Charles Kime & Thomas Kaminski
© 2008 Pearson Education, Inc.
Boolean Algebra
and Logic Gates
Logic and Computer Design Fundamentals
Boolean Algebra and Logic Gates 2
Binary Logic and Gates
Binary variables take on one of two values. Logical operators operate on binary values and
binary variables. Basic logical operators are the logic functions
AND, OR and NOT. Logic gates implement logic functions. Boolean Algebra: a useful mathematical system
for specifying and transforming logic functions. We study Boolean algebra as a foundation for
designing and analyzing digital systems!
Boolean Algebra and Logic Gates 3
Binary Variables Recall that the two binary values have
different names:• True/False• On/Off• Yes/No• 1/0
We use 1 and 0 to denote the two values. Variable identifier examples:
• A, B, y, z, or X1 for now• RESET, START_IT, or ADD1 later
Boolean Algebra and Logic Gates 4
Logical Operations The three basic logical operations are:
• AND • OR• NOT
AND is denoted by a dot (·). OR is denoted by a plus (+). NOT is denoted by an overbar ( ¯ ), a
single quote mark (') after, or (~) before the variable.
Boolean Algebra and Logic Gates 5
Examples:• is read “Y is equal to A AND B.”• is read “z is equal to x OR y.”• is read “X is equal to NOT A.”
Notation Examples
Note: The statement: 1 + 1 = 2 (read “one plus one equals two”)
is not the same as1 + 1 = 1 (read “1 or 1 equals 1”).
BAY yxz
AX
Boolean Algebra and Logic Gates 6
Operator Definitions
Operations are defined on the values "0" and "1" for each operator:
AND
0 · 0 = 00 · 1 = 01 · 0 = 01 · 1 = 1
OR
0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 1
NOT
1001
Boolean Algebra and Logic Gates 7
01
10
X
NOT
XZ
Truth Tables
Tabular listing of the values of a function for all possible combinations of values on its arguments
Example: Truth tables for the basic logic operations:
111
001
010
000
Z = X·YYX
AND OR
X Y Z = X+Y
0 0 0
0 1 1
1 0 1
1 1 1
Boolean Algebra and Logic Gates 8
Truth Tables – Cont’d
Used to evaluate any logic function Consider F(X, Y, Z) = X Y + Y Z
X Y Z X Y Y Y Z F = X Y + Y Z0 0 0 0 1 0 00 0 1 0 1 1 10 1 0 0 0 0 00 1 1 0 0 0 01 0 0 0 1 0 01 0 1 0 1 1 11 1 0 1 0 0 11 1 1 1 0 0 1
Boolean Algebra and Logic Gates 9
Using Switches• Inputs:
logic 1 is switch closed logic 0 is switch open
• Outputs: logic 1 is light on logic 0 is light off.
• NOT input: logic 1 is switch open logic 0 is switch closed
Logic Function Implementation
Switches in series => AND
Switches in parallel => OR
CNormally-closed switch => NOT
Boolean Algebra and Logic Gates 10
Example: Logic Using Switches
Light is on (L = 1) for L(A, B, C, D) =
and off (L = 0), otherwise.
Useful model for relay and CMOS gate circuits, the foundation of current digital logic circuits
Logic Function Implementation – cont’d
B
A
D
C
A (B C + D) = A B C + A D
Boolean Algebra and Logic Gates 11
Logic Gates
In the earliest computers, switches were opened
and closed by magnetic fields produced by
energizing coils in relays. The switches in turn
opened and closed the current paths.
Later, vacuum tubes that open and close current
paths electronically replaced relays.
Today, transistors are used as electronic
switches that open and close current paths.
Boolean Algebra and Logic Gates 12
Logic Gate Symbols and Behavior
Logic gates have special symbols:
And waveform behavior in time as follows:
X 0 0 1 1
Y 0 1 0 1
X · Y(AND) 0 0 0 1
X + Y(OR) 0 1 1 1
(NOT) X 1 1 0 0
OR gate
X
YZ = X + Y
X
YZ = X · Y
AND gate
X Z= X
NOT gate orinverter
Boolean Algebra and Logic Gates 13
Logic Diagrams and Expressions
Boolean equations, truth tables and logic diagrams describe the same function!
