Question

Let$\mathit{A}\mathbf{,}\mathit{B}$ be subsets of$\mathit{U}$ . Prove De Morgan's laws:

(a)$\mathit{U}\mathbf{-}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{U}\mathbf{-}\mathbf{A}\mathbf{)}\mathbf{\cup }\mathbf{(}\mathbf{U}\mathbf{-}\mathbf{B}\mathbf{)}$

(b)$\mathit{U}\mathbf{-}\mathbf{(}\mathbf{A}\mathbf{\cup }\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{U}\mathbf{-}\mathbf{A}\mathbf{)}\mathbf{\cup }\mathbf{(}\mathbf{U}\mathbf{-}\mathbf{B}\mathbf{)}$

Short answer

(a) De Morgan’s law$U-(A\cap B)=(U-A)\cup (U-B)$ is proved.

(b) De Morgan’s law$U-(A\cup B)=(U-A)\cup (U-B)$ is proved.

Step-by-step solution

Write the given data from the question.

The$A,B$ be the subset of $U$.

Prove the statement U-(A∩B)=(U-A)∪(U-B).

(a)

Let assume$x\in U-(A\cap B)$ .

If $x\in A,B$then$x\in (A\cap B)$ but$x\notin A$ and$x\notin B$ .

Therefore$x\in U-A$, and$x\in U-B$ then$x\in (U-A)\cup (U-B)$ .

Similarly,$x\in (U-A)\cup (U-B)$ ,$x\in (U-A)$ and $x\in (U-B)$.

If$x\in A,B$ then$x\notin (U-A)$ and $x\notin (U-B)$therefore, $x\notin (U-A)\cup (U-B)$.

Therefore $x$is neither$x\notin (A\cap B)$ . It is implying that $x\in U-(A\cap B)$.

Hence De Morgan’s law$U-(A\cap B)=(U-A)\cup (U-B)$ is proved.

Prove the statement U-(A∪B)=(U-A)∪(U-B).

(b)

Let assume$x\in U-(A\cap B)$ .

If $x\in A,B$then $x\in (A\cup B)$but $x\notin A$and$x\notin B$ .

Therefore,$x\in U-A$ and $x\in U-B$then $x\in (U-A)\cap (U-B)$.

Similarly, $x\in (U-A)\cap (U-B)$, $x\in (U-A)$and$x\in (U-B)$ .

If $x\in A,B$then $x\notin (U-A)$and$x\notin (U-B)$therefore, $x\notin (U-A)\cap (U-B)$.

Therefore $x$is neither$x\notin (A\cap B)$ . It is implying that$x\in U-(A\cup B)$ .

Hence De Morgan’s law$U-(A\cup B)=(U-A)\cup (U-B)$ is proved.