Proof Using Algebra of Sets
Second De Morgan’s law: (A ∩ B)’ = A’ ∪ B’
Let X = (A ∩ B)’ and Y = A’ ∪ B’
Let p be any element of X, then p ∈ X ⇒ p ∈ (A ∩ B)’
⇒ p ∉ (A ∩ B)
⇒ p ∉ A and p ∉ B
⇒ p ∈ A’ or p ∈ B’
⇒ p ∈ A’ ∪ B’
⇒ p ∈ Y
∴ X ⊂ Y ————–(i)
Again, let q be any element of Y, then q ∈ Y ⇒ q ∈ A’ ∪ B’
⇒ q ∈ A’ or q ∈ B’
⇒ q ∉ A and q ∉ B
⇒ q ∉ (A ∩ B)
⇒ q ∈ (A ∩ B)’
⇒ q ∈ X
∴ Y ⊂ X ————–(ii)
From (i) and (ii) X = Y
(A ∩ B)’ = A’ ∪ B’
Proof Using Venn Diagram
Venn Diagram for (A ∩ B)’
Venn diagram for A’ ∪ B’
From both diagrams, we can clearly say
(A ∩ B)’ = A’ ∪ B’
That is the Second De Morgan’s Law.
De Morgan’s Law – Theorem, Proofs, Formula & Examples
De Morgan’s law is the most common law in set theory and Boolean algebra as well as set theory. In this article, we will learn about De Morgan’s law, De Morgan’s law in set theory, and De Morgan’s law in Boolean algebra along with its proofs, truth tables, and logic gate diagrams. The article also includes the solved De Morgan’s Law Example and FAQs on De Morgan’s law. Let us learn about De Morgan’s law.
Table of Content
- What is De Morgan’s Law
- De Morgan’s Law in Set Theory
- First De Morgan’s Law
- Second De Morgan’s Law
- Proof Using Algebra of Sets
- De Morgan’s Law in Boolean Algebra
- De Morgan’s Law Formula
- Solved Examples on De Morgan’s Law
- Logic Applications of De Morgan’s Law