2. Chapter 2 - Part 1 2
Overview
Part 1 – Gate Circuits and Boolean Equations
• Binary Logic and Gates
• Boolean Algebra
• Standard Forms
Part 2 – Circuit Optimization
• Two-Level Optimization
• Map Manipulation
• Multi-Level Circuit Optimization
Part 3 – Additional Gates and Circuits
• Other Gate Types
• Exclusive-OR Operator and Gates
• High-Impedance Outputs
3. Chapter 2 - Part 1 3
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 foundation for
designing and analyzing digital systems!
4. Chapter 2 - Part 1 4
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
5. Chapter 2 - Part 1 5
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.
6. Chapter 2 - Part 1 6
Examples:
• is read “Y is equal to AAND 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 as
1 + 1 = 1 (read “1 or 1 equals 1”).
B
A
Y
y
x
z
A
X
7. Chapter 2 - Part 1 7
Operator Definitions
Operations are defined on the values
"0" and "1" for each operator:
AND
0 · 0 = 0
0 · 1 = 0
1 · 0 = 0
1 · 1 = 1
OR
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 1
NOT
1
0
0
1
8. Chapter 2 - Part 1 8
0
1
1
0
X
NOT
X
Z
Truth Tables
Truth table a 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:
1
1
1
0
0
1
0
1
0
0
0
0
Z = X·Y
Y
X
AND OR
X Y Z = X+Y
0 0 0
0 1 1
1 0 1
1 1 1
9. Chapter 2 - Part 1 9
Using Switches
• For inputs:
logic 1 is switch closed
logic 0 is switch open
• For outputs:
logic 1 is light on
logic 0 is light off.
• NOT uses a switch such
that:
logic 1 is switch open
logic 0 is switch closed
Logic Function Implementation
Switches in series => AND
Switches in parallel => OR
C
Normally-closed switch => NOT
10. Chapter 2 - Part 1 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 circuits and for CMOS
gate circuits, the foundation of current digital
logic technology
Logic Function Implementation (Continued)
B
A
D
C
11. Chapter 2 - Part 1 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.
12. Chapter 2 - Part 1 12
Logic Gates (continued)
Implementation of logic gates with transistors (See
Reading Supplement CMOS Circuits)
Transistor or tube implementations of logic functions are called logic gates or just
gates
Transistor gate circuits can be modeled by switch circuits
•
F
+V
X
Y
+V
X
+V
X
Y
•
•
•
•
•
• •
•
• •
•
•
(a) NOR
G = X +Y
(b) NAND (c) NOT
X .Y
X
•
•
•
13. Chapter 2 - Part 1 13
Logic Gate Symbols and Behavior
Logic gates have special symbols:
And waveform behavior in time as follows:
(b) Timing diagram
X 0 0 1 1
Y 0 1 0 1
X · Y
(AND) 0 0 0 1
X 1 Y
(OR) 0 1 1 1
(NOT) X 1 1 0 0
(a) Graphic symbols
OR gate
X
Y
Z= X+ Y
X
Y
Z= X · Y
AND gate
X Z= X
NOT gate or
inverter
14. Chapter 2 - Part 1 14
Logic Diagrams and Expressions
Boolean equations, truth tables and logic diagrams describe
the same function!
X
Y F
Z
Logic Diagram
Equation
Z
Y
X
F
Truth Table
1
1 1 1
1
1 1 0
1
1 0 1
1
1 0 0
0
0 1 1
0
0 1 0
1
0 0 1
0
0 0 0
X Y Z Z
Y
X
F
15. Chapter 2 - Part 1 15
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
An algebraic structure defined on a set of at least two elements, B,
together with three binary operators (denoted +, · and ) that
satisfies the following basic identities:
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
16. Chapter 2 - Part 1 16
The identities above are organized into pairs. These pairs
have names as follows:
1-4 Existence of 0 and 1 5-6 Idempotence
7-8 Existence of complement 9 Involution
10-11 Commutative Laws 12-13 Associative Laws
14-15 Distributive Laws 16-17 DeMorgan’s Laws
Some Properties of Identities & the Algebra
The dual of an algebraic expression is obtained by
interchanging + and · and interchanging 0’s and 1’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.
