Key Concepts and Formulas

• Implication: $p \to q \equiv \sim p \vee q$
• Tautology: A statement that is always true, regardless of the truth values of its components.
• Fallacy (Contradiction): A statement that is always false, regardless of the truth values of its components.
• De Morgan's Laws:
  • $\sim (p \wedge q) \equiv \sim p \vee \sim q$
  • $\sim (p \vee q) \equiv \sim p \wedge \sim q$
• Law of Excluded Middle: $p \vee \sim p \equiv T$ (Tautology)

Step-by-Step Solution

Step 1: Analyze Statement S1

• What we are doing: Rewrite the implication using the equivalence $p \to q \equiv \sim p \vee q$.
• Math: $(p \to q) \vee (\sim q \to p) \equiv (\sim p \vee q) \vee (\sim (\sim q) \vee p)$
• Reasoning: We are replacing the implication symbols with their equivalent disjunction form.

Step 2: Simplify S1

• What we are doing: Simplify the expression obtained in Step 1, using the double negation rule $\sim(\sim q) \equiv q$.
• Math: $(\sim p \vee q) \vee (q \vee p) \equiv \sim p \vee q \vee q \vee p$
• Reasoning: Applying the double negation rule simplifies the expression.

Step 3: Rearrange and Simplify S1 Further

• What we are doing: Rearrange the terms and use the fact that $p \vee \sim p$ is a tautology.
• Math: $\sim p \vee p \vee q \vee q \equiv (\sim p \vee p) \vee q \equiv T \vee q$
• Reasoning: We group $\sim p$ and $p$ together to form a tautology. Also, $q \vee q \equiv q$.

Step 4: Final Simplification of S1

• What we are doing: Simplify the expression to its final form.
• Math: $T \vee q \equiv T$
• Reasoning: Since T (True) OR anything is always True, the statement simplifies to a tautology. Therefore, S1 is a tautology.

Step 5: Analyze Statement S2

• What we are doing: Analyze the second statement: $(p \wedge \sim q) \wedge (\sim p \wedge q)$.
• Math: $(p \wedge \sim q) \wedge (\sim p \wedge q)$
• Reasoning: We are setting up the expression to be analyzed.

Step 6: Rearrange S2

• What we are doing: Rearrange the terms using the associative and commutative properties of the $\wedge$ operator.
• Math: $(p \wedge \sim q) \wedge (\sim p \wedge q) \equiv (p \wedge \sim p) \wedge (q \wedge \sim q)$
• Reasoning: The associative and commutative properties allow us to rearrange the terms.

Step 7: Simplify S2

• What we are doing: Simplify the expression further, using the fact that $p \wedge \sim p$ is a fallacy (always False).
• Math: $(p \wedge \sim p) \wedge (q \wedge \sim q) \equiv F \wedge F$
• Reasoning: $p \wedge \sim p$ is always false, and $q \wedge \sim q$ is always false.

Step 8: Final Simplification of S2

• What we are doing: Simplify to its final form.
• Math: $F \wedge F \equiv F$
• Reasoning: Since False AND False is always False, the statement is a fallacy. Therefore, S2 is a fallacy.

Common Mistakes & Tips

• Carefully apply De Morgan's Laws and the implication equivalence. A sign error can change the entire result.
• Remember that $p \vee T \equiv T$ and $p \wedge F \equiv F$.
• When simplifying, rearrange terms to group complementary pairs like $p$ and $\sim p$.

Summary

We analyzed the two statements S1 and S2 using truth tables and logical equivalences. We found that S1 simplifies to a tautology (always true), and S2 simplifies to a fallacy (always false). Therefore, only statement S1 is true.

Final Answer

The final answer is \boxed{only (S1) is true}, which corresponds to option (A).