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Laws of Boolean Algebra

Commutative Laws:

There are two commutative laws 

Associative Laws:

Distributive Laws:

There are two types of distributive laws

Identity:

Inverse:

Idempotence Laws:

Idempotence law indicates same value.

Complementation Laws:

Absorption Laws:

Transposition Law:

Consensus Theorem:

We use this theorem to remove redundant term. Conditions of this theorem as follows:

Conditions:

1. Three variable (Expression must be constituted by three variables)

Example: AC + A’B + BC

2. Each variable twice.  (Each variable must come twice in this expression)

Example: AB + A’C + BC

3. Only one variable is in complemented and uncomplemented form.

So, this theorem generally use in Sum of Product (SOP) and Product of Sum (POS) also.

Now we explain it by an example:

In that case we remove ‘BC’ because as per theorem one variable is complemented or uncomplemented so; we only consider these terms which is related to this variable (A) in complemented and uncomplemented form. 

Here ‘BC’ term is redundant and we remove ‘BC’ because B has nocomplemented and uncomplemented form.

So, after minimization the term will 

Now, we can prove it:

LHS:

Some more example of consensus theorem in SOP term:

Here B is complemented and uncomplemented variable 

So, only consider those terms which associated with B complemented and uncomplemented form (AB + B’C). 

And we remove AC because A has no complemented and uncomplemented form.

Here only one variable is complemented and uncomplemened which is ‘A’.

More examples on consensus theorem in POS term

Here only A is complemented and uncomplemented.

We can write:

One variable B is complemented and uncomplemented.

De Morgan’s Theorem:

Theorem:

We can use all laws/Theorems to simplify Boolean expression.

Example 1:

Example 2:

Operator Precedence:

Now here we discuss operator precedence for evaluating Boolean expression as follows:

1. Parentheses ( )

2. NOT (complement)

3. AND (.)

4. OR (+)

In Boolean expression we first evaluated all expression inside the parentheses before all other operations. After that next operation will be performed as precedence is the complement, then follows the AND and finally OR operation.

Example:

Here we first evaluate all operations inside the parentheses (B + C.D). So, as per precedence we solve ANDing ‘C.D’ first and then OR with B. After that we evaluate complement of B follows the AND operation with (B + C.D). Finally we evaluate OR operator with A.