17. Chapter 2 - Part 1 17
Unless it happens to be self-dual, the dual of an
expression does not equal the expression itself.
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 =
Are any of these functions self-dual?
Some Properties of Identities & the Algebra
(Continued)
18. Chapter 2 - Part 1 18
There can be more that 2 elements in B, i. e.,
elements other than 1 and 0. What are some
common useful Boolean algebras with more
than 2 elements?
1.
2.
If B contains only 1 and 0, then B is called the
switching algebra which is the algebra we use
most often.
Some Properties of Identities & the Algebra
(Continued)
Algebra of Sets
Algebra of n-bit binary vectors
19. Chapter 2 - Part 1 19
Boolean Operator Precedence
The order of evaluation in a Boolean
expression is:
1. Parentheses
2. NOT
3. AND
4. OR
Consequence: Parentheses appear
around OR expressions
Example: F = A(B + C)(C + D)
20. Chapter 2 - Part 1 20
Example 1: Boolean Algebraic Proof
A + A·B = A (Absorption Theorem)
Proof Steps Justification (identity or theorem)
A + A·B
= A · 1 + A · B X = X · 1
= A · ( 1 + B) X · Y + X · Z = X ·(Y + Z)(Distributive Law)
= A · 1 1 + X = 1
= A X · 1 = X
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.
21. Chapter 2 - Part 1 21
AB + AC + BC = AB + AC (Consensus Theorem)
Proof Steps Justification (identity or
theorem)
AB + AC + BC
= AB + AC + 1 · BC ?
= AB +AC + (A + A) · BC ?
=
Example 2: Boolean Algebraic Proofs
22. Chapter 2 - Part 1 22
Example 3: Boolean Algebraic Proofs
Proof Steps Justification (identity or
theorem)
=
Y
X
Z
)
Y
X
(
)
Z
X
(
X
Z
)
Y
X
(
Y Y
23. Chapter 2 - Part 1 23
x y
y
Useful Theorems
n
inimizatio
M
y
y
y
x
y
y
y
x
tion
Simplifica
y
x
y
x
y
x
y
x
Absorption
x
y
x
x
x
y
x
x
Consensus
z
y
x
z
y
z
y
x
z
y
x
z
y
z
y
x
Laws
s
DeMorgan'
x
x
x x
x x
x x
x x
y x
y
24. Chapter 2 - Part 1 24
Proof of Simplification
y
y
y
x
y
y
y
x
x
x
25. Chapter 2 - Part 1 25
Proof of DeMorgan’s Laws
y
x x
y
y
x y
x
26. Chapter 2 - Part 1 26
Boolean Function Evaluation
x y z F1 F2 F3 F4
0 0 0 0 0
0 0 1 0 1
0 1 0 0 0
0 1 1 0 0
1 0 0 0 1
1 0 1 0 1
1 1 0 1 1
1 1 1 0 1
z
x
y
x
F4
x
z
y
x
z
y
x
F3
x
F2
xy
F1
z
yz
y
27. Chapter 2 - Part 1 27
Expression Simplification
An application of Boolean algebra
Simplify to contain the smallest number
of literals (complemented and
uncomplemented variables):
28. Chapter 2 - Part 1 28
Complementing Functions
Use DeMorgan's Theorem to complement
a function:
1. Interchange AND and OR operators
2. Complement each constant value and
literal
Example: Complement F =
F = (x + y + z)(x + y + z)
Example: Complement G = (a + bc)d + e
G =
x
z
y
z
y
x
29. Chapter 2 - Part 1 29
Overview – Canonical Forms
What are Canonical Forms?
Minterms and Maxterms
Index Representation of Minterms and
Maxterms
Sum-of-Minterm (SOM) Representations
Product-of-Maxterm (POM) Representations
Representation of Complements of Functions
Conversions between Representations
30. Chapter 2 - Part 1 30
Canonical Forms
It is useful to specify Boolean functions in
a form that:
• Allows comparison for equality.
• Has a correspondence to the truth tables
Canonical Forms in common usage:
• Sum of Minterms (SOM)
• Product of Maxterms (POM)
31. Chapter 2 - Part 1 31
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)produce
2 x 2 = 4 combinations:
(both normal)
(X normal, Y complemented)
(X complemented, Y normal)
(both complemented)
Thus there are four minterms of two variables.
Y
X
XY
Y
X
Y
X
x
32. Chapter 2 - Part 1 32
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) produce
2 x 2 = 4 combinations:
(both normal)
(x normal, y complemented)
(x complemented, y normal)
(both complemented)
Y
X
Y
X
Y
X
Y
X
33. Chapter 2 - Part 1 33
Examples: Two variable minterms and
maxterms.
The index above is important for describing
which variables in the terms are true and which
are complemented.
Maxterms and Minterms
Index Minterm Maxterm
0 x y x + y
1 x y x + y
2 x y x + y
3 x y x + y
34. Chapter 2 - Part 1 34
Standard Order
Minterms and maxterms are designated with a subscript
The subscript is a number, corresponding to a binary pattern
The bits in the pattern represent the complemented or
normal state of each variable listed in a standard order.
All variables will be present in a minterm or maxterm and
will be listed in the same order (usually alphabetically)
Example: For variables a, b, c:
• Maxterms: (a + b + c), (a + b + c)
• Terms: (b + a + c), a c b, and (c + b + a) are NOT in
standard order.
• Minterms: a b c, a b c, a b c
• Terms: (a + c), b c, and (a + b) do not contain all
variables
35. Chapter 2 - Part 1 35
Purpose of the Index
The index for the minterm or maxterm,
expressed as a binary number, is used to
determine whether the variable is shown in the
true form 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”.
36. Chapter 2 - Part 1 36
Index Example in Three Variables
Example: (for three variables)
Assume the variables are called X, Y, and Z.
The standard order is X, then Y, then Z.
The Index 0 (base 10) = 000 (base 2) for three
variables). All three variables are complemented
for minterm 0 ( ) and no variables are
complemented for Maxterm 0 (X,Y,Z).
• Minterm 0, called m0
is .
• Maxterm 0, called M0
is (X + Y + Z).
• Minterm 6 ?
• Maxterm 6 ?
Z
,
Y
,
X
Z
Y
X
37. Chapter 2 - Part 1 37
Index Examples – Four Variables
Index Binary Minterm Maxterm
i Pattern mi Mi
0 0000
1 0001
3 0011
5 0101
7 0111
10 1010
13 1101
15 1111
d
c
b
a d
c
b
a
d
c
b
a
d
c
b
a
d
c
b
a d
c
b
a
d
c
b
a
d
c
b
a d
c
b
a
d
b
a
d
c
b
a d
c
b
a
?
?
?
?
c
38. Chapter 2 - Part 1 38
Review: DeMorgan's Theorem
and
Two-variable example:
and
Thus M2 is the complement of m2 and vice-versa.
Since DeMorgan's Theorem holds for n variables,
the above holds for terms of n variables
giving:
and
Thus Mi is the complement of mi.
Minterm and Maxterm Relationship
y
x
y
·
x
y
x
y
x
y
x
M2
y
x·
m2
i m
M i i
i M
m
39. Chapter 2 - Part 1 39
Function Tables for Both
Minterms of Maxterms of
2 variables 2 variables
Each column in the maxterm function table is the
complement of the column in the minterm function
table since Mi is the complement of mi.
x y m0 m1 m2 m3
0 0 1 0 0 0
0 1 0 1 0 0
1 0 0 0 1 0
1 1 0 0 0 1
x y M0 M1 M2 M3
0 0 0 1 1 1
0 1 1 0 1 1
1 0 1 1 0 1
1 1 1 1 1 0
40. Chapter 2 - Part 1 40
Observations
In the function tables:
• Each minterm has one and only one 1 present in the 2n terms (a minimum of
1s). All other entries are 0.
• Each maxterm has one and only one 0 present in the 2n terms All other
entries are 1 (a maximum of 1s).
We can implement any function by "ORing" the minterms
corresponding to "1" entries in the function table. These are called
the minterms of the function.
We can implement any function by "ANDing" the maxterms
corresponding to "0" entries in the function table. These are called
the maxterms of the function.
This gives us two canonical forms:
• Sum of Minterms (SOM)
• Product of Maxterms (POM)
for stating any Boolean function.
41. Chapter 2 - Part 1 41
x y z index m1 + m4 + m7 = F1
0 0 0 0 0 + 0 + 0 = 0
0 0 1 1 1 + 0 + 0 = 1
0 1 0 2 0 + 0 + 0 = 0
0 1 1 3 0 + 0 + 0 = 0
1 0 0 4 0 + 1 + 0 = 1
1 0 1 5 0 + 0 + 0 = 0
1 1 0 6 0 + 0 + 0 = 0
1 1 1 7 0 + 0 + 1 = 1
Minterm Function Example
Example: Find F1 = m1 + m4 + m7
F1 = x y z + x y z + x y z
43. Chapter 2 - Part 1 43
Maxterm Function Example
Example: Implement F1 in maxterms:
F1 = M0 · M2 · M3 · M5 · M6
)
z
y
z)·(x
y
·(x
z)
y
(x
F1
z)
y
x
)·(
z
y
x
·(
x y z i M0 M2 M3 M5 M6 = F1
0 0 0 0 0 1 1 1 = 0
0 0 1 1 1 1 1 1 1 = 1
0 1 0 2 1 0 1 1 1 = 0
0 1 1 3 1 1 0 1 1 = 0
1 0 0 4 1 1 1 1 1 = 1
1 0 1 5 1 1 1 0 1 = 0
1 1 0 6 1 1 1 1 0 = 0
1 1 1 7 1
1 1 1 1 = 1
1
44. Chapter 2 - Part 1 44
Maxterm Function Example
F(A, B,C,D) =
14
11
8
3 M
M
M
M
)
D
,
C
,
B
,
A
(
F
45. Chapter 2 - Part 1 45
Canonical Sum of Minterms
Any Boolean function can be expressed as a
Sum of Minterms.
• For the function table, the minterms used are the
terms corresponding to the 1's
• For expressions, expand all terms first to explicitly
list all minterms. Do this by “ANDing” any term
missing a variable v with a term ( ).
Example: Implement as a sum of
minterms.
First expand terms:
Then distribute terms:
Express as sum of minterms: f = m3 + m2 + m0
y
x
x
f
y
x
)
y
y
(
x
f
y
x
y
x
xy
f
v
v
46. Chapter 2 - Part 1 46
Another SOM Example
Example:
There are three variables, A, B, and C which we
take to be the standard order.
Expanding the terms with missing variables:
Collect terms (removing all but one of duplicate
terms):
Express as SOM:
C
B
A
F
47. Chapter 2 - Part 1 47
Shorthand SOM Form
From the previous example, we started with:
We ended up with:
F = m1+m4+m5+m6+m7
This can be denoted in the formal shorthand:
Note that we explicitly show the standard
variables in order and drop the “m”
designators.
)
7
,
6
,
5
,
4
,
1
(
)
C
,
B
,
A
(
F m
C
B
A
F
48. Chapter 2 - Part 1 48
Canonical Product of Maxterms
Any Boolean Function can be expressed as a Product of
Maxterms (POM).
• For the function table, the maxterms used are the terms
corresponding to the 0's.
• For an expression, expand all terms first to explicitly list all
maxterms. Do this by first applying the second distributive law ,
“ORing” terms missing variable v with a term equal to and then
applying the distributive law again.
