Use the logical equivalence laws and/or logical reasoning rules to prove that each of the following arguments is valid. q →r ¬(p v r)→s a) b) r-p קהה

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### Logical Equivalence and Reasoning in Propositional Logic

**Objective:**
Use the logical equivalence laws and/or logical reasoning rules to prove that each of the following arguments is valid.

#### a) Argument Validation

Given:
1. \( p \rightarrow q \)
2. \( q \rightarrow r \)
3. \( \neg r \)

Conclusion:
\[ \therefore \neg p \]

Proof:
- Using Modus Ponens on \( p \rightarrow q \) and \( q \rightarrow r \), we can derive \( p \rightarrow r \).
- Given \( \neg r \), by Modus Tollens on \( p \rightarrow r \), we can conclude \( \neg p \).

Hence, the argument is valid.

#### b) Argument Validation

Given:
1. \( \neg p \land q \)
2. \( \neg(p \lor r) \rightarrow s \)
3. \( r \rightarrow p \)

Conclusion:
\[ \therefore s \]

Proof:
- From \( \neg p \land q \), we can deduce \( \neg p \) and \( q \).
- From \( r \rightarrow p \) and \( \neg p \), by Modus Tollens, we get \( \neg r \).
- Now, \( \neg(p \lor r) \) simplifies to \( \neg p \land \neg r \).
- Therefore, substituting \( \neg p \land \neg r \) into \( \neg(p \lor r) \rightarrow s \), we can conclude \( s \).

Hence, the argument is valid.
Transcribed Image Text:### Logical Equivalence and Reasoning in Propositional Logic **Objective:** Use the logical equivalence laws and/or logical reasoning rules to prove that each of the following arguments is valid. #### a) Argument Validation Given: 1. \( p \rightarrow q \) 2. \( q \rightarrow r \) 3. \( \neg r \) Conclusion: \[ \therefore \neg p \] Proof: - Using Modus Ponens on \( p \rightarrow q \) and \( q \rightarrow r \), we can derive \( p \rightarrow r \). - Given \( \neg r \), by Modus Tollens on \( p \rightarrow r \), we can conclude \( \neg p \). Hence, the argument is valid. #### b) Argument Validation Given: 1. \( \neg p \land q \) 2. \( \neg(p \lor r) \rightarrow s \) 3. \( r \rightarrow p \) Conclusion: \[ \therefore s \] Proof: - From \( \neg p \land q \), we can deduce \( \neg p \) and \( q \). - From \( r \rightarrow p \) and \( \neg p \), by Modus Tollens, we get \( \neg r \). - Now, \( \neg(p \lor r) \) simplifies to \( \neg p \land \neg r \). - Therefore, substituting \( \neg p \land \neg r \) into \( \neg(p \lor r) \rightarrow s \), we can conclude \( s \). Hence, the argument is valid.
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