1. Boolean Algebra

2. Boolean Algebra Invented by George Boole in 1854
An algebraic structure defined by a set {0, 1}, together with two binary operators (+ and ·) and a unary operator ( )
Identity element 1. 3. 5. 7. 9. X
+ 0 = + 1
X + X
X = X 2. 4. 6. 8. X
. 1 =
. 0
X . X
Idempotence
Complement
Involution
10.
12.
14.
16.
X + Y
Y + X =
(X + Y) Z +
X + (Y Z)
X(Y + XY XZ
X . Y
11.
13.
15.
17. XY YX =
(XY) Z
X(Y Z) X
+ YZ
(X + Y)
(X + Z)
X . Y
X + Y
Commutative
Associative
Distributive
DeMorgan ’ s

3. Some Properties of Boolean Algebra
A two-valued Boolean algebra is also know as Switching Algebra.
The dual of an algebraic expression is obtained by interchanging + and · and interchanging 0’s and 1’s.
Sometimes, the dot symbol ‘’ (AND operator) is not written when the meaning is clear

4. Dual of a Boolean Expression
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 =
(X+Y) · (W · Z) = (X+Y) · (W+Z)
(A+B) · (A+C) · (B+C)

5. 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 · B = 1
= A Identity element

6. Boolean Algebraic Proof – Example 2
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 AC X = 1
= AB + AC Identity element

7. Useful Theorems Minimization X Y + X Y = Y Absorption X + X Y = X
Simplification
X + 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

8. Truth Table to Verify DeMorgan’s
X + Y = X · Y
X · Y = X + Y X Y
X·Y
X+Y
X · Y 1
Generalized DeMorgan’s Theorem:
X1 + X2 + … + Xn = X1 · X2 · … · Xn
X1 · X2 · … · Xn = X1 + X2 + … + Xn
We can also use algebraic properties in doing this proof. 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)’.
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
= X = 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’ x’) X . X’ = 0
= (0 + 0) X = 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.

9. 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 + z y
G' = ((a (b' + c'))+ d ) e' = (a (b' + c') + d) e'

10. 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) + D C B A

11. Canonical Forms Minterms and Maxterms
1- Sum-of-Minterm (SOM) Canonical Form
2- Product-of-Maxterm (POM) Canonical Form

12. 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. x Y X XY X Y

13. 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 +

14. Minterms & Maxterms for 2 variables
Two variable minterms and maxterms.
The minterm mi should evaluate for each combination of x and y.
The maxterm is the complement of the minterm x y
Index
Minterm
Maxterm
m0 = x y
M0 = x + y 1
m1 = x y
M1 = x + y 2
m2 = x y
M2 = x + y 3
m3 = x y
M3 = x + y

15. Minterms & Maxterms for 3 variables
M3 = x + y + z
m3 = x y z 3 1
M4 = x + y + z
m4 = x y z 4
M5 = x + y + z
m5 = x y z 5
M6 = x + y + z
m6 = x y z 6 y x z
M7 = x + y + z
m7 = x y z 7
M2 = x + y + z
m2 = x y z 2
M1 = x + y + z
m1 = x y z
M0 = x + y + z
m0 = x y z
Maxterm
Minterm
Index
Maxterm Mi is the complement of minterm mi
Mi = mi and mi = Mi

16. Purpose of the Minterms and Maxterms
Minterms and Maxterms are designated with an index
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”.

17. 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

18. 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 d
+ a b c d
+ a b c d
+ a b c d
+ a b c d
F(A,B,C,D,E) = A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE
a b c d
+ a b c d
+ a b c d
+ a b c d

19. Product-Of-Maxterm Examples
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
(a+b+c+d)
(a+b+c+d)
(a+b+c+d)
(a+b+c+d)
F(A,B,C,D,E) = A’B’C’DE’ + A’BC’D’E + AB’C’D’E + AB’CDE
(a+b+c+d)
(a+b+c+d)
(a+b+c+d)
(a+b+c+d)

20. Standard Forms A B C + A B C + B
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 A B C + A B C + B (A + B) (A + B + C ) C (A B + C) (A + C) A B C + A C (A + B)

21. Standard Sum-of-Products (SOP)
A sum of minterms form for n variables can be written down directly from a truth table.
This form often can be simplified so that the corresponding circuit is simpler.

22. Standard Sum-of-Products (SOP)
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 + A
F = B C + A
Simplified F contains 3 literals compared to 15 ) 7 , 6 5 4 1 ( C B A F S =

23. AND/OR Two-Level Implementation
The two implementations for F are shown below
It is quite apparent which is simpler!

24. 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