Example: Convert to product of maxterms:
Apply the distributive law:
Add missing variable z:
Express as POM: f = M2 · M3
y
x
x
)
z
,
y
,
x
(
f
y
x
)
y
(x
1
)
y
)(x
x
(x
y
x
x
z
y
x
)
z
y
x
(
z
z
y
x
v
v
49. Chapter 2 - Part 1 49
Convert to Product of Maxterms:
Use x + y z = (x+y)·(x+z) with ,
and to get:
Then use to get:
and a second time to get:
Rearrange to standard order,
to give f = M5
· M2
Another POM Example
B
A
C
B
C
A
C)
B,
f(A,
B
z
)
B
C
B
C
)(A
A
C
B
C
(A
f
y
x
y
x
x
)
B
C
C
)(A
A
BC
C
(
f
)
B
C
)(A
A
B
C
(
f
C)
B
)(A
C
B
A
(
f
A
y
C),
B
(A
x
C
50. Chapter 2 - Part 1 50
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 forms.
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 )
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F m
)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F m
)
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F M
51. Chapter 2 - Part 1 51
Conversion Between Forms
To convert between sum-of-minterms and product-of-
maxterms form (or vice-versa) we follow these steps:
• Find the function complement by swapping terms in the list
with terms not in the list.
• Change from products to sums, or vice versa.
Example:Given F as before:
Form the Complement:
Then use the other form with the same indices – this
forms the complement again, giving the other form of
the original function:
)
7
,
5
,
3
,
1
(
)
z
,
y
,
x
(
F m
)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F m
)
6
,
4
,
2
,
0
(
)
z
,
y
,
x
(
F M
52. Chapter 2 - Part 1 52
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
53. Chapter 2 - Part 1 53
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,
and
• The second level is a single OR gate (with
fewer than 2n
inputs).
This form often can be simplified so that
the corresponding circuit is simpler.
54. Chapter 2 - Part 1 54
A Simplification Example:
Writing the minterm expression:
F = A B C + A B C + A B C + ABC + ABC
Simplifying:
F =
Simplified F contains 3 literals compared to 15 in
minterm F
Standard Sum-of-Products (SOP)
)
7
,
6
,
5
,
4
,
1
(
m
)
C
,
B
,
A
(
F
55. Chapter 2 - Part 1 55
AND/OR Two-level Implementation
of SOP Expression
The two implementations for F are shown
below – it is quite apparent which is simpler!
F
B
C
A
56. Chapter 2 - Part 1 56
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 two-level
implementations
Questions:
• How can we attain a “simplest” expression?
• Is there only one minimum cost circuit?
• The next part will deal with these issues.
58. Chapter 1 58
This Summary is an Online Content from this Book:
Morris Mano, DIGITAL DESIGN, 4th Edition, Prentice Hall, 2007
It is edited for
Logic Analysis and Design Course
6803213-3
by:
T.Mariah Sami Khayat
Teacher Assistant @ Adam University College
For Contacting:
mskhayat@uqu.edu.sa
Kingdom of Saudi Arabia
Ministry of Education
Umm AlQura University
Adam University College
Computer Science Department
السعودية العربية المملكة
التعليم وزارة
القرى أم جامعة
أضم الجامعية الكلية
اآللي الحاسب قسم
Editor's Notes
#10:L (A, B, C, D) = A ((B C') + D) = A B C' + A D
Complementary metal–oxide–semiconductor = CMOS = is a technology for constructing integrated circuits.
أكسيد معدني متمم أشباه الموصلات
#12:The transistor without the “bubble” on its input is an N-type field effect
transistor. It acts like a closed switch between its top and bottom terminals
with an H (1) applied to its input on its left. It acts like an open switch with
an L (0) applied to its input. The transistor with the “bubble” on its input is a
P-type field effect transistor. The +V at the top provides an H (1) and the
Ground symbol at the bottom provides an L (0). By modeling the two types
of field effect transistors as switches, one can see how the series and
parallel interconnections can produce 1’s and 0’s on the outputs on the
right in response to applied 1’s and 0’s on the inputs on the left. NOR and
NAND are OR and AND, each followed by a NOT respectively.
#17:Dual G = ((X+Y) · (W · Z)') = ((X+Y) ·(W' + Z')
Dual H = (A + B)(A + C)(B + C). Using the Boolean identities,
= (A +BC) (B+C) = AB + AC + BC. So H is self-dual.
