cin' + A B Cin
= B Cin + A (1) Cin + A B Cin' + A B Cin
= B Cin + A Cin + A B (Cin' + Cin)
= B Cin + A Cin + A B (1)
= B Cin + A Cin + A B
associative

55. Chapter 3: Boolean Algebra/Logic Gates
Z = (x’ + y’) w’ + (xy)’
= ( x’ + y’ ) w’ + ( x’ + y’ )
= ( x’ + y’ ) w’ + ( x’ + y’ ) . 1
= ( x’ + y’ ) ( w’ + 1)
= ( x’ + y’ ) (1)
= x’ + y’
CS1103
Chapter 3: Boolean Algebra/Logic Gates

56. Chapter 3: Boolean Algebra/Logic Gates
M = (A’ + B’C’)’
= (A’)’ (B’)’(C’)’
= ABC
S = (A’B’(C’ + D’))’
= (A’B’)’ + (C’ + D’)’
= (A’)’ + (B’)’ + (C’)’ (D’)’
= A + B + CD
CS1103
Chapter 3: Boolean Algebra/Logic Gates

57. Complement of Functions
Given a function, F, the complement of this function, F', is obtained by interchanging 1 with 0 in the function’s output values.
Example: F1 = xyz'
Complement:
F1' = (x.y.z')'
= x' + y' + (z')' DeMorgan
= x' + y' + z Involution

58. Complement of Functions
More general DeMorgan’s theorems useful for obtaining complement functions:
(A + B + C Z)' = A' . B' . C' … . Z'
(A . B . C Z)' = A' + B' + C' + … + Z'

59. Algebraic Forms of Switching Functions (1)
Literal: A variable, complemented or uncomplemented.
Product term: A literal or literals ANDed together.
Sum term: A literal or literals ORed together.
SOP (Sum of Products):
ORing product terms
f(A, B, C) = ABC + A'C + B'C
POS (Product of Sums)
ANDing sum terms
f (A, B, C) = (A' + B' + C')(A + C')(B + C')

60. Algebraic Forms of Switching Functions (2)
A minterm is a product term in which all the variables appear exactly once either complemented or uncomplemented.
Canonical Sum of Products (canonical SOP):
Represented as a sum of minterms only.
Example: f1(A,B,C) = A'BC' + ABC' + A'BC + ABC (2.1)
Minterms of three variables:

61. Algebraic Forms of Switching Functions (3)
Compact form of canonical SOP form:
f1(A,B,C) = m2 + m3 + m6 + m7 (2.2)
A further simplified form:
f1(A,B,C) = S m (2,3,6,7) (minterm list form) (2.3)
The order of variables in the functional notation is important.
Deriving truth table of f1(A,B,C) from minterm list:

62. Algebraic Forms of Switching Functions (4)
Example: Given f(A,B,Q,Z) = A'B'Q'Z' + A'B'Q'Z + A'BQZ' + A'BQZ, express f(A,B,Q,Z) and f '(A,B,Q,Z) in minterm list form.
f(A,B,Q,Z) = A'B'Q'Z' + A'B'Q'Z + A'BQZ' + A'BQZ
= m0 + m1 + m6 + m7
= S m(0, 1, 6, 7)
f '(A,B,Q,Z) = m2 + m3 + m4 + m5 + m8 + m9 + m10 + m11 + m12
+ m13 + m14 + m15
= S m(2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14, 15)
(2.6)
AB + (AB)' = 1 and AB + A' + B' = 1, but AB + A'B' ¹ 1.

63. Algebraic Forms of Switching Functions (5)
A maxterm is a sum term in which all the variables appear exactly once either complemented or uncomplemented.
Canonical Product of Sums (canonical POS):
Represented as a product of maxterms only.
Example: f2(A,B,C) = (A+B+C)(A+B+C')(A'+B+C)(A'+B+C') (2.7)
Maxterms of three variables:

64. Algebraic Forms of Switching Functions (6)
f2(A,B,C) = M0M1M4M5 (2.8)
= PM(0,1,4,5) (maxterm list form) (2.9)
The truth table for f2(A,B,C):

65. Algebraic Forms of Switching Functions (7)
Truth tables of f1(A,B,C) of Eq. (2.3) and f2(A,B,C) of Eq. (2.7) are identical.
Hence, f1(A,B,C) = S m (2,3,6,7)
= f2(A,B,C)
= PM(0,1,4,5) (2.10)
Example: Given f(A,B,C) = ( A+B+C')(A+B'+C')(A'+B+C')(A'+B'+C'), construct the truth table and express in both maxterm and minterm form.
f(A,B,C) = M1M3M5M7 = PM(1,3,5,7) = S m (0,2,4,6)

66. Algebraic Forms of Switching Functions (8)
Relationship between minterm mi and maxterm Mi:
For f(A,B,C), (m1)' = (A'B'C)' = A + B + C' = M1
In general, (mi)' = Mi (2.11)
(Mi)' = ((mi)')' = mi (2.12)

67. Algebraic Forms of Switching Functions (9)
Example: Relationship between the maxterms for a function and its complement.
For f(A,B,C) = ( A+B+C')(A+B'+C')(A'+B+C')(A'+B'+C')
The truth table is:

68. Algebraic Forms of Switching Functions (10)
From the truth table
f '(A,B,C) = PM(0,2,4,6) and f(A,B,C) = PM(1,3,5,7)
Since f(A,B,C) × f '(A,B,C) = 0,
(M0M2M4M6)(M1M3M5M7) = 0 or
In general, (2.13)
Another observation from the truth table:
f(A,B,C) = S m (0,2,4,6) = PM(1,3,5,7)
f '(A,B,C) = S m (1,3,5,7) = PM(0,2,4,6)

69. Derivation of Canonical Forms (1)
Derive canonical POS or SOP using switching algebra.
Theorem 10. Shannon's expansion theorem
(a). f(x1, x2, …, xn) = x1 f(1, x2, …, xn) + (x1)' f(0, x2, …, xn)
(b). f(x1, x2, …, xn) = [x1 + f(0, x2, …, xn)] [(x1)' + f(1, x2, …, xn)]
Example: f(A,B,C) = AB + AC' + A'C
f(A,B,C) = AB + AC' + A'C = A f(1,B,C) + A' f(0,B,C)
= A(1×B + 1×C' + 1'×C) + A'(0×B + 0×C' + 0'×C) = A(B + C') + A'C
f(A,B,C) = A(B + C') + A'C = B[A(1+C') + A'C] + B'[A(0 + C') + A'C]
= B[A + A'C] + B'[AC' + A'C] = AB + A'BC + AB'C' + A'B'C
f(A,B,C) = AB + A'BC + AB'C' + A'B'C
= C[AB + A'B×1 + AB'×1' + A'B'×1] + C'[AB + A'B×0 + AB'×0' + A'B'×0]
= ABC + A'BC + A'B'C + ABC' + AB'C'

70. Derivation of Canonical Forms (2)
Alternative: Use Theorem 6 to add missing literals.
Example: f(A,B,C) = AB + AC' + A'C to canonical SOP form.
AB = ABC' + ABC = m6 + m7
AC' = AB'C' + ABC' = m4 + m6
A'C = A'B'C + A'BC = m1 + m3
Therefore,
f(A,B,C) = (m6 + m7) + (m4 + m6) + (m1 + m3) = m(1, 3, 4, 6, 7)
Example: f(A,B,C) = A(A + C') to canonical POS form.
A = (A+B')(A+B) = (A+B'+C')(A+B'+C)(A+B+C')(A+B+C)
= M3M2M1M0
(A+C')= (A+B'+C')(A+B+C') = M3M1
f(A,B,C) = (M3M2M1M0)(M3M1) = M(0, 1, 2, 3)

71. Incompletely Specified Functions
A switching function may be incompletely specified.
Some minterms are omitted, which are called don't-care minterms.
Don't cares arise in two ways:
Certain input combinations never occur.
Output is required to be 1 or 0 only for certain combinations.
Don't care minterms: di Don't care maxterms: Di
Example: f(A,B,C) has minterms m0, m3, and m7 and don't-cares d4 and d5.
Minterm list is: f(A,B,C) = m(0,3,7) + d(4,5)
Maxterm list is: f(A,B,C) = M(1,2,6)·D(4,5)
f '(A,B,C) = m(1,2,6) + d(4,5) = M(0,3,7)·D(4,5)
f (A,B,C)= A'B'C' + A'BC + ABC + d(AB'C' + AB'C)
= B'C' + BC (use d4 and omit d5)

72. Standard Forms Exampes: x, x.(y+z'), (x+y').(x'+y+z), (A+B).(A'+B')
Sum Term: a single literal or a logical sum (OR) of several literals.
Examples: x, x+y+z', A'+B, A+B
Sum-of-Products (SOP) Expression: a product term or a logical sum (OR) of several product terms.
Examples: x, x+y.z', x.y'+x‘.y.z, A.B+A'.B'
Product-of-Sums (POS) Expression: a sum term or a logical product (AND) of several sum terms.
Exampes: x, x.(y+z'), (x+y').(x'+y+z), (A+B).(A'+B')

73. Standard Forms POS: (x + y).(x + y).(x + z)
Every Boolean expression can either be expressed as sum-of-products or product-of-sums expression.
Examples:
SOP: x.y + x.y + x.y.z
POS: (x + y).(x + y).(x + z)
both: x + y + z or x.y.z
neither: x.(w + y.z) or z + w.x.y + v.(x.z + w)

74. Shorthand Notation for
Two Level Canonical Forms
Sum of Products
product term / minterm: A 1 B 1 C 1
Minterms
ANDed product of literals in which each
variable appears exactly once, in true or
complemented form (but not both!)
A B C = m
A B C = m 1
A B C = m 2
A B C = m 3
F in canonical form:
A B C = m 4
F(A,B,C) = m(3,4,5,6,7)
A B C = m 5
= m3 + m4 + m5 + m6 + m7
= A' B C + A B' C' + A B' C
+ A B C' + A B C
A B C = m 6
A B C = m 7
Shorthand Notation for
Minterms of 3 Variables
canonical form/minimal form
F = A B' (C + C') + A' B C + A B (C' + C)
= A B' + A' B C + A B
= A (B' + B) + A' B C
= A + A' B C
= A + B C B C F A
F’ = (A + B C)' = A' (B' + C') = A' B' + A' C'
2-Level AND/OR
Realization

75. 2 Level Canonical Forms Maxterm: Maxterm form:
Product of Sums / Conjunctive Normal Form / Maxterm Expansion
Maxterm:
ORed sum of literals in which each
variable appears exactly once in either
true or complemented form, but not both!
Maxterm form:
Find truth table rows where F is 0
0 in input column implies true literal
1 in input column implies complemented
literal
Maxterm Shorthand Notation
for a Function of Three Variables
F(A,B,C) = M(0,1,2)
= (m(0,1,2))’ = (A’B’C’ + A’B’C + A’BC’)’
= (A + B + C) (A + B + C') (A + B' + C)
F’(A,B,C) = M(3,4,5,6,7)
= (A + B' + C') (A' + B + C) (A' + B + C') (A' + B' + C) (A' + B' + C')

76. Two Level Canonical Forms
F' = A' B' C' + A' B' C + A' B C'
Apply DeMorgan's Law to obtain F:
(F')' = (A' B' C' + A' B' C + A' B C')'
F = (A + B + C) (A + B + C') (A + B' + C)
F' = (A + B' + C') (A' + B + C) (A' + B + C') (A' + B' + C) (A' + B' + C')
Apply DeMorgan's Law to obtain F:
(F')' = {(A + B' + C') (A' + B + C) (A' + B + C')
(A' + B' + C) (A' + B' + C')}'
F = A' B C + A B' C' + A B' C + A B C' + A B C