CHAPTER 6
BOOLEAN ALGEBRA
Boolean Operation
The complement of an element, denoted with a bar, is defined by 0̅ = 1and 1̅ = 0. The Boolean sum,
denoted by + or by OR, has the following values:
1 + 1 = 1,1 + 0 = 1, 0 + 1 = 1, 0 + 0 = 0.
The Boolean product, denoted by ・or by AND, has the following values:
1 ・1 = 1, 1 ・0 = 0, 0 ・1 = 0, 0 ・0 = 0.
Example:
Find the value of 1 ・0 + ̅̅̅̅̅̅̅̅̅̅̅
(0 + 1).
Solution:
Using the definitions of complementation, the Boolean sum, and the Boolean product,it follows that
̅̅̅̅̅̅̅̅̅̅̅
1 ・0 + (0
+ 1) = 0 + 1̅ = 0 + 0 = 0.
Boolean Identities
Example:
Find the sum-of-products expansion for the function𝐹(𝑥, 𝑦, 𝑧) = (𝑥 + 𝑦)𝑧̅.
Solution:
We will find the sum-of-products expansion of F(x, y, z) in two ways. First, we will use Boolean identities
to expand the product and simplify. We find that
𝐹(𝑥, 𝑦, 𝑧) = (𝑥 + 𝑦)𝑧̅
𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑧̅ + 𝑦𝑧̅
𝐹(𝑥, 𝑦, 𝑧) = 𝑥1𝑧̅ + 1𝑦𝑧̅
𝐹(𝑥, 𝑦, 𝑧) = 𝑥(𝑦 + 𝑦̅)𝑧̅ + (𝑥 + 𝑥̅ )𝑦𝑧̅
𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥𝑦𝑧̅ + 𝑥̅ 𝑦𝑧̅
𝐹(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧̅.
Distributive law
Identity law
Unit property
Distributive law
Idempotent law
Logic Gates
Example:
Construct circuits that produce the following outputs: (𝑎) (𝑥 + 𝑦)𝑥, (𝑏) 𝑥 (𝑦 + 𝑧), 𝑎𝑛𝑑 (𝑐) (𝑥 + 𝑦 +
𝑧)(𝑥 𝑦 𝑧).
Solution:
Minimization
of Circuit
Summation of Minterms
Example:
Find the summation of minterms for function 𝑭 and 𝑮of Table 1.
Solution:
𝑭 = ∑ 𝑚(0,2,4,5) = 𝑥̅ 𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥𝑦̅𝑧
𝑮 = ∑ 𝑚(0,2,4,6) = 𝑥̅ 𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥𝑦𝑧̅
Karnaugh Map
Two – Variable K- Map
Example:
Find the K-maps for (𝑎) 𝑥𝑦 + 𝑥̅ 𝑦, (𝑏) 𝑥𝑦̅ + 𝑥̅ 𝑦, and (𝑐) 𝑥𝑦̅ + 𝑥̅ 𝑦 + 𝑥̅ 𝑦̅.
Solution:
We can identify minterms that can be combined from the K-map. Whenever there are 1sin two adjacent
cells in the K-map, the minterms represented by these cells can be combinedinto a product involving just one of
the variables. For instance, 𝑥𝑦and 𝑥𝑦are represented byadjacent cells and can be combined into 𝑦, because 𝑥𝑦 +
𝑥𝑦 = (𝑥 + 𝑥)𝑦 = 𝑦.The grouping of minterms is shown above using the K-maps for these
expansions.Minimal expansions for these sums-of-products are (𝑎) 𝑦, (𝑏) 𝑥𝑦̅ + 𝑥̅ 𝑦, 𝑎𝑛𝑑 (𝑐) 𝑥̅ + 𝑦̅.
Three – Variable K- Map
Example:
Use K-maps to minimize these sum-of-products expansions.
(𝑎) 𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧 + 𝑥̅ 𝑦̅𝑧̅
(𝑏) 𝑥𝑦̅𝑧 + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧 + 𝑥̅ 𝑦̅𝑧 + 𝑥̅ 𝑦̅𝑧̅
(𝑐) 𝑥𝑦𝑧 + 𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧 + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧 + 𝑥̅ 𝑦̅𝑧 + 𝑥̅ 𝑦̅𝑧̅
(𝑑)𝑥𝑦𝑧̅ + 𝑥𝑦̅𝑧̅ + 𝑥̅ 𝑦̅𝑧 + 𝑥̅ 𝑦̅𝑧̅
Solution:
The K-maps for these sum-of-products expansions are shown below. The groupingof blocks shows that
minimal expansions into Boolean sums of Boolean products are(𝑎) 𝑥𝑧̅ + 𝑦̅𝑧̅ + 𝑥̅ 𝑦𝑧, (𝑏) 𝑦̅ + 𝑥̅ 𝑧, (𝑐) 𝑥 + 𝑦̅ +
𝑧, and (𝑑) 𝑥𝑧̅ + 𝑥̅ 𝑦̅.
Four – Variable K- Map
Example:
Use K-maps to simplify these sum-of-products expansions.
