CT455:
Computer Organization
Boolean Algebra
Lecture 3: Boolean Algebra
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Digital circuits
Boolean Algebra
Two-Valued Boolean Algebra
Boolean Algebra Postulates
Precedence of Operators
Truth Table & Proofs
Duality
Lecture 3: Boolean Algebra
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Basic Theorems of Boolean Algebra
Boolean Functions
Complement of Functions
Standard Forms
Minterm & Maxterm
Canonical Forms
Conversion of Canonical Forms
Binary Functions
Digital Circuits
 Digital circuit can be represented by a black-box with
inputs on one side, and outputs on the other.
inputs
:
Digital
circuit
:
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.
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.
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”.
Boolean Algebra
a
F
F
T
T
b
F
T
F
T
a AND b
F
F
F
T
a
F
F
T
T
b
F
T
F
T
a OR b
F
T
T
T
a
F
T
NOT a
T
F
Later, Shannon introduced switching algebra (two-valued Boolean
algebra) to represent bi-stable switching circuit.
Two-valued Boolean Algebra
 Set of Elements: {0,1}
 Set of Operations: { ., + , ¬ }
x
0
0
1
1
x
y
y
0
1
0
1
x
0
0
1
1
x.y
0
0
0
1
x.y
y
0
1
0
1
x+y
0
1
1
1
Sometimes denoted by ’,
for example a’
x
0
1
¬x
1
0
x
y
x+y
Signals: High = 5V = 1; Low = 0V = 0
x
x'
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
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)
Boolean Algebra Postulates
 Complement: For every x in B, there
exists an element x' in B such that
x + x' = 1
x . x' = 0
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)
+
Precedence of Operators
 Use brackets to overwrite precedence.
 Examples:
a . (b + c)
(a + b)' . c
Rules for Boolean Algebra
1. A + 0 = A
2. A + 1 = 1
3. A . 0
= 0
4. A . 1
= A
5. A + A = A
6. A + A’ = 1
7. A . A = A
8. A . A’ = 0
9. (A’)’
= A
10. A + AB = A
11. A + A’B = A + B
12. (A + B)(A + C) = A+BC
10. A + AB
A + AB
A
0
0
1
1
= A
= A(1 + B)
DISTRIBUTIVE LAW
= A . 1
RULE 2
= A
RULE 4
B
0
1
0
1
AB
0
0
0
1
A+AB
0
0
1
1
Theorem 10 (Absorption)
(a) a + ab = a
(b) a(a + b) = a
Examples:
(X + Y) + (X + Y)Z
AB'(AB' + B'C)
= X + Y
= AB'
[T4(a)]
[T4(b)]
11. A + A’B
A + A’B
= 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
0
0
1
1
B
0
1
0
1
A’B
0
1
0
0
A+A’B
0
1
1
1
A+B
0
1
1
1
Theorem 11
(a) a + a'b = a + b
(b) a(a' + b) = ab
Examples:
B + AB'C'D
(X + Y)((X + Y)' + Z)
= B + AC'D
[T11(a)]
= (X + Y)Z [T11(b)]
11. A + A’B = A + B
เปลีย
่ นเป็นรูปอืน
่ ๆได้
A + A’B’
= A + B’
A’ + AB
= A+ B
A + A’BC
= A + BC
(AB)’ + ABCD
= (AB)’ + CD
12. (A + B)(B + C)
= AA + AC + AB + BC
= A + AC + AB + BC
= A(1+C) + AB + BC
= A.1 + AB + BC
= A(1+B) +BC
= A.1 +BC
= A . BC
ABC
A+B
A+C
000
001
010
011
100
101
110
111
0
0
1
1
1
1
1
1
0
1
0
1
1
1
1
1
(A+B)(A+C) BC
0
0
0
1
1
1
1
1
0
0
0
1
0
0
0
1
A+BC
0
0
0
1
1
1
1
1
Theorem
(a) ab + ab' = a
Examples:
ABC + AB'C = AC
(b) (a + b)(a + b') = a
[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')
Theorem 7
(a) ab + ab'c = ab + ac
Examples:
wy' + wx'y + wxyz + wxz'
(x'y' + z)(w + x'y' + z')
(b) (a + b)(a + b' + c) = (a + b)(a + c)
= 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')
