hw14.3

Construct a proof for the argument: P ↔ ¬Q ∴ ¬(P ↔ Q)




1 P ↔ ¬Q




hw14.4

Construct a proof for the argument: ¬(P ↔ Q) ∴ P → ¬Q




1 ¬(P ↔ Q)




hw14.6

Construct a proof for the argument: Fb ∨ ∀xGx ∴ ∀x(Fb ∨ Gx)




1 Fb ∨ ∀xGx




Using the checker:

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




Let's break down the problems one by one and construct formal proofs for each.

### **Problem 14.3**  
**Argument:**  
\( P \leftrightarrow \neg Q \) ∴ \( \neg(P \leftrightarrow Q) \)

**Goal:** Construct a proof for \( \neg(P \leftrightarrow Q) \).

#### Proof for 14.3:

1. \( P \leftrightarrow \neg Q \) (Premise)  
2. \( P \leftrightarrow Q \) assumes that \( P \) and \( Q \) have the same truth value (Assumption for reductio ad absurdum).  
3. By the biconditional elimination on line 2, we have \( P \rightarrow Q \) and \( Q \rightarrow P \).  
4. From \( P \leftrightarrow \neg Q \) (line 1), we also have \( P \rightarrow \neg Q \) and \( \neg Q \rightarrow P \).  
5. Using modus ponens on lines 2 and 4, \( Q \) contradicts \( \neg Q \), leading to a contradiction.  
6. Since assuming \( P \leftrightarrow Q \) leads to a contradiction, we conclude \( \neg(P \leftrightarrow Q) \). (RAA)  

Thus, the argument is valid.  

---

### **Problem 14.4**  
**Argument:**  
\( \neg(P \leftrightarrow Q) \) ∴ \( P \rightarrow \neg Q \)

**Goal:** Construct a proof for \( P \rightarrow \neg Q \).

#### Proof for 14.4:

1. \( \neg(P \leftrightarrow Q) \) (Premise)  
2. Assume \( P \) (Assumption for conditional derivation).  
3. From line 1, \( P \leftrightarrow Q \) is false, meaning \( P \) and \( Q \) have opposite truth values.  
4. If \( P \) is true, then \( Q \) must be false, i.e., \( \neg Q \).  
5. Therefore, \( P \rightarrow \neg Q \) by conditional derivation \( (\rightarrow I) \).

Thus, the argument is valid.  

---

### **Problem 14.6**  
**Argument:**  
\( Fb \lor \forall x Gx \) ∴ \( \forall x (Fb \lor Gx) \)

**Goal:** Construct a proof for \( \forall x (Fb \lor Gx) \).

#### Proof for 14.6:

1. \( Fb \lor \forall x Gx \) (Premise)  
2. To derive \( \forall x (Fb \lor Gx) \), assume \( x \) is arbitrary.  
3. Since \( Fb \) holds for a particular \( b \), it follows that \( Fb \lor Gx \) holds for any \( x \).  
4. From line 3, \( Fb \lor Gx \) holds for arbitrary \( x \).  
5. By universal derivation (AI), \( \forall x (Fb \lor Gx) \).

Thus, the argument is valid.  

---

These proofs use standard logical rules, such as biconditional elimination, reductio ad absurdum, conditional derivation, and universal instantiation.

### Questions:

1. How does reductio ad absurdum work in proofs like 14.3?
2. Can the biconditional elimination rule always lead to a contradiction?
3. Why is universal derivation key in proving arguments like 14.6?
4. How can modus tollens help simplify contradictions in proofs?
5. What is the difference between conditional derivation and modus ponens?

### Tip:  
When working with biconditionals, remember they imply two conditional statements in both directions. Always check both implications for consistency.

Construct a proof for the argument: P ↔ ¬Q ∴ ¬(P ↔ Q), ¬(P ↔ Q) ∴ P → ¬Q, Fb ∨ ∀xGx ∴ ∀x(Fb ∨ Gx)

Learn how to construct formal proofs for logical arguments using biconditional elimination, reductio ad absurdum, and universal generalization. This guide covers proofs for arguments like P ↔ ¬Q ∴ ¬(P ↔ Q) and Fb ∨ ∀xGx ∴ ∀x(Fb ∨ Gx). Suitable for undergraduate logic courses.

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