0

932views

State and Prove DeMorgan's Laws.

**1 Answer**

0

932views

State and Prove DeMorgan's Laws.

0

27views

written 3.1 years ago by |

DeMorgan's Theorems:-

- A mathematician named DeMorgan developed a pair of important rules regarding group complementation in Boolean algebra.

Theorem 1:-

- DeMorgan's First theorem states that for any two elemnts A and B in a boolean algebra, the complemeny of a product is equal to the sum of complements.

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

- The LHS of this theorem represents a NAND gate with inputs A and B, whereas the RHS represents an OR gate with inverted inputs.
- This OR gate is called as Bubbled OR.
- Thus according to DeMorgan's First theorem NAND gate is equaivalent to Bubbled OR gate.
- The circuit representation is given below

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:-

- DeMorgan's Second theorem states that for any two elemnts A and B in a boolean algebra, the complemeny of a sum is equal to the product of complements.

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

- The LHS of this theorem represents a NOR gate with inputs A and B, whereas the RHS represents an AND gate with inverted inputs.
- This AND gate is called as Bubbled AND.
- Thus according to DeMorgan's Second theorem NOR gate is equaivalent to Bubbled AND gate.
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 |

ADD COMMENT
EDIT

Please log in to add an answer.