Question

State and prove De-morgans theorem, i.e $\overline{A+B}=\overline{A}.\overline{B}$ and $\overline{A}.\overline{B}=\overline{A+B}$

Answer 1

We have,

Prove that:-

$\overline{A+B}=\overline{A}.\overline{B}$

$\overline{A}.\overline{B}=\overline{A+B}$

Proof:-

$\overline{A+B}=\overline{A}.\overline{B}$

Since each variable can have a value either $0or1$, then the four cases arises:-

(1).

$When,A=0,B=0,$

$Then,$

$\overline{A+B}=\overline{0+0}=\overline{0}=1$

$\overline{A}.\overline{B}=\overline{0}.\overline{0}=1.1=1$

$Hence,\overline{A+B}=\overline{A}.\overline{B}$

(2).

$A=1,B=0$

$Then,$

$\overline{A+B}=\overline{1+0}=\overline{1}=0$

$\overline{A}.\overline{B}=\overline{1}.\overline{0}=0.1=0$

$Hence,$

$\overline{A+B}=\overline{A}.\overline{B}$

(3)

$A=0,B=1$

$Then,$

$\overline{A+B}=\overline{0+1}=\overline{1}=0$

$\overline{A}.\overline{B}=\overline{0}.\overline{1}=1.0=0$

$Hence,$

$\overline{A+B}=\overline{A}.\overline{B}$

(4).

$A=1,B=1$

$Then,$

$\overline{A+B}=\overline{1+1}=\overline{1}=0$

$\overline{A}.\overline{B}=\overline{1}.\overline{1}=0.0=0$

$Hence,$

$\overline{A+B}=\overline{A}.\overline{B}$

Hence proved,

Similarly, We can prove that,

$\overline{A}.\overline{B}=\overline{A+B}$