Simplification of Boolean Expressions by Using Boolean Algebra

Simplification is an approach in which a Boolean expression is minimized or reduced into an equivalent expression by using some Boolean identities. Boolean algebra is a mathematic applied to binary number system. It was developed by George Boole, an English mathematician, to simplify complex logical operation to their simplest form.

The simplification of a Boolean function is important because it reduces the number of logic devices/gates required to implement a logic function. This in turn reduces the hardware cost and complexity of the circuit. Also, it increases the reliability of the system.

In this tutorial, we will understand the procedure of simplifying or reducing or minimizing a complex Boolean function into its simplest form. Also, we will discuss some examples for better understanding of the concept.

Procedure of Simplifying Boolean Expression using Boolean Algebra

A complex Boolean expression can be simplified or minimized by following steps given below −

Step 1 − Multiply all the variables required to remove parentheses ‘()’.

Step 2 − Find all the identical terms in the expression. Only one of those terms will be retained and all other are dropped. For example −

$$\mathrm{ABC+ABC+ABC=ABC}$$

Step 3 − Find a variable and its negation in the same term. This term will be also dropped. For example −

$$\mathrm{AB\overline{B}C=A.0.C=AC}$$

Step 4 − Identify pairs of terms which are identical except for one variable that may be missing in one of the terms. In this case, the larger term will be dropped.

For example −

$$\mathrm{AB\overline{C}+AB=AB(\overline{C}+1)=AB.1=AB}$$

Step 5 − Identify those pairs of terms that have the same variables with one or more variables complemented. If a variable in one term of such a pair is complemented and in the second term it is not complemented. Then, such terms are combined into a single term with that variable dropped.

For example −

$$\mathrm{AB\overline{C}+ABC=AB(\overline{C}+C)=AB.1=AB}$$

Now, let us take some examples to understand the concept in depth.

Example 1

Simplify the following Boolean expression by using Boolean algebra.

$$\mathrm{\mathit{f}=A[B+C\lgroup \overline{A\overline{B}+AC\rgroup}}]$$

Solution

Given Boolean expression is,

$$\mathrm{\mathit{f}=A[B+C\lgroup \overline{A\overline{B}+AC\rgroup}}]$$

Simplifying term $\mathrm{\lgroup\overline{A\overline{B}+AC\rgroup }}$, by applying De Morgan’s law $\mathrm{\lgroup\overline{A+B}=\overline{A}\:\overline{B}\rgroup}$, we get,

$$\mathrm{\mathit{f}=A[B+C\lgroup \overline{A\overline{B}}+\overline{AC\rgroup}}]$$

Using De Morgan’s theorem, $\mathrm{\lgroup \overline{AB}=\overline{A}+\overline{C}\rgroup}$, we get,

$$\mathrm{\mathit{f}=A[B+C\lgroup \overline{A}+B\rgroup\lgroup A+\overline{C}\rgroup}]$$

Multiplying $\mathrm{ [\lgroup\overline{A}+B\rgroup\lgroup\overline{A}+\overline{C} \rgroup]}$, we get,

$$\mathrm{\mathit{f}=A[B+C\lgroup\overline{A}\:\overline{A}+\overline{A}B+\overline{A}\:\overline{C}+B\overline{C}\rgroup]}$$

Simplifying,

$$\mathrm{\mathit{f}=A\lgroup B+C\overline{A}\:\overline{A}+C\overline{A}B+C\overline{A}\:\overline{C}+CB\overline{C}\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=A\lgroup B+\overline{A}C+\overline{A}BC+0+0\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=A\lgroup B+\overline{A}C+\overline{A}BC\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=A[ B+\overline{A}C\lgroup 1+B\rgroup]}$$

$$\mathrm{\Rightarrow \mathit{f}=A\lgroup B+\overline{A}C\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=AB+A\overline{A}C}$$

$$\mathrm{\Rightarrow \mathit{f}=AB+0}$$

$$\mathrm{\therefore \mathit{f}=AB}$$

This is the simplified or minimized form of the given Boolean function.

