Axioms of Boolean Algebra

• Difficulty Level : Medium
• Last Updated : 10 Feb, 2022

In this article, we are going to discuss Axioms of Boolean Algebra; these axioms/Theorems are important as these will be used in many different topics of Digital Electronics like Sequential Circuit Designing and Combinational Circuit Designing as well. These Axioms are the building blocks of Digital Electronics. So understand these axioms carefully. Now we will look into these Axioms one by one.

Axiom

The sets of logical expressions are known as Axioms or postulates of Boolean Algebra. An axiom is nothing more than the definition of three basic logic operations (AND, OR, and NOT).

Here,

• + denotes the Logical OR operation
• . denotes the Logical AND operation
• ! denotes the Logical NOT operation
• 0 and 1 denotes the Logical False and True respectively
• 0.0 = 0
• 0.1 = 0
• 1.0 = 0
• 1.1 = 1
• 0+0 = 0
• 0+1 = 1
• 1+0 = 1
• 1+1 = 1
• ! 0 = 1
• ! 1 = 0

Now, as we have discussed basic Axioms of boolean algebra let’s try to generalize them:

• 0.A = 0 (If A = 0, then 0.0 = 0 and when A=1, 0.1 = 0, Hence the expression will always be 0 regardless of the value of A)
• 1+A =1 (If A = 0 then 1+0 =1 and when A=1, 1+1 =1,Hence the expression will always be 1 regardless of the value of A)
• 0+A = A (If A = 0. then 0+0 = 0 and when A=1, 0+1=1. Hence the expression will be equal to value of A always)
• 1.A = A (If A = 0 then 1.0 = 0 and when A=1, 1.1=1.Hence the expression will be equal to value of A always)
• !A = A (If !A = 0 then A = 1 and !A = 1 then A = 0)
• A+A = A (If A = 0 then 0+0 = 0 and when A =1 then 1+1 = 1)
• A.A = A (If A = 0 then 0.0 = 0 and when A = 1 then 1.1 = 1)

These generalized expressions are very important as they are used to simplify many Boolean Functions and expressions. Minimizing the boolean function is useful in eliminating variables and Gate Level Minimization.

Comparison of Boolean algebra with Arithmetic algebra:

• In Arithmetic algebra, we have four basic operations that are addition, subtraction, multiplication, and division. Whereas in boolean algebra we have three basic operations that are AND, OR, NOT.
• In boolean algebra, we have only two kinds of values/end results that are either true or false. But in arithmetic algebra, the answer can be of any value, it can be positive, negative, zero, or any value that we may think about.

Postulates/Laws of Boolean Algebra:

1. Commutative Law:

• A+B = B+A
• A.B = B.A

For example:

```0 + 1 = 1 and 1 + 0 = 1 (that is A+B = B+A)
0.1 = 0 and 1.0 = 0 (that is A.B = B.A)```

2. Associative Law:

• (A+B)+C = A+(B+C)
• (A.B).C = A.(B.C)

For example:

```(0 + 1)+1 = 0 + (1+1) =1 (that it (A+B)+C = A+(B+C)
Similarly you can try for (A.B).C = A.(B.C)```

3. Distributive Law:

• A(B+C) = AB + AC
• A + BC = (A+B). (A+C)

Proof of A +BC = (A+B).(A+C)

Let’s try to simplify RHS of the expression:

```(A+B).(A+C) = A.A + A.C + B.A + B.C
We know that A.A = A  and B.A = A.B so the expression becomes:
A + AC + A.B + B.C```

On further Simplifying A.(1+C) + A.B + B.C ,now 1+C = 1 also, A.1 = A, so the expression becomes:

```A +  A.B + B.C
A(1+B) + B.C = A.1 + B.C = A + BC (1+B = 1 and A.1 = 1 )
Hence LHS = RHS  ```

4. Idempotance Law:

• A+A=A
• A.A = A

These Laws were discussed earlier

5. Absorption Law:

• A + A.B = A
• A.(A+B) = A

Let’s Simplify the LHS of both the expressions to obtain RHS:

```A + A.B = A(1+B) = A.1 = A
A.(A+B) = A.A + A.B
A + A.B = A(1+B)
A.1 = A ```
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