Truth tables are unique, but expressions and logic diagrams are not. This gives flexibility in implementing functions.
X
Y F
Z
Logic Diagram
Logic Equation
ZY X F
Truth Table
11 1 1
11 1 0
11 0 1
11 0 0
00 1 1
00 1 0
10 0 1
00 0 0
X Y Z Z Y X F
Boolean Algebra and Logic Gates 14
1.
3.
5.
7.
9.
11.
13.
15.
17.
Commutative
Associative
Distributive
DeMorgan’s
2.
4.
6.
8.
X . 1 X=
X . 0 0=
X . X X=
0=X . X
Boolean Algebra
10.
12.
14.
16.
X + Y Y + X=
(X + Y) Z+ X + (Y Z)+=X(Y + Z) XY XZ+=
X + Y X . Y=
XY YX=
(XY) Z X(YZ)=
X + YZ (X + Y) (X + Z)=
X . Y X + Y=
X + 0 X=
+X 1 1=
X + X X=
1=X + X
X = X
Invented by George Boole in 1854 An algebraic structure defined by a set B = {0, 1}, together with two
binary operators (+ and ·) and a unary operator ( )
Idempotence
Complement
Involution
Identity element
Boolean Algebra and Logic Gates 15
Some Properties of Boolean Algebra
Boolean Algebra is defined in general by a set B that can have more than two values
A two-valued Boolean algebra is also know as Switching Algebra. The Boolean set B is restricted to 0 and 1. Switching circuits can be represented by this algebra.
The dual of an algebraic expression is obtained by interchanging + by· and interchanging 0’s by1’s.
The identities appear in dual pairs. When there is only one identity on a line the identity is self-dual, i. e., the dual expression = the original expression.
Sometimes, the dot symbol ‘’ (AND operator) is not written when the meaning is clear
Boolean Algebra and Logic Gates 16
Example: F = (A + C) · B + 0
dual F = (A · C + B) · 1 = A · C + B Example: G = X · Y + (W + Z)
dual G = Example: H = A · B + A · C + B · C
dual H = Unless it happens to be self-dual, the dual of an
expression does not equal the expression itself Are any of these functions self-dual?
(A+B)(A+C)(B+C)=(A+BC)(B+C)=AB+AC+BC
Dual of a Boolean Expression
(X+Y) · (W · Z) = (X+Y) · (W+Z)
(A+B) · (A+C) · (B+C)
H is self-dual
Boolean Algebra and Logic Gates 17
Boolean Operator Precedence
The order of evaluation is:
1. Parentheses2. NOT3. AND4. OR
Consequence: Parentheses appear around OR expressions
Example: F = A(B + C)(C + D)
Boolean Algebra and Logic Gates 18
Boolean Algebraic Proof – Example 1
A + A · B = A (Absorption Theorem)Proof Steps Justification A + A · B= A · 1 + A · B Identity element: A · 1 = A
= A · ( 1 + B) Distributive
= A · 1 1 + B = 1
= A Identity element
Our primary reason for doing proofs is to learn:• Careful and efficient use of the identities and theorems of
Boolean algebra, and• How to choose the appropriate identity or theorem to apply
to make forward progress, irrespective of the application.
Boolean Algebra and Logic Gates 19
AB + AC + BC = AB + AC (Consensus Theorem)Proof Steps Justification= AB + AC + BC= AB + AC + 1 · BC Identity element= AB + AC + (A + A) · BC Complement= AB + AC + ABC + ABC Distributive= AB + ABC + AC + ACB Commutative= AB · 1 + ABC + AC · 1 + ACB Identity element= AB (1+C) + AC (1 + B) Distributive= AB . 1 + AC . 1 1+X = 1= AB + AC Identity element
Boolean Algebraic Proof – Example 2
Boolean Algebra and Logic Gates 20
Useful Theorems
MinimizationX Y + X Y = Y
AbsorptionX + X Y = X
SimplificationX + X Y = X + Y
DeMorgan’s X + Y = X · Y
Minimization (dual)(X+Y)(X+Y) = Y
Absorption (dual)X · (X + Y) = X
Simplification (dual)X · (X + Y) = X · Y
DeMorgan’s (dual) X · Y = X + Y
Boolean Algebra and Logic Gates 21
Truth Table to Verify DeMorgan’s
X Y X·Y X+Y X Y X+Y X · Y X·Y X+Y
0 0 0 0 1 1 1 1 1 10 1 0 1 1 0 0 0 1 11 0 0 1 0 1 0 0 1 11 1 1 1 0 0 0 0 0 0
X + Y = X · Y X · Y = X + Y
Generalized DeMorgan’s Theorem:
X1 + X2 + … + Xn = X1 · X2 · … · Xn
X1 · X2 · … · Xn = X1 + X2 + … + Xn
Boolean Algebra and Logic Gates 22
Complementing Functions
Use DeMorgan's Theorem:
1. Interchange AND and OR operators
2. Complement each constant and literal
Example: Complement F =
F = (x + y + z)(x + y + z) Example: Complement G = (a + bc)d + e
G = (a (b + c) + d) e
x zyzyx
Boolean Algebra and Logic Gates 23
Expression Simplification An application of Boolean algebra Simplify to contain the smallest number
of literals (variables that may or may not be complemented)
= AB + ABCD + A C D + A C D + A B D
= AB + AB(CD) + A C (D + D) + A B D
= AB + A C + A B D = B(A + AD) +AC
= B (A + D) + A C (has only 5 literals)
DCBADCADBADCABA
Boolean Algebra and Logic Gates 24
Next … Canonical Forms
Minterms and Maxterms
Sum-of-Minterm (SOM) Canonical Form
Product-of-Maxterm (POM) Canonical Form
Representation of Complements of Functions
Conversions between Representations
Boolean Algebra and Logic Gates 25
Minterms
Minterms are AND terms with every variable present in either true or complemented form.
Given that each binary variable may appear normal (e.g., x) or complemented (e.g., ), there are 2n minterms for n variables.
Example: Two variables (X and Y) produce2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented)
Thus there are four minterms of two variables.
YXXY
YXYX
x
Boolean Algebra and Logic Gates 26
Maxterms
Maxterms are OR terms with every variable in true or complemented form.
Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2n maxterms for n variables.
Example: Two variables (X and Y) produce2 x 2 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented)
YX YX YX YX
Boolean Algebra and Logic Gates 27
Two variable minterms and maxterms.
The minterm mi should evaluate to 1 for each combination of x and y.
The maxterm is the complement of the minterm
Minterms & Maxterms for 2 variables
x y Index Minterm Maxterm
0 0 0 m0 = x y M0 = x + y
0 1 1 m1 = x y M1 = x + y
1 0 2 m2 = x y M2 = x + y
1 1 3 m3 = x y M3 = x + y
Boolean Algebra and Logic Gates 28
Minterms & Maxterms for 3 variables
M3 = x + y + zm3 = x y z3110M4 = x + y + zm4 = x y z4001M5 = x + y + zm5 = x y z5101M6 = x + y + zm6 = x y z6011
1
100y
1
000x
1
010z
M7 = x + y + zm7 = x y z7
M2 = x + y + zm2 = x y z2
M1 = x + y + zm1 = x y z1
M0 = x + y + zm0 = x y z0MaxtermMintermIndex
Maxterm Mi is the complement of minterm mi
Mi = mi and mi = Mi
Boolean Algebra and Logic Gates 29
Purpose of the Index
Minterms and Maxterms are designated with an index
The index number corresponds to a binary pattern
The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true or complemented form
For Minterms:• ‘1’ means the variable is “Not Complemented” and
• ‘0’ means the variable is “Complemented”.
For Maxterms:• ‘0’ means the variable is “Not Complemented” and
• ‘1’ means the variable is “Complemented”.
