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The image contains two logical equivalences that are requested to be proved without using a truth table. Let's go through them one by one.

### i. \( (P \rightarrow R) \land (Q \rightarrow R) \equiv (P \lor Q) \rightarrow R \)

#### Left-hand side (LHS):
The expression is \( (P \rightarrow R) \land (Q \rightarrow R) \). To simplify it, we can use the definition of implication:
- \( P \rightarrow R \equiv \neg P \lor R \)
- \( Q \rightarrow R \equiv \neg Q \lor R \)

Thus, the LHS becomes:
\[
(\neg P \lor R) \land (\neg Q \lor R)
\]

We can now use the distributive property of logic:
\[
(\neg P \land \neg Q) \lor R
\]

#### Right-hand side (RHS):
The expression is \( (P \lor Q) \rightarrow R \). Again, using the definition of implication:
\[
(P \lor Q) \rightarrow R \equiv \neg (P \lor Q) \lor R
\]

By De Morgan's law, \( \neg (P \lor Q) \equiv \neg P \land \neg Q \), so:
\[
\neg (P \lor Q) \lor R \equiv (\neg P \land \neg Q) \lor R
\]

#### Conclusion:
Both sides simplify to \( (\neg P \land \neg Q) \lor R \), so:
\[
(P \rightarrow R) \land (Q \rightarrow R) \equiv (P \lor Q) \rightarrow R
\]
The equivalence is proven.

---

### ii. \( \neg (P \lor (Q \land R)) \equiv \neg P \land (\neg Q \lor \neg R) \)

#### Left-hand side (LHS):
The expression is \( \neg (P \lor (Q \land R)) \). Using De Morgan's law:
\[
\neg (P \lor (Q \land R)) \equiv \neg P \land \neg (Q \land R)
\]
Again, using De Morgan's law for \( \neg (Q \land R) \):
\[
\neg (Q \land R) \equiv \neg Q \lor \neg R
\]
Thus, the LHS becomes:
\[
\neg P \land (\neg Q \lor \neg R)
\]

#### Right-hand side (RHS):
The RHS is already in the same form:
\[
\neg P \land (\neg Q \lor \neg R)
\]

#### Conclusion:
Both sides are identical, so the equivalence is proven:
\[
\neg (P \lor (Q \land R)) \equiv \neg P \land (\neg Q \lor \neg R)
\]

---

Would you like more details on these steps or further examples?

Here are some related questions:
1. What are the key laws of logic used in equivalence proofs?
2. How does the distributive property work in logical expressions?
3. Can you explain De Morgan's laws in more depth?
4. How does the implication \(P \rightarrow R\) relate to disjunction?
5. What other methods can be used to prove logical equivalences without truth tables?

**Tip:** When proving logical equivalences, De Morgan's laws and the definition of implication are commonly used for simplifying complex expressions.

Prove the following logical equivalences without using the truth table: 
1. (P → R) ∧ (Q → R) ≡ (P ∨ Q) → R 
2. ¬(P ∨ (Q ∧ R)) ≡ ¬P ∧ (¬Q ∨ ¬R)

Learn how to prove the logical equivalences (P → R) ∧ (Q → R) ≡ (P ∨ Q) → R and ¬(P ∨ (Q ∧ R)) ≡ ¬P ∧ (¬Q ∨ ¬R) without using truth tables. This guide covers important concepts such as implication, logical equivalence, and De Morgan's laws, providing a step-by-step solution. Suitable for students in Grades 11-12.

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