ITPEC FE Morning October 2019 Question 1
ITPEC FE Morning October 2019 — Question 1 of 80
Tautology identification — find the expression that is always true.
Check all four combinations of p and q (00, 01, 10, 11):
- •
+= OR,·= AND,p̄= NOT p
Option (b): (p · q) + (p̄ + (p · q̄))
| p | q | p·q | p·q̄ | p̄ | p̄ + p·q̄ | Result |
|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 1 |
All rows = 1 → tautology.
Why not others:
- (a) p̄ + (p · q) — p=1, q=0 → 0 + 0 = 0
- (c) (p · q̄) · (p̄ + q) — p=0, q=0 → 0 · 1 = 0
- (d) ((p̄ · q) · (p · q)) · q̄ — inner (p̄ · q) · (p · q) is always 0 (requires both p and p̄ to be 1)
Key rule: For 2 variables, brute-force all 4 input combinations. A tautology must return 1 for every row.
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