Boolean Algebra and Logic Gates 30
Standard Order All variables should be present in a minterm or
maxterm and should be listed in the same order (usually alphabetically)
Example: For variables a, b, c:
• Maxterms (a + b + c), (a + b + c) are in standard order
• However, (b + a + c) is NOT in standard order
(a + c) does NOT contain all variables
• Minterms (a b c) and (a b c) are in standard order
• However, (b a c) is not in standard order
(a c) does not contain all variables
Boolean Algebra and Logic Gates 31
Sum-Of-Minterm (SOM) Sum-Of-Minterm (SOM) canonical form:
Sum of minterms of entries that evaluate to ‘1’
x y z F Minterm0 0 0 0
0 0 1 1 m1 = x y z
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1 m6 = x y z
1 1 1 1 m7 = x y z
F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z
Focus on the ‘1’ entries
Boolean Algebra and Logic Gates 32
+ a b c d
Sum-Of-Minterm Examples
F(a, b, c, d) = ∑(2, 3, 6, 10, 11)
F(a, b, c, d) = m2 + m3 + m6 + m10 + m11
G(a, b, c, d) = ∑(0, 1, 12, 15)
G(a, b, c, d) = m0 + m1 + m12 + m15
+ a b c da b c d + a b c d+ a b c d + a b c d
+ a b c da b c d + a b c d
Boolean Algebra and Logic Gates 33
Product-Of-Maxterm (POM) Product-Of-Maxterm (POM) canonical form:
Product of maxterms of entries that evaluate to ‘0’
x y z F Maxterm0 0 0 1
0 0 1 1
0 1 0 0 M2 = (x + y + z)
0 1 1 1
1 0 0 0 M4 = (x + y + z)
1 0 1 1
1 1 0 0 M6 = (x + y + z)
1 1 1 1
Focus on the ‘0’ entries
F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)
Boolean Algebra and Logic Gates 34
F(a, b, c, d) = ∏(1, 3, 6, 11)
F(a, b, c, d) = M1 · M3 · M6 · M11
G(a, b, c, d) = ∏(0, 4, 12, 15)
G(a, b, c, d) = M0 · M4 · M12 · M15
Product-Of-Maxterm Examples
(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)
(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)
Boolean Algebra and Logic Gates 35
Observations
We can implement any function by "ORing" the minterms corresponding to the ‘1’ entries in the function table. A minterm evaluates to ‘1’ for its corresponding entry.
We can implement any function by "ANDing" the maxterms corresponding to ‘0’ entries in the function table. A maxterm evaluates to ‘0’ for its corresponding entry.
The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Product-of-Maxterms (POM).
If a Boolean function has fewer ‘1’ entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer ‘0’ entries then the POM form will have fewer literals than SOM.
Boolean Algebra and Logic Gates 36
Converting to Sum-of-Minterms Form
A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table
Consider F = y + x z
x y z F Minterm0 0 0 1 m0 = x y z
0 0 1 1 m1 = x y z
0 1 0 1 m2 = x y z
0 1 1 0
1 0 0 1 m4 = x y z
1 0 1 1 m5 = x y z
1 1 0 0
1 1 1 0
F = ∑(0, 1, 2, 4, 5) =
m0 + m1 + m2 + m4 + m5 =
x y z + x y z + x y z +
x y z + x y z
Boolean Algebra and Logic Gates 37
Converting to Product-of-Maxterms Form
A function that is not in the Product-of-Maxterms form can be converted to that form by means of a truth table
Consider again: F = y + x z
x y z F Maxterm0 0 0 1
0 0 1 1
0 1 0 1
0 1 1 0 M3 = (x+y+z)
1 0 0 1
1 0 1 1
1 1 0 0 M6 = (x+y+z)
1 1 1 0 M7 = (x+y+z)
F = ∏(3, 6, 7) =
M3 · M6 · M7 =
(x+y+z) (x+y+z) (x+y+z)
Boolean Algebra and Logic Gates 38
Conversions Between Canonical Forms
F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) =
x y z + x y z + x y z + x y z + x y z
F = M0 · M4 · M6 = ∏(0, 4, 6) =(x+y+z)(x+y+z)(x+y+z)
x y z F Minterm Maxterm0 0 0 0 M0 = (x + y + z)
