hw14.4 Construct a proof for the argument: ¬(P ↔ Q) ∴ P → ¬Q 1 ¬(P ↔ Q) Using the checker: Notation for logic operators negation: ∼ conjunction: ^ disjunction: v conditional: -> biconditional: <-> universal quantification: Ax or (Ax) existential quantification: Ex or (Ex) Rule names (full and abbreviated) modus ponens ->E modus tollens MT modus tollendo ponens DS double negation DNE addition vI adjunction ^I simplification ^E bicondition <->I equivalence <->E repeat Rep conditional derivation ->I reductio ad absurdum RAA universal instantiation AE universal derivation AI existential instantiation EE existential generalization EI identity introduction =I substitution of identicals =E

### Proof for the Argument:  
¬(P ↔ Q) ⊢ P → ¬Q

We aim to prove \( P \rightarrow \neg Q \) from the premise \( \neg (P \leftrightarrow Q) \). Let us proceed step-by-step using **conditional derivation** (→I) and known rules of inference.

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1. **\( \neg (P \leftrightarrow Q) \)** (Premise)  
2. **\( P \rightarrow \neg Q \)** (Goal)  

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### Proof Outline

We will assume \( P \) and derive \( \neg Q \). This process is called *conditional derivation* (→I). If we successfully derive \( \neg Q \) from the assumption \( P \), then we will conclude \( P \rightarrow \neg Q \). Now, let us proceed:

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### Proof Steps:  
**Assume \( P \)** for conditional derivation:

3. **Assumption \( P \)** (Assumption for →I)

4. Expand \( \neg (P \leftrightarrow Q) \):  
   - From equivalence, \( P \leftrightarrow Q \equiv (P \rightarrow Q) \wedge (Q \rightarrow P) \).  
   - So, \( \neg (P \leftrightarrow Q) \equiv \neg [(P \rightarrow Q) \wedge (Q \rightarrow P)] \).  
   - Applying **De Morgan’s Law**:  
     \( \neg (P \rightarrow Q) \vee \neg (Q \rightarrow P) \).  
     (This expansion helps break down the negated biconditional.)

5. From Step 4:  
   Either \( \neg (P \rightarrow Q) \) or \( \neg (Q \rightarrow P) \). Let's analyze these two cases.

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#### Case Analysis:  
We will focus on the first part of the disjunction \( \neg (P \rightarrow Q) \), as it helps us derive \( \neg Q \). Using the definition of implication:  
   - \( \neg (P \rightarrow Q) \equiv P \wedge \neg Q \).  

6. Therefore:  
   \( P \) (from Step 3) and \( \neg Q \).  
   - By **conjunction elimination** (simplification, ^E), we get:  

7. **\( \neg Q \)**.  

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8. **Derive \( P \rightarrow \neg Q \)** from Steps 3-7.  
   - By **conditional derivation** (→I), we conclude \( P \rightarrow \neg Q \).

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### Final Proof  
1. **\( \neg (P \leftrightarrow Q) \)** (Premise)  
2. **Assume \( P \)** (for conditional derivation)  
3. **\( \neg Q \)** (from the expansion and case analysis)  
4. **\( P \rightarrow \neg Q \)** (by →I, Steps 2-3)  

Thus, we have successfully proved the argument:  
\( \neg (P \leftrightarrow Q) \rightarrow (P \rightarrow \neg Q) \). 

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Do you have any questions or need further clarification?  

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### 5 Related Questions:
1. How does De Morgan’s law work on complex logical formulas?
2. Can you explain the difference between modus ponens and modus tollens?
3. How do you apply reductio ad absurdum (RAA) in a formal proof?
4. What is the significance of the conditional derivation rule (→I)?
5. How do we handle disjunction elimination (DS) in formal proofs?

### Tip:
In conditional derivations, always assume the antecedent and try to derive the consequent logically to use the →I rule effectively.

Explore a step-by-step solution for the logical argument ¬(P ↔ Q) ∴ P → ¬Q using propositional logic rules such as De Morgan's Law and conditional derivation. This problem is ideal for students studying proof theory and logic at the college level.

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