Math Problem Statement

Show that ¬(p ∨ (¬p ∧ q)) and ¬p ∧ ¬q are logically equivalent by developing a series of logical equivalences

Solution

To show that ¬(p(¬pq))\neg (p \vee (\neg p \wedge q)) and ¬p¬q\neg p \wedge \neg q are logically equivalent, we will develop a series of logical equivalences.

Step 1: Apply De Morgan’s Law

¬(p(¬pq))\neg (p \vee (\neg p \wedge q)) Using De Morgan’s Law: ¬(AB)¬A¬B\neg(A \vee B) \equiv \neg A \wedge \neg B, we rewrite: ¬p¬(¬pq)\neg p \wedge \neg (\neg p \wedge q)

Step 2: Apply De Morgan’s Law Again

Using De Morgan’s Law: ¬(AB)¬A¬B\neg(A \wedge B) \equiv \neg A \vee \neg B, we rewrite: ¬p(¬(¬p)¬q)\neg p \wedge (\neg (\neg p) \vee \neg q)

Since ¬(¬p)p\neg (\neg p) \equiv p, this simplifies to: ¬p(p¬q)\neg p \wedge (p \vee \neg q)

Step 3: Apply Distribution

Using the Distributive Law: A(BC)(AB)(AC)A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C), we expand: (¬pp)(¬p¬q)(\neg p \wedge p) \vee (\neg p \wedge \neg q)

Since ¬pp\neg p \wedge p is always false (FF), we get: F(¬p¬q)F \vee (\neg p \wedge \neg q)

Since FXXF \vee X \equiv X, we obtain: ¬p¬q\neg p \wedge \neg q

Conclusion:

Thus, we have shown that: ¬(p(¬pq))¬p¬q\neg (p \vee (\neg p \wedge q)) \equiv \neg p \wedge \neg q

This proves that the two statements are logically equivalent.


Would you like a more detailed breakdown or another example? Here are some related questions to explore:

  1. Can you prove that p(qr)p \vee (q \wedge r) is equivalent to (pq)(pr)(p \vee q) \wedge (p \vee r)?
  2. How does the principle of double negation apply in logical proofs?
  3. Can you show the equivalence of pqp \Rightarrow q and ¬pq\neg p \vee q?
  4. What is the difference between a tautology and a contradiction in logic?
  5. How do truth tables help verify logical equivalences?

Tip: When working with logical equivalences, always apply De Morgan’s Laws and distribution carefully to simplify expressions systematically.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Propositional Logic
Logical Equivalence
Boolean Algebra

Formulas

De Morgan's Laws: ¬(A ∨ B) ≡ ¬A ∧ ¬B, ¬(A ∧ B) ≡ ¬A ∨ ¬B
Distributive Law: A ∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C)

Theorems

De Morgan's Laws
Distributive Law
Double Negation

Suitable Grade Level

Undergraduate (Discrete Mathematics) or Advanced High School (Grade 11-12)