[T7(b)]
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)
Laws of Boolean Algebra
DeMorgan's Law
(X + Y)' = X'  Y'
NOR is equivalent to AND
with inputs complemented
(X  Y)' = X' + Y'
NAND is equivalent to OR
with inputs complemented
X
0
0
1
1
Y
0
1
0
1
X
1
1
0
0
Y X+Y X•Y
1
1
1
0
0
0
1
0
0
0
0
0
X
0
0
1
1
Y
0
1
0
1
X
1
1
0
0
Y
1
0
1
0
X•Y X+Y
1 1
1 1
1 1
0 0
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’
= {(A’B’C)’  (A’BC)’  (AB’C)’  (ABC’)’ } ‘
AND - OR
NAND - NAND
Z' = (A + B + C')  (A + B' + C')  (A' + B + C')  (A' + B' + C)
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)'
= a'(b' + c')
= a'b' + a'c'
Note: (a + bc)' ¹ a'b' + c'
[T8(a)]
[T8(b)]
[P5(b)]
More Examples for DeMorgan's Theorem

(a(b + z(x + a')))' = a' + (b + z(x + a'))'
= a' + b' (z(x + a'))'
= a' + b' (z' + (x + a')')
= a' + b' (z' + x'(a')')
= a' + b' (z' + x'a)
= a' + b' (z' + x')

(a(b + c) + a'b)' = (ab + ac + a'b)'
= (b + ac)'
= b'(ac)'
= b'(a' + c')
[T8(b)]
[T8(a)]
[T8(b)]
[T8(a)]
[T3]
[T5(a)]
[P5(b)]
[T6(a)]
[T8(a)]
[T8(b)]
Truth Table
 Provides a listing of every possible combination
of inputs and its corresponding outputs.
INPUTS OUTPUTS
…
…
…
…
 Example (2 inputs, 2 outputs):
x
0
0
1
1
y
0
1
0
1
x.y
0
0
0
1
x+y
0
1
1
1
Switching Functions
Switching algebra: Boolean algebra with the set of elements K
= {0, 1}
If there are n variables, we can define 2 2 switching functions.
Sixteen functions of two variables (Table 2.3):
n
AB f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15
00
01
10
11
0
0
0
0
1
0
0
0
0
1
0
0
1
1
0
0
0
0
1
0
1
0
1
0
0
1
1
0
1
1
1
0
0
0
0
1
1
0
0
1
0
1
0
1
1
1
0
1
0
0
1
1
1
0
1
1
0
1
1
1
1
1
1
1
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 0'1'0 = 0 + 1 = 1.
Truth Tables (1)
Shows the value of a function for all possible input
combinations.
Truth tables for OR, AND, and NOT (Table 2.4):
ab
f(a,b)=a+b
ab
f(a,b)=ab
a
f(a)=a'
00
0
00
0
0
1
01
1
01
0
1
0
10
1
10
0
11
1
11
1
Truth Tables (2)
Truth tables for f(A,B,C) = AB + A'C + AC'
(Table 2.5)
ABC
f(A,B,C)
ABC
f(A,B,C)
000
0
FFF
F
001
010
011
100
101
110
111
1
0
1
1
0
1
1
FFT
FTF
FTT
TFF
TFT
TTF
TTT
T
F
T
T
F
T
T
Truth Table
 Example (3 inputs, 2 outputs):
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
y+z
0
1
1
1
0
1
1
1
x.(y + z)
0
0
0
0
0
1
1
1
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.
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
y+z
0
1
1
1
0
1
1
1
x.(y + z)
0
0
0
0
0
1
1
1
x.y
0
0
0
0
0
0
1
1
x.z
0
0
0
0
0
1
0
1
(x.y)+(x.z)
0
0
0
0
0
1
1
1
(ii) Check that LHS = RHS
Postulate is SATISFIED because output column 2 & 5 (for
LHS & RHS expressions) are equal for all cases.
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).1
(identity)
= (x + x).(x + x') (complementarity)
= x + x.x'
(distributivity)
=x+0
(complementarity)
=x
(identity)
Basic Theorems of Boolean Algebra
2. Null elements for + and . operators.
(a) x + 1 = 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
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.1
(Theorem 2a)
=x
(identity)
 By duality, theorem 4b:
x.(x+y) = x
 Try prove this by algebraic manipulation.
