Circuit Simplification: Boolean
Algebra
Digital Electronics
© 2014 Project Lead The Way, Inc.
What is Boolean Algebra ?
Boolean Algebra is a mathematical technique that provides
the ability to algebraically simplify logic expressions. These
simplified expressions will result in a logic circuit that is
equivalent to the original circuit, yet requires fewer gates.
Z  BC  AB  AC
Simplification
With Boolean
Algebra
Z  A  BC
2
George Boole
My name is George Boole and I lived in
England in the 19th century. My work
on mathematical logic, algebra, and the
binary number system has had a
unique influence upon the
development of computers. Boolean
Algebra is named after me.
3
Boolean Theorems (1 of 7)
Single Variable - AND Function
Theorem #1
Theorem #2
Theorem #3
Theorem #4
X0  0
X 1 X
XX X
XX 0
X
0
0
X
1
X
X
X
X
X
0
X
X Y
Z
X Y
Z
X Y
Z
X Y
Z
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
1
0
1
0
0
1
0
0
1
0
0
1
0
0
1
1
1
1
1
1
1
1
1
1
1
1
4
Boolean Theorems (2 of 7)
Single Variable - OR Function
Theorem #5
Theorem #6
Theorem #7
Theorem #8
X0 X
X  1 1
XXX
X  X 1
X
0
X
X
1
1
X
X
X
X
1
X
X Y
Z
X Y
Z
X Y
Z
X Y
Z
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
0
1
1
0
1
1
1
0
1
1
0
1
1
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
5
Boolean Theorems (3 of 7)
Single Variable - Invert Function
Theorem #9
XX
X
X
X
X X X
0
1
0
1
0
1
6
Example #1: Boolean Algebra
Example
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in
SOP form.
F1  A A B  C C D
7
Example #1: Boolean Algebra
Example
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in
SOP form.
F1  A A B  C C D
Solution
F1  A A B  C C D
F1  A B  C C D
; Theorem #3
F1  A B  0 D
; Theorem #4
F1  A B  0
; Theorem #1
F1  A B
; Theorem #5
F1  A B
8
Example #2: Boolean Algebra
Example
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in
SOP form.
F2  B B C  B C C  A B 1  A B 0
9
Example #2: Boolean Algebra
Example
Simplify the following Boolean expression and note the Boolean
theorem used at each step. Put the answer in SOP form.
F2  B B C  B C C  A B 1  A B 0
Solution
F2  B B C  B C C  A B 1  A B 0
F2  B C  B C  A B 1  A B 0 ; Theorem #3 (twice)
F2 
BC
 A B 1  A B 0 ; Theorem #7
F2 
BC
 A B  A B 0 ; Theorem #2
F2 
BC
AB 
F2 
BC
AB
F2  B C  A B
0
; Theorem #1
; Theorem #5
10
Boolean Theorems (4 of 7)
Commutative Law
Theorem #10 – AND Function
X Y  Y  X
Theorem #11 – OR Function
X YY X
11
Boolean Theorems (5 of 7)
Associative Law
Theorem #12– AND Function
X (Y Z)  (X Y) Z
Theorem #13 – OR Function
X  (Y  Z)  (X  Y)  Z
12
Boolean Theorems (6 of 7)
Distributive Law
Theorem #14 – AND Function
X (Y  Z)  X Y  X Z
Theorem #15 – OR Function
( X  Y)(W  Z)  XW  XZ  YW YZ
13
Example #3: Boolean Algebra
Example
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in
SOP form.
F3  R T  (R  S)(R  T)
14
Example #3: Boolean Algebra
Example
Simplify the following Boolean expression and note the Boolean
theorem used at each step. Put the answer in SOP form.
F3  R T  (R  S)(R  T)
Solution
F3  R T  R  S R  T



F3  R T  R R  R T  S R  S T ; Theorem #15
F3  R T  0  R T  S R  S T
; Theorem #4
F3  R T  R T  S R  S T
; Theorem #5
F3  T R  R  S  S R
; Theorem #14
F3
; Theorem #8


 T 1  S   S R
F3  T1  S R
; Theorem #6
F3  T  S R
; Theorem #2
15
Boolean Theorems (7 of 7)
Consensus Theorem
Theorem #16
XXYXY
Theorem #18
XXYXY
Theorem #17
XXYXY
Theorem #19
XXYXY
16
Example #4: Boolean Algebra
Example
Simplify the following Boolean expression and note the
Boolean theorem used at each step. Put the answer in
SOP form.
F4  P S  P Q S  P Q S
17
Example #4: Boolean Algebra
Example
Simplify the following Boolean expression and note the Boolean
theorem used at each step. Put the answer in SOP form.
F4  P S  P Q S  P Q S
Solution
F4  P S  P Q S  P Q S


 P S  Q   P Q S
; Theorem #17
F4  P S  PQ  P Q S
; Theorem #14
F4  PS 1  Q   P Q
; Theorem #14
F4  PS 1   P Q
; Theorem #6
F4  PS  P Q
; Theorem #2
F4  P S  Q S  P Q S ; Theorem #14
F4
18
Multiple methods, same solution
• Were you able to reach the same solution,
but used a different method?
• Simplifying Boolean expressions can
involved many detailed steps or a few
combined steps.
• Let’s revisit Example #4 to see another
method of simplifying the expression
19
Example #4: Boolean Algebra
Example
Simplify the following Boolean expression and note the Boolean
theorem used at each step. Put the answer in SOP form.
F4  P S  P Q S  P Q S
Solution 1
F4  P S  P Q S  P Q S
Solution 2
F4  P S  P Q S  P Q S


 P S  Q   P Q S
; Theorem #14
; Theorem #17
F4  PS 1  P Q S
; Theorem #6
F4  P S  PQ  P Q S
; Theorem #14
F4  PS  P Q S
; Theorem #2
; Theorem #14
F4  P S  Q S
; Theorem #14
F4  P S  Q S  P Q S
F4
F4  PS 1  Q   P Q
F4  PS 1  Q   P Q S ; Theorem #14


F4  PS 1   P Q
; Theorem #6
F4  PS  Q
F4  PS  P Q
; Theorem #2
F4  PS  P Q

; Theorem #17
; Theorem #14
20
Summary
Boolean Theorems
1)
X 0  0
10) X  Y  Y  X
2)
X 1 X
11) X  Y  Y  X
3)
XX X
4)
XX0
5)
X0 X
6)
X  1 1
15) X  Y W  Z   XW  XZ  YW  YZ
7)
XXX
16) X  XY  X  Y
8)
X  X 1
17) X  X Y  X  Y
9)
XX
18) X  XY  X  Y
12) XYZ   XY Z
Commutative
Law
13) X  Y  Z    X  Y   Z
Associative
Law
14) XY  Z   XY  XZ
Distributive
Law
Consensus
Theorem
19) X  X Y  X  Y
21