Example 2

Reduce the following Boolean expression by using Boolean algebra.

$$\mathrm{\mathit{f}=\lgroup A+B\rgroup[AC+\lgroup B+\overline{C}\rgroup]}$$

Solution

The given Boolean expression is,

$$\mathrm{\mathit{f}=\lgroup A+B\rgroup\overline{[AC+\lgroup B+\overline{C}\rgroup]}}$$

Simplifying the term $\mathrm{\overline{\lgroup AC+\lgroup B+\overline{C}\rgroup\rgroup}}$, using De Morgan’s theorem, $\mathrm{\lgroup\overline{A+B}=\overline{A}\:\overline{B}\rgroup}$, we get,

$$\mathrm{\mathit{f}=\lgroup A+B\rgroup\overline{[AC.\lgroup B+\overline{C}\rgroup]}}$$

Simplifying $\mathrm{\overline{[ AC}.\overline{\lgroup B+\overline{C}\rgroup}]}$, using De Morgan’s theorem, we get,

$$\mathrm{\mathit{f}=\lgroup A+B\rgroup[\lgroup \overline{A}+\overline{C}\rgroup.\lgroup \overline{B}C\rgroup]}$$

$$\mathrm{\Rightarrow \mathit{f}=\lgroup A+B\rgroup\lgroup \overline{A}\:\overline{B}C+\overline{C}\:\overline{B}C\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=\lgroup A+B\rgroup\lgroup \overline{A}\:\overline{B}C+0\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=\lgroup A+B\rgroup\lgroup \overline{A}\:\overline{B}C\rgroup}$$

$$\mathrm{\Rightarrow \mathit{f}=\lgroup A\overline{A}\:\overline{B}C+B\overline{A}\:\overline{B}C\rgroup}$$

On rearranging, we get,

$$\mathrm{\Rightarrow \mathit{f}=\lgroup A\overline{A}\:\overline{B}C+\overline{A}\:\overline{B}BC\rgroup}$$

$$\mathrm{\therefore \mathit{f}=0+0=0}$$

Hence, on simplifying the given Boolean function, we obtained a 0 as the reduced expression. This indicates that the given function cannot be realized.

Example 3

Simplify the following Boolean expression by using Boolean algebra.

$$\mathrm{\mathit{f}=\lgroup A+AB\rgroup\lgroup A+\overline{A}B\rgroup\lgroup A+C\rgroup}$$

Solution

The given Boolean expression is,

$$\mathrm{\mathit{f}=\lgroup A+AB\rgroup\lgroup A+\overline{A}B\rgroup\lgroup A+C\rgroup}$$

Multiplying the first two terms, we get,

$$\mathrm{\mathit{f}=\lgroup AA+A\overline{A}B+ABA+AB\overline{A}B\rgroup\lgroup A+C\rgroup}$$

Rearranging,

$$\mathrm{\mathit{f}=\lgroup AA+A\overline{A}B+AAB+A\overline{A}BB\rgroup\lgroup A+C\rgroup}$$

Reducing,

$$\mathrm{\mathit{f}=\lgroup A+0+AB+0\rgroup\lgroup A+C\rgroup}$$

Taking common factor,

$$\mathrm{\mathit{f}= A\lgroup 1+B\rgroup\lgroup A+C\rgroup}$$

Reducing,

$$\mathrm{\mathit{f}= A.1\lgroup A+C\rgroup=A\lgroup A+C\rgroup}$$

Expanding,

$$\mathrm{\mathit{f}=\lgroup AA+AC\rgroup}$$

Reducing,

$$\mathrm{\mathit{f}=\lgroup A+AC\rgroup}$$

Simplifying,

$$\mathrm{\mathit{f}= A\lgroup 1+C\rgroup=A}$$

Hence, this is the simplified or minimized form of the given Boolean expression.

Conclusion

In this tutorial, we discussed simplification of Boolean expression by using Boolean algebra. Where, we discussed the step by step procedure to simplify a complex Boolean expression. Also, we discussed several solved examples to understand the concept thoroughly.

In logic circuit implementation, the circuit complexity and hardware cost play significant role. The circuit size also governs the speed of the circuit. Therefore, to achieve ideality as much as possible, we need to realize a logic circuit using minimum number of hardware parts, i.e., logic gates. This can be accomplished by reducing the Boolean expression into its minimized or simplified form.

Several techniques have been developed to simplify a complex Boolean expression. Boolean algebra is one of the most fundamental technique. As it is clear from the above examples that Boolean algebra provides a set of rules that can be directly applied to simplify or minimize a Boolean expression into its reduced form.