0 0 1 1 m1 = x y z
0 1 0 1 m2 = x y z
0 1 1 1 m3 = x y z
1 0 0 0 M4 = (x + y + z)
1 0 1 1 m5 = x y z
1 1 0 0 M6 = (x + y + z)
1 1 1 1 m7 = x y z
Boolean Algebra and Logic Gates 39
Algebraic Conversion to Sum-of-Minterms
Expand all terms first to explicitly list all minterms AND any term missing a variable v with (v + v) Example 1: f = x + x y (2 variables)
f = x (y + y) + x yf = x y + x y + x yf = m3 + m2 + m0 = ∑(0, 2, 3)
Example 2: g = a + b c (3 variables)g = a (b + b)(c + c) + (a + a) b cg = a b c + a b c + a b c + a b c + a b c + a b cg = a b c + a b c + a b c + a b c + a b cg = m1 + m4 + m5 + m6 + m7 = ∑ (1, 4, 5, 6, 7)
Boolean Algebra and Logic Gates 40
Algebraic Conversion to Product-of-Maxterms
Expand all terms first to explicitly list all maxterms OR any term missing a variable v with v · v Example 1: f = x + x y (2 variables)
Apply 2nd distributive law:
f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M1
Example 2: g = a c + b c + a b (3 variables)
g = (a c + b c + a) (a c + b c + b) (distributive)
g = (c + b c + a) (a c + c + b) (x + x y = x + y)
g = (c + b + a) (a + c + b) (x + x y = x + y)
g = (a + b + c) (a + b + c) = M5 . M2 = ∏ (2, 5)
Boolean Algebra and Logic Gates 41
Function Complements
The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical form
Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices
Example: Given F(x, y, z) = ∑ (1, 3, 5, 7)
F(x, y, z) = ∑ (0, 2, 4, 6)
F(x, y, z) = ∏ (1, 3, 5, 7)
Boolean Algebra and Logic Gates 42
Summary of Minterms and Maxterms
There are 2n minterms and maxterms for Boolean functions with n variables.
Minterms and maxterms are indexed from 0 to 2n – 1
Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms
The complement of a function contains those minterms not included in the original function
The complement of a sum-of-minterms is a product-of-maxterms with the same indices
Boolean Algebra and Logic Gates 43
Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms
Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms
Examples:• SOP:
• POS:
These “mixed” forms are neither SOP nor POS•
•
Standard Forms
B C B A C B A C · )C B(A · B) (A
C) (A C) B (A B) (A C A C B A
Boolean Algebra and Logic Gates 44
Standard Sum-of-Products (SOP)
A sum of minterms form for n variables can be written down directly from a truth table.
• Implementation of this form is a two-level network of gates such that:
• The first level consists of n-input AND gates
• The second level is a single OR gate
This form often can be simplified so that the corresponding circuit is simpler.
Boolean Algebra and Logic Gates 45
A Simplification Example: Writing the minterm expression:
F = A B C + A B C + A B C + ABC + ABC Simplifying:
F = A B C + A (B C + B C + B C + B C)F = A B C + A (B (C + C) + B (C + C))F = A B C + A (B + B)F = A B C + AF = B C + A
Simplified F contains 3 literals compared to 15
Standard Sum-of-Products (SOP)
)7,6,5,4,1()C,B,A(F
Boolean Algebra and Logic Gates 46
AND/OR Two-Level Implementation
The two implementations for F are shown below
F
B
C
A
It is quite apparent which
is simpler!
Boolean Algebra and Logic Gates 47
SOP and POS Observations The previous examples show that:
• Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard forms (SOP, POS) differ in complexity
• Boolean algebra can be used to manipulate equations into simpler forms
• Simpler equations lead to simpler implementations