consider (𝑞𝑞 → 𝑝𝑝) ∧ ~(𝑟𝑟 ∧ 𝑞𝑞), Simplify this statement, explaining all your steps (make sure to use the correct names of the
logical equivalences)

We are tasked with simplifying the logical statement \((q \rightarrow p) \land \neg(r \land q)\).

### Step 1: Use Implication Law
The implication \(q \rightarrow p\) can be rewritten using the **Implication Law**:
\[
q \rightarrow p \equiv \neg q \lor p
\]
So, the original expression becomes:
\[
(\neg q \lor p) \land \neg(r \land q)
\]

### Step 2: Apply De Morgan's Law to Negation
Next, let's simplify \(\neg(r \land q)\) using **De Morgan's Law**. This law states that:
\[
\neg(r \land q) \equiv \neg r \lor \neg q
\]
So, the expression becomes:
\[
(\neg q \lor p) \land (\neg r \lor \neg q)
\]

### Step 3: Apply the Distributive Law
Now, distribute \((\neg q \lor p)\) over \((\neg r \lor \neg q)\) using the **Distributive Law**:
\[
(\neg q \lor p) \land (\neg r \lor \neg q) \equiv (\neg q \land (\neg r \lor \neg q)) \lor (p \land (\neg r \lor \neg q))
\]
Simplify the two parts of the expression:

1. \(\neg q \land (\neg r \lor \neg q)\):
   Since \(\neg q\) appears in both terms, we simplify this part to just \(\neg q\) (by **Idempotent Law**).

2. \(p \land (\neg r \lor \neg q)\):
   Distribute \(p\) over \(\neg r \lor \neg q\):
   \[
   p \land (\neg r \lor \neg q) \equiv (p \land \neg r) \lor (p \land \neg q)
   \]

### Step 4: Combine the Results
Now we combine the results from the two parts:
\[
\neg q \lor ((p \land \neg r) \lor (p \land \neg q))
\]
We can simplify this further using the **Associative Law** and **Distributive Law**:
\[
\neg q \lor (p \land \neg r) \lor (p \land \neg q)
\]

### Final Simplified Expression
Thus, the simplified form of the original expression is:
\[
\neg q \lor (p \land \neg r) \lor (p \land \neg q)
\]

### Recap of Steps:
1. Applied **Implication Law**: \(q \rightarrow p \equiv \neg q \lor p\).
2. Applied **De Morgan's Law**: \(\neg(r \land q) \equiv \neg r \lor \neg q\).
3. Applied **Distributive Law** to distribute \(\neg q \lor p\) over \(\neg r \lor \neg q\).
4. Simplified the result using the **Idempotent Law** and **Associative Law**.

---

Do you want more details on any of these steps or the logical laws?

Here are five related questions for further exploration:
1. How would you simplify \((p \lor q) \land \neg(p \land r)\)?
2. What is the negation of \((q \rightarrow p) \land r\)?
3. Can you simplify \(p \rightarrow (q \rightarrow r)\) using logical equivalences?
4. How do De Morgan's Laws apply to three terms, e.g., \(\neg(p \land q \land r)\)?
5. What is the equivalent disjunctive normal form (DNF) of the simplified expression?

**Tip:** Always apply De Morgan’s laws when dealing with negations over conjunctions or disjunctions.

Consider (q → p) ∧ ~(r ∧ q). Simplify this statement, explaining all your steps (make sure to use the correct names of the logical equivalences).

Learn how to simplify the logical expression (q → p) ∧ ¬(r ∧ q) step by step, using key logical equivalences like Implication Law, De Morgan's Law, and the Distributive Law. This tutorial is suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation