De Morgan's First Law

De Morgan’s Laws:

A well-defined collection of objects or elements is known as a set. Various operations like complement of a set, union and intersection can be performed on two sets. These operations and their usage can be further simplified using a set of laws known as De Morgan’s Laws. These are very easy and simple laws.

Any set consisting of all the objects or elements related to a particular context is defined as a universal set. Consider a universal set U such that A and B are the subsets of this universal set.

According to the De Morgan’s first law, the complement of union of two sets A and B is equal to the intersection of the complement of the sets A and B.

(A∪B)’= A’∩ B’      —–(1)

Where complement of a set is defined as

A’= {x:x ∈ U and x ∉ A}

Where A’ denotes the complement.

This law can be easily visualized using Venn Diagrams.

The L.H.S of the equation 1 represents the complement of union of two sets A and B. First of all, union of two setsA and B is defined as the set of all elements which lie eitherin set A or in set B. It can be visualized using Venn Diagrams as shown:

Figure 1 Union of Sets

The highlighted or the green colored portion denotes A∪B. The complement of union of A and B i.e., (A∪B)’is set of all those elements which are not in A∪B. This can be visualized as follows:

Figure 2 Complement of sets

Similarly, R.H.S of equation 1 can be represented using Venn Diagrams as well, the first part i.e., A’ can be depicted as following:

Figure 3 Complement of set A

The portion in black indicates set A and blue part denotes its complement i.e., A’.
Similarly, B’ is represented as:

Figure 4 Complement of set B

The portion in black indicates set B and yellow part denotes its complement i.e., B’.

If fig. 3 and 4 are superimposed on one another, we get the figure similar to that of complement of sets.

Figure 5 Intersection of complements of sets

Hence L.H.S = R.H.S


As, A∪B= either in A or in B

(A∪B)’= L.H.S = neither in A nor in B

Also, A’= Not in A

B’= Not in B

A’∩ B’= Not in A and not in B

⇒(A∪B)’= A’∩ B’