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.
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
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
F1
0
0
0
0
0
0
1
0
F2
0
1
0
0
1
1
1
1
F3
0
1
0
1
1
1
0
0
From the truth table, F3=F4.
Can you also prove by algebraic manipulation that F3=F4?
F4
0
1
0
1
1
1
0
0
Simplifying logic expressions
CS1103
Chapter 3: Boolean Algebra/Logic
Gates
EXAMPLE
• A(A+B)
• A(A’ + B)
• ( AB + A’B’)’
=
AA + AB
=
A + AB
=
A(1 + B)
=
A
=
AA’ + AB
=
AB
= (A’ + B’)(A + B)
= A’A + A’B + AB’ + BB’
= AB’ + A’B
X
= ABC + ABC’
= AB ( C + C’)
= AB.1
= AB
A
= X(X’YZ + X’YZ’)
= XX’YZ + XX’YZ’
= 0.YZ + 0.YZ’
= 0
CS1103
Chapter 3: Boolean Algebra/Logic
Gates
EXAMPLE
AB’ + A’B
= AB’+AA’+A’B+BB’
= A(B’+A’) + B(A’+B’)
= (A’+B’)(A+B)
= (AB)’(A+B)
A
B
0
0
1
1
0
1
0
1
AB’ A’B (AB)’ A+B
0
0
1
0
0
1
0
0
1
1
1
0
0
1
1
1
AB’+A’B (AB)’(A+B)
0
0
1
1
1
1
0
0
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’
W
= AB(A’C + B)
= ABA’C + ABB
= 0.BC + AB
= 0 + AB
= AB
CS1103
Chapter 3: Boolean Algebra/Logic
Gates
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
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’
Z = X + X’.Y + Y’.X’
Z = X + X’. (Y + Y’)
Z = X + X’. (1)
Z=1
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= 0+0
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
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
EXAMPLE
SIMPLIFY THE EXPRESSION
AB + A(B +C) + B(B + C) USING BOOLEAN ALGEBRA
1.
AB + AB + AC +BB + BC
2.
AB + AB + AC + B + BC
3.
AB + AC + B + BC
4.
AB + AC + B(1 + C)
5.
AB + AC + B.1
6.
AB + AC + B
7.
B(A + 1) +AC
8.
B.1 + AC
9.
B + AC
SIMPLIFY THE EXPRESSION
[AB’(C + BD)+ A’B’]C USING BOOLEAN ALGEBRA
1.
[AB’(C + BD)+ A’B’]C
2. (AB’C + AB’BD + A’B’)C
3. (AB’C + A.0.D + A’B’)C
4. (AB’C + 0 + A’B’)C
5. (AB’C + A’B’)C
6. AB’CC + A’B’C
7. AB’C + A’B’C
8. B’C(A + A’)
9. B’C.1
10.B’C
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’
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
associative
B Cin + A Cin + A B (Cin' + Cin)
B Cin + A Cin + A B (1)
B Cin + A Cin + A B
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
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
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.
x y z F1 F1'
0 0 0 0
1
Example: F1 = xyz'
0 0 1 0
1
0 1 0 0
1
Complement:
0 1 1 0
1
F1' = (x.y.z')'
1
= x' + y' + (z')' DeMorgan 1 0 0 0
1
= x' + y' + z
Involution 1 0 1 0
1 1 0 1
0
1 1 1 0
1
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'
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')
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:
Minterm
A'B'C'
A'B'C
A'BC'
A'BC
AB'C'
AB'C
ABC'
ABC
Minterm Code
000
001
010
011
100
101
110
111
Minterm Number
m0
m1
m2
m3
m4
m5
m6
m7
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:
Row No. Inputs
(i)
ABC
0
000
1
001
2
010
3
011
4
100
5
101
6
110
7
111
Outputs
f1(A,B,C)=Sm(2,3,6,7)
0
0
1 m2
1 m3
0
0
1 m6
1 m7
Complement
f1'(A,B,C)=Sm(0,1,4,5)
1 m0
1 m1
0
0
1 m4
1 m5
0
0
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 n 1
m
i
1
(2.6)
AB + (AB)' = 1 and AB + A' + B' = 1, but AB + A'B'  1.
