Under Digital ElectronicsWe have known the basic operation of binary arithmetic such as binary addition, binary subtraction, binary multiplication and binary division. Now we will look through the most important part of binary arithmetic on which a lot of Boolean algebra stands, that is 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.
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.
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.
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