Show that ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) and ¬𝑝 ∧ ¬𝑞 are logically equivalent by:a) Using a truth table (3 POINTS)b) Using logical equivalences
Question
Show that ¬(𝑝 ∨ (¬𝑝 ∧ 𝑞)) and ¬𝑝 ∧ ¬𝑞 are logically equivalent by:
a) Using a truth table (3 POINTS)
b) Using logical equivalences
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.
Similar Questions
Show that each of these conditional statements is a tautology using logical equivalences.a) (𝑝 ∧ 𝑞) → 𝑝b) ¬𝑝 → (𝑝 → 𝑞)c) (𝑝 ∧ 𝑞) → (𝑝 → 𝑞)
Using laws of logic prove that following are equivalenti)~ ( ) and ~ p q p q →
Using Boolean laws simplify each of the following propositions:a. a ∧ b ∨ a ∧ ¬b
Match the truth tables to its corresponding propositional logicAnswer 1 Question 13Answer 2 Question 13Answer 3 Question 13
De Morgan's law is used in finding the equivalence of a logic expression using other logical functions.Question 11Select one:TrueFalse
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.