i 0
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:
Maxterm
A+B+C
A+B+C'
A+B'+C
A+B'+C'
A'+B+C
A'+B+C'
A'+B'+C
A'+B'+C'
Maxterm Code
000
001
010
011
100
101
110
111
Maxterm Number
M0
M1
M2
M3
M4
M5
M6
M7
Algebraic Forms of Switching Functions (6)
f2(A,B,C) = M0M1M4M5
= PM(0,1,4,5) (maxterm list form)
The truth table for f2(A,B,C):
Rwo No. Inputs
M0
M1
M4
M5
Outputs
(i)
ABC A+B+C A+B+C' A'+B+C A'+B+C' f2(A,B,C)
0
000
0
1
1
1
0
1
001
1
0
1
1
0
2
010
1
1
1
1
1
3
011
1
1
1
1
1
4
100
1
1
0
1
0
5
101
1
1
1
0
0
6
110
1
1
1
1
1
7
111
1
1
1
1
1
(2.8)
(2.9)
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)
Row No.
(i)
0
1
2
3
4
5
6
7
Inputs
ABC
000
001
010
011
100
101
110
111
Outputs
f(A,B,C)= PM(1,3,5,7) = Sm(0,2,4,6)
1
m0
0
M1
1
m2
0
M3
1
m4
0
M5
1
m6
0
M7
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
(Mi)' = ((mi)')' = mi
(2.11)
(2.12)
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:
Row No. Inputs
(i)
ABC
0
000
1
001
2
010
3
011
4
100
5
101
6
110
7
111
Outputs
f (A,B,C)
1
0
1
0
1
0
1
0
Outputs
f '(A,B,C)= PM(0,2,4,6)
0
M0
1
0
M2
1
0
M4
1
0
M6
1
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  M  0
In general,  M  0
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)
2 3 1
i
i 0
2 1
n


i 0
i
(2.13)
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'
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) = Sm(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

Therefore,
f(A,B,C) = (M3M2M1M0)(M3M1) = PM(0, 1, 2, 3)
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) = Sm(0,3,7) + d(4,5)

Maxterm list is: f(A,B,C) = PM(1,2,6)·D(4,5)

f '(A,B,C) = Sm(1,2,6) + d(4,5) = PM(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)
Standard Forms
 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')
Standard Forms
 Every Boolean expression can either be expressed as
sum-of-products or product-of-sums expression.
Examples:
SOP:
POS:
both:
neither:
x.y + x.y + x.y.z
(x + y).(x + y).(x + z)
x + y + z or x.y.z
x.(w + y.z) or z + w.x.y + v.(x.z + w)
Two Level Canonical Forms
Sum of Products
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
Minterms
ABC =m
A B C = m0
A B C = m1
A B C = m2
A B C = m3
A B C = m4
A B C = m5
A B C = m6
7
Shorthand Notation for
Minterms of 3 Variables
B
C
F
A
2-Level AND/OR
Realization
product term / minterm:
ANDed product of literals in which each
variable appears exactly once, in true or
complemented form (but not both!)
F in canonical form:
F(A,B,C) = Sm(3,4,5,6,7)
= m3 + m4 + m5 + m6 + m7
= A' B C + A B' C' + A B' C
+ A B C' + A B C
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 + BC
F’ = (A + B C)' = A' (B' + C') = A' B' + A' C'
2
Level
Canonical
Forms
Product of Sums / Conjunctive Normal Form / Maxterm Expansion
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
Maxterms
A + B + C = M0
A + B + C = M1
A + B + C = M2
A + B + C = M3
A + B + C = M4
A + B + C = M5
A + B + C = M6
A + B + C = M7
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) = PM(0,1,2)
= (Sm(0,1,2))’ = (A’B’C’ + A’B’C + A’BC’)’
= (A + B + C) (A + B + C') (A + B' + C)
F’(A,B,C) = PM(3,4,5,6,7)
= (A + B' + C') (A' + B + C) (A' + B + C') (A' + B' + C) (A' + B' + C')
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