# Boolean Algebra Simplification Rules

~ on ~ 0 comments**Boolean Algebra Simplification Rules with Truth Table:**The following rules will help us in the simplification (or reduction) of logic functions. Simplifying a complex logic expression using rules and laws is known as

*straight simplification*. We first give two basic sets of rules used for logic simplification using the straight simplification method. It may be found that these basic rules can be proved only by using truth tables.

**Rule 1(a):**

**Let**

*x*be a Boolean variable and

*x*' its complement. Then Rule 1(a) states that

*x*+

*x*′ =

**1**(2.12a)

Equation (2.12a) explains that

*the sum of a variable and its complement is equal to binary*. This also suggests that when we add a variable to its complement, it gets eliminated. It may be noted that In this text, we will be using English alphabets a,b,x,y, etc to represent Boolean variables.**1****Proof:**To prove Rule 1(a), we first draw the Table 2.8, which has to test the validity of Eq. (2.12a). This truth table is drawn by assigning values of 0 and 1 to variables

*x*and

*x*'. In this case, there are four possible values to be tested. The third column from the left shows the entry of binary

**1**in

**each of its rows, proving the rule**

*x*+

*x*′ =

**1**.

**Rule 1(b):**

**This is the complementary of Rule 1(a) and states that**

*x*

*x*′ =

**0**

*(2.12b)*

**Proof:**

**This relation is also proved in Table 2.8. The results shown in the fourth (last) column from the left shows the entry of binary**

**0**in each row, which proves Rule 1(b).

**Table 2.8**Truth

**table**

**to prove**

*x*+

*x*′=

**1**and

*xx*′=

**0**

x |
x' |
x + x' |
x x' |

01 |
10 |
11 |
00 |

**Rule 2(a):**

**Let**

*x*be a Boolean variable. Then Rule 2(a) states that

**1**+

*x*

**=****1**(2.13a)

In words, this rule explains that

*the sum of a binary*itself. This rule also suggests that a Boolean variable is eliminated when it is added to**1**and any Boolean variable results in binary**1****1**.**Proof:**

**This relation is proved using Table 2.9. The results shown in the second column from the left shows the entry of**

**1**in each row, which proves Rule 2(a).

**Rule 2(b):**

**Let**

*x*be a Boolean variable. Then Rule 2(b) states that

**0**+

*x*(2.13b)

**= x**
In words, this rule explains that

*a binary***0**added to any Boolean variable results in the variable*itself*. This rule also suggests that a Boolean variable does not change its value if a**0**is added to it.**Proof:**

**This relation is also proved in Table 2.9. The result shown in the third column from the left shows the entries of**

**0**and

**1**in the respective rows, which proves Rule 2(b).

**Rule 2(c):**

**As in the above cases, let**

*x*be a Boolean variable. Then Rule 2(c) states that

*x*

*×***1**

*= x*(2.13c)

In words, this rule explains that

*the multiplication of any Boolean variable with a binary*This rule also suggests that a Boolean variable does not change its value if a binary**1**produces that variable itself.**1**is used to multiply it.**Proof:**

**The proof of this relation is also given in Table 2.9. The result shown in the fourth column from the left shows the entries of**

**0**and

**1**in the respective rows, which proves Rule 2(c).

**Rule 2(d):**

**As in the previous cases, assume that**

*x*is a Boolean variable. Rule 2(d) states that

*x*

*×*

**0**

*=*

**0**(2.13d)

In words, this rule explains that

*the multiplication of any Boolean variable with a binary*.**0**results in the cancellation of that variable**Proof:**

**Equation (2.13d) is also proved**

**in Table 2.9. The result shown in the fifth column from the left shows the entries of**

**0**and

**0**in the respective rows, which proves Rule 2(d).

**Table 2.9**Truth table to prove Eqs. (2.13a) to (2.13d)

x |
1+x |
0 + x |
1× x |
0× x |

01 |
11 |
01 |
01 |
00 |

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