Show that ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) and ¬𝑝 ∧ ¬𝑞 are logically equivalent by:a) Using a truth table (3 POINTS)b) Using logical equivalences
Question
Solution
Sure, let's start with the truth table:
| p | q | ¬p | ¬q | ¬p ∧ q | p ∨ (¬p ∧ q) | ¬(p ∨ (¬p ∧ q)) | ¬p ∧ ¬q |
|---|---|---|---|---|---|---|---|
| T | T | F | F | F | T | F | F |
| T | F | F | T | F | T | F | F |
| F | T | T | F | T | T | F | F |
| F | F | T | T | F | F | T | T |
As you can see from the truth table, ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) and ¬𝑝 ∧ ¬𝑞 have the same truth values in all possible scenarios, so they are logically equivalent.
Now, let's prove this using logical equivalences:
- ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) = ¬𝑝 ∧ ¬(¬𝑝 ∧ 𝑞) (De Morgan's Law)
- ¬𝑝 ∧ ¬(¬𝑝 ∧ 𝑞) = ¬𝑝 ∧ (𝑝 ∨ ¬𝑞) (De Morgan's Law)
- ¬𝑝 ∧ (𝑝 ∨ ¬𝑞) = (¬𝑝 ∧ 𝑝) ∨ (¬𝑝 ∧ ¬𝑞) (Distributive Law)
- (¬𝑝 ∧ 𝑝) ∨ (¬𝑝 ∧ ¬𝑞) = F ∨ (¬𝑝 ∧ ¬𝑞) (Law of Contradiction)
- F ∨ (¬𝑝 ∧ ¬𝑞) = ¬𝑝 ∧ ¬𝑞 (Identity Law)
So, ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) and ¬𝑝 ∧ ¬𝑞 are logically equivalent.
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