-Distributive Identity then
AB + AC + BBC + BCC = //X.X = X
AB + AC + BC + BC = //X+X = X
AB + AC + BC
#18:Algebra of Sets correspondence to switching algebra: Set - variable,
Union - OR, Intersect AND ., Universe - 1, Empty set - 0,
Complement of Set - NOT, Subset – AND of variables and variable complements.
Algebra of n-bit binary vectors correspondence to switching algebra:
{n-bit binary vectors} – {0,1}, bitwise OR - OR, bitwise AND - AND,
bitwise NOT - NOT, Vector of all 1s – 1, Vector of all 0s – 0.
#20:Absorption = نظرية الامتصاص
Justification مبرر أو مسوغ =
Identities = المسلمات
#21: Justification 1: 1 . X = X
Justification 2: X + X’ = 1
= AB + A’C + ABC + A’BC X(Y + Z) = XY + XZ (Distributive Law)
= AB + ABC + A’C + A’BC X + Y = Y + X (Commutative Law)
= AB . 1 + ABC + A’C . 1 + A’C . B X . 1 = X, X . Y = Y . X (Commutative Law)
= AB (1 + C) + A’C (1 + B) X(Y + Z) = XY +XZ (Distributive Law)
= AB . 1 + A’C . 1 = AB + A’C X . 1 = X
إجماع = Consensus
#22: = X’ Y’ Z + X Y’ (A + B)’ = A’ . B’ (DeMorgan’s Law)
= Y’ X’ Z + Y’ X A . B = B . A (Commutative Law)
= Y’ (X’ Z + X) A(B + C) = AB + AC (Distributive Law)
= Y’ (X’ + X)(Z + X) A + BC = (A + B)(A + C) (Distributive Law)
= Y’ . 1 . (Z + X) A + A’ = 1
= Y’ (X + Z) 1 . A = A, A + B = B + A (Commutative Law)
#24: x . y + x’ . y
= (x + x’) . y X(Y + Z) = XY + XZ (Distributive Law)
= 1 . y X + X’ = 1
= y X . 1 = X
x + y . x’ + y X+(YZ) = XY + XZ (Distributive Law)
= (x . x’) + y X . X’ = 0
= 0 + y X + 0 = X
= y
#25:It is important that we do not USE DeMorgan’s Laws in doing this proof.
This requires a different proof method. We will show that, x’ . y’, satisfies
the definition of the complement of (x + y), defined as (x + y)’ by
DeMorgan’s Law.
To show this we need to show that A + A’ = 1 and A.A’ = 0 with
A = x + y and A’ = x’. y’. This proves that x’. y’ = (x + y)’.
(if x+y is a complement of x’.y’, then (x+y)’ is equal to x’.y’ ) That’s the idea ;)
Part 1: Show x + y + x’. y’ = 1.
x + y + x’. y’
= (x + y + x’) (x + y + y’) X + YZ = (X + Y)(X + Z) (Distributive Law)
= (x + x’ + y) (x + y + y’) X + Y = Y + X (Commutative Law)
= (1 + y)(x + 1) X + X’ = 1
= 1 . 1 1 + X = 1
= 1 1 . X = 1
Part 2: Show (x + y) . x’. y’ = 0.
(x + y) . x’. y’
= (x . x’. y’ + y . x’. y’) X (Y + Z) = XY + XZ (Distributive Law)
= (x . x’. y’ + y . y’ . x’) XY = YX (Commutative Law)
= (0 . y’ + 0 . x’) X . X’ = 0
= (0 + 0) 0 . X = 0
= 0 X + 0 = X (With X = 0)
Based on the above two parts, x’y’ = (x + y)’
The second DeMorgans’ law is proved by duality. Note that DeMorgan’s
Law, given as an identity is not an axiom in the sense that it can be
proved using the other identities.
#26: F3 is 1 for x’y’z’, x’yz, xy’z’ and xy’z => F3 = 1,0,0,1,1,1,0,0
F4 is 1 for xy’z’, xy’z, x’y’z and x’y z => F4 = 0,1,0,1,1,1,0,0