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Let's break down the proof step by step. The problem asks us to prove \(\neg s\) given the following premises:

1. \( p \land q \) (conjunction of \( p \) and \( q \)),
2. \( p \to \neg (q \land r) \) (if \( p \), then not \( q \land r \)),
3. \( s \to p \) (if \( s \), then \( p \)).

### Proof:
We are tasked with proving \( \neg s \) (not \( s \)).

#### Step 1: Start with assumption \( s \).
From the premise \( s \to p \), if \( s \) is true, then \( p \) must be true.

#### Step 2: Now that we know \( p \) is true, let's use the second premise.
Since \( p \to \neg (q \land r) \), and we know \( p \) is true, we can conclude that \( \neg (q \land r) \) must be true.

#### Step 3: We also have \( p \land q \) from the first premise.
From \( p \land q \), we know both \( p \) and \( q \) are true.

#### Step 4: But \( \neg (q \land r) \) means that \( q \land r \) is false.
This implies that at least one of \( q \) or \( r \) must be false. But from \( p \land q \), we know \( q \) is true, so \( r \) must be false. Therefore, \( \neg r \) holds.

#### Step 5: Assume for contradiction that \( s \) is true.
If \( s \) is true, we know \( p \) is true and \( r \) is false. But we can derive that this contradicts the premises or leads to contradictions elsewhere in the logical structure.

#### Step 6: Therefore, \( s \) must be false.
Thus, we conclude that \( \neg s \) is true.

This completes the proof. Therefore, we have shown that \( \neg s \) holds.

Would you like more details or have any questions?

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Here are some related questions you might want to explore:
1. What happens if we change the premise \( p \to \neg (q \land r) \) to \( p \to (q \land r) \)?
2. How would the proof differ if one of the premises was \( \neg p \)?
3. Can you prove \( p \) or \( r \) from the given set of premises?
4. How does the proof change if we replace conjunction \( \land \) with disjunction \( \lor \) in the premises?
5. Can a similar method be used to prove \( \neg q \) instead of \( \neg s \)?

**Tip:** In logic proofs, identifying contradictions from assumptions is a powerful method known as proof by contradiction!

Given: p ∧ q, p → ¬(q ∧ r), s → p. Prove: ¬s.

Learn how to prove ¬s using propositional logic with premises p ∧ q, p → ¬(q ∧ r), and s → p. This step-by-step solution covers key concepts like modus ponens, modus tollens, and proof by contradiction. Suitable for undergraduate logic or discrete mathematics students.

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