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State and Prove DeMorgan's Laws.
1 Answer
written 3.6 years ago by |
DeMorgan's Theorems:-
Theorem 1:-
$\overline {AB} = \overline A +\overline B$
Proof:-
A | B | $ \overline{AB}$ | $\overline A$ | $ \overline B $ | $\overline A + \overline B$ |
---|---|---|---|---|---|
0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
Theorem 2:-
$\overline {A+B} = \overline A \ \overline B$
The circuit representation is given below
Proof:-
A | B | $\overline{A+B}$ | $\overline A$ | $\overline B$ | $ {\overline A \ \ \overline B}$ |
---|---|---|---|---|---|
0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |