### 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 ValidationGiven: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 ValidationGiven: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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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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Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

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Question

### 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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