**De-Morgan's Theorem**which is called

**De-Morgan's Laws**often.

Before discussing **De-Morgan's theorems** we should know about complements. Complements are the reverse value of the existing value. We are trying to say that as there are only two digits in binary number system 0 & 1. Now if A = 0 then complement of A will be 1 or A’ = 1

There are actually two theorems that were put forward by De-Morgan. On the basis ofDE Morgan’s laws much Boolean algebra are solved. Solving these types of algebra with De-Morgan's theorem has a major application in the field of digital electronics. De Morgan’s theorem can be stated as follows:-

Theorem 1:

The compliment of the product of two variables is equal to the sum of the compliment of each variable.

Thus according to **De-Morgan's laws** or De-Morgan's theorem if A and B are the two variables or Boolean numbers. Then accordingly

(A.B)’ = A’ + B’

Theorem 2:

The compliment of the sum of two variables is equal to the product of the compliment of each variable.

Thus according to De Morgan’s theorem if A and B are the two variables then.

(A + B)’ = A’.B’

De-Morgan's laws can also be implemented in Boolean algebra in the following steps:-

(1) While doing Boolean algebra at first replace the given operator. That is if (+) is there then replace it with (.) and if (.) is there then replace it with (+).

(2) Next compliment of each of the term is to be found.

De-Morgan's theorem can be proved by the simple induction method from the table given below.

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|

A | B | A’ | B | A+B | A.B | (A+B)’ | A’.B’ | (A.B)’ | A’+B’ |

0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |

1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |

Now look at the table very carefully in each row. Firstly the value of A = 0 and the value of B = 0. Now for this values A’ = 1, B’ = 1. Again A+B = 0 and A.B = 0. Thus (A+B)’ = 1 and (A.B)’ = 1, A’ + B’ = 1 and A’.B’ = 1. From this table you can therefore see that the value of column no 7 and 8 are equal and column no 9 and 10 are also equal which proves the De-Morgan's theorem.

Again different values of A and B we see the same thing i.e. column no 7 and 8 are equal to each other and 9 and 10 are equal to each other. Thus by this truth table we can prove De-Morgan's theorem.

Some examples given below can make your idea clear.

Let, Solve AB + A’ + B’

AB + A’ + B’

= AB + (AB)’ [since accordingly (AB)' = A' + B' which is a De-Morgan's law]

= 1 [as in Boolean algebra A+A’=1]

Therefore, AB + A’ + B’ = 1. With the help of De-Morgan's theorem our calculation become much easier.

Let other example be,

Solve A’ + B’ + (A + B)’

A’ + B’ + (A + B)’

= A’ + B’ + A’.B’ [since according to theorem (A+B)' = A'.B']

= A’(1 + B’)+ B’ [since 1+B' = 1]

= A’+ B’

Therefore. A’ + B’ + (A + B)’ = A’ + B’.

In both the equations we have suitably used De-Morgan's laws to make our calculation much easier.