(a) State De Morgan's theorem for the following expressions. (i) NOT (A AND B) = ? [1] (ii) NOT (A OR B) = ? [1] (b) Using De Morgan's theorem, simplify the...
Question 1 Report
(a) State De Morgan's theorem for the following expressions.
(i) NOT (A AND B) = ?
[1]
(ii) NOT (A OR B) = ?
[1]
(b) Using De Morgan's theorem, simplify the following expression:
NOT (NOT P OR NOT Q)
[2]
(c) Complete the truth table to verify that NOT (A AND B) is equivalent to (NOT A) OR (NOT B).
| A | B | A AND B | NOT(A AND B) | NOT A | NOT B | (NOT A) OR (NOT B) |
|---|---|---|---|---|---|---|
| 0 | 0 | |||||
| 0 | 1 | |||||
| 1 | 0 | |||||
| 1 | 1 |
[3]
Answer Details
(a) De Morgan's theorem provides equivalences for negated compound expressions:
(i) NOT (A AND B) = (NOT A) OR (NOT B) [1]
Negating an AND expression converts it to an OR of the individual negations.
(ii) NOT (A OR B) = (NOT A) AND (NOT B) [1]
Negating an OR expression converts it to an AND of the individual negations.
(b) Simplifying NOT (NOT P OR NOT Q):
- Apply De Morgan's theorem: NOT (NOT P OR NOT Q) = NOT(NOT P) AND NOT(NOT Q). [1]
- NOT(NOT P) = P, and NOT(NOT Q) = Q (double negation cancels out).
- Result: P AND Q. [1]
(c) Verification truth table:
| A | B | A AND B | NOT(A AND B) | NOT A | NOT B | (NOT A) OR (NOT B) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 0 |
Row 1: A AND B = 0, NOT(A AND B) = 1. NOT A = 1, NOT B = 1, (NOT A) OR (NOT B) = 1. Columns match. [1]
Row 2: A AND B = 0, NOT(A AND B) = 1. NOT A = 1, NOT B = 0, 1 OR 0 = 1. Match. [1]
Row 3: A AND B = 0, NOT(A AND B) = 1. NOT A = 0, NOT B = 1, 0 OR 1 = 1. Match.
Row 4: A AND B = 1, NOT(A AND B) = 0. NOT A = 0, NOT B = 0, 0 OR 0 = 0. Match. [1]
The fourth and seventh columns are identical in every row, confirming De Morgan's theorem: NOT (A AND B) = (NOT A) OR (NOT B).
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