Question

prove that (AUB )'=A'nB'​

Answer 1

$\textbf{Concept:}$

$x\in\,A\;\implies\;x\notin\,A$

$x\notin\,A\;\implies\;x\in\,A$

$\text{To prove: \bf(A{\cup}B)'=A'{\cap}B' }$

​$\text{Let\, x\in(A{\cup}B)' }$

$\iff\;x\notin\,A{\cup}B$

$\iff\;x\notin\,A\;\&\;x\notin\,B$

$\iff\;x\in\,A'\;\&\;x\in\,B'$

$\iff\;x\in\,A'{\cap}B'$

$\therefore\boxed{\bf(A{\cup}B)'=A'{\cap}B'}$

Answer 2

Given:

$\bold{(A \cup B)'= A'\cap B' }$

To find:

proving

Solution:

$\bold{(A \cup B)'= A'\cap B' }$

To solve the above equation we apply the De Morgan’s theorem:

Let

$x = (A \cup B)' \\y = A' \cap B'$

Let z be an arbitrary element of x then  z ∈ x ⇒ z ∈ (A U B)'

$\Rightarrow z \notin (A \cup B)\\\\\Rightarrow z \notin A\ \ and \ \ z \notin B\\\\\Rightarrow z \in A' \ \ and \ \ z \in B'\\\\ \Rightarrow z \in A' \cap B'\\\\\Rightarrow z \in y\\\\\therefore, \ \ x \subset y ....... (i)$

Again, let z1 be an arbitrary element of y then z1 ∈ y ⇒ z1 ∈ A' ∩ B'

$\Rightarrow z1 \in A' \ \ and \ \ z1 \in B'\\\\\Rightarrow z1\notin A \ \ and \ \ z1 \notin B\\\\\Rightarrow z1 \notin (A \cup B)\\\\\Rightarrow z1 \in (A \cup B)' \\\\\Rightarrow z1 \in P\\\\\therefore, Q \subset P ....(ii)\\\\ \ to \ combine \ (i) \ and \ (ii) \ we \ get; \ x = y \ we \get \ (A \cup B)' = A' \cap B'$