(𝑎) 𝑤𝑥𝑦𝑧 + 𝑤𝑥𝑦𝑧̅ + 𝑤𝑥𝑦̅𝑧̅ + 𝑤𝑥̅ 𝑦𝑧 + 𝑤𝑥̅ 𝑦̅𝑧 + 𝑤𝑥̅ 𝑦̅𝑧̅ + 𝑤
̅𝑥𝑦̅𝑧 + 𝑤
̅𝑥̅ 𝑦𝑧 + 𝑤
̅𝑥̅ 𝑦𝑧̅
(𝑏) 𝑤𝑥𝑦̅𝑧̅ + 𝑤𝑥̅ 𝑦𝑧 + 𝑤𝑥̅ 𝑦𝑧̅ + 𝑤𝑥̅ 𝑦̅𝑧̅ + 𝑤
̅𝑥𝑦̅𝑧̅ + 𝑤
̅𝑥̅ 𝑦𝑧̅ + 𝑤
̅𝑥̅ 𝑦̅𝑧̅
(𝑐)𝑤𝑥𝑦𝑧̅ + 𝑤𝑥𝑦̅𝑧̅ + 𝑤𝑥̅ 𝑦𝑧 + 𝑤𝑥̅ 𝑦𝑧̅ + 𝑤𝑥̅ 𝑦̅𝑧̅ + 𝑤
̅𝑥𝑦𝑧 + 𝑤
̅𝑥𝑦𝑧̅ + 𝑤
̅𝑥𝑦̅𝑧̅ + 𝑤
̅𝑥𝑦̅𝑧 + 𝑤
̅𝑥̅ 𝑦𝑧̅ +
𝑤
̅𝑥̅ 𝑦̅𝑧̅
Solution:
The K-maps for these expansions are shown below. Using the blocks shownleads to the sum of products
(𝑎) 𝑤𝑦𝑧 + 𝑤𝑥𝑧̅ + 𝑤𝑥̅ 𝑦̅ + 𝑤
̅𝑥̅ 𝑦 + 𝑤
̅𝑥𝑦̅𝑧, (𝑏) 𝑦̅𝑧̅ + 𝑤𝑥̅ 𝑦 + 𝑥̅ 𝑧̅, and(𝑐) 𝑧̅ + 𝑤
̅𝑥 + 𝑤𝑥̅ 𝑦.
12.4.2.4 Don’t Cares Conditions
Example:
One way to code decimal expansions using bits is to use the four bits of the binary expansionof each digit
in the decimal expansion. For instance, 873 is encoded as 100001110011. Thisencoding of a decimal expansion
is called a binary coded decimal expansion. Because thereare 16 blocks of four bits and only 10 decimal digits,
there are six combinations of four bits thatare not used to encode digits. Suppose that a circuit is to be built that
produces an output of 1 ifthe decimal digit is 5 or greater and an output of 0 if the decimal digit is less than 5.
How canthis circuit be simply built using OR gates, AND gates, and inverters?
Solution:
Let 𝐹(𝑤, 𝑥, 𝑦, 𝑧)denote the output of the circuit, where𝑤𝑥𝑦𝑧is a binary expansionof a decimal digit. The
values of F are shown in Table 1. The K-map for 𝐹, with ds in thedon’t care positions, is shown in K-map (a).
We can either include or exclude squares withds from blocks. This gives us many possible choices for the blocks.
For example, excludingall squares with ds and forming blocks, as shown in K-map (b), produces the
expression𝑤𝑥̅ 𝑦̅ + 𝑤
̅𝑥𝑦 + 𝑤
̅𝑥𝑧. Including some of the ds and excluding others and forming blocks, as shown in
K-map (c), produces the expression 𝑤𝑥̅ + 𝑤
̅𝑥𝑦 + 𝑥𝑦̅𝑧. Finally, including all the dsand using the blocks shown
in K-map (d) produces the simplest sum-of-products expansionpossible, namely, 𝐹(𝑥, 𝑦, 𝑧) = 𝑤 + 𝑥𝑦 + 𝑥𝑧.
12.4.3 Quine – McKluskey Method
Example 1:
We will show how the Quine–McCluskey method can be used to find a minimal expansionequivalent to
Hence, the final answer is 𝑧 + 𝑥̅ 𝑦̅
Example 2:
Use the Quine–McCluskey method to simplify the sum-of-products expansion 𝑤𝑥𝑦𝑧̅ + 𝑤𝑥̅ 𝑦𝑧 +
𝑤
̅𝑥𝑦𝑧 + 𝑤𝑥̅ 𝑦𝑧̅ + 𝑤
̅𝑥𝑦̅𝑧 + 𝑤
̅𝑥̅ 𝑦𝑧 + 𝑤
̅𝑥̅ 𝑦̅𝑧.
Hence, the final answer are 𝑤
̅𝑧 + 𝑤𝑦𝑧̅ + 𝑤𝑥̅ 𝑦 𝑜𝑟𝑤
̅𝑧 + 𝑤𝑦𝑧̅ + 𝑥̅ 𝑦𝑧.