# De Morgan Theorem and Demorgans Laws

Published on 24/2/2012 & updated on 25/8/2018HOME / DIGITAL ELECTRONICS / ARITHMETIC

**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 and 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 of **DE 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

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

- 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 (+).
- 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 |

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. Therefore,

With the help of De-Morgan's theorem our calculation become much easier.
Let other example be,
In both the equations we have suitably used **De-Morgan's laws** to make our calculation much easier.

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