Key Concepts and Formulas

• 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.
• Conditional Statement ($p \to q$): This is false only when $p$ is true and $q$ is false. Otherwise, it is true.
• Biconditional Statement ($p \leftrightarrow q$): This is true when $p$ and $q$ have the same truth value (both true or both false). Otherwise, it is false.
• Disjunction ($p \vee q$): This is true if either $p$ or $q$ (or both) are true. It is false only when both $p$ and $q$ are false.
• Conjunction ($p \wedge q$): This is true only when both $p$ and $q$ are true. Otherwise, it is false.
• Negation ($\sim p$): This is true when $p$ is false, and false when $p$ is true.

Step-by-Step Solution

Step 1: Analyze Statement $S_1$: $(q \vee p) \to (p \leftrightarrow \sim q)$

• We need to determine if $(q \vee p) \to (p \leftrightarrow \sim q)$ is a tautology. We'll construct a truth table to evaluate the statement for all possible truth values of $p$ and $q$.
• We'll build the truth table column by column.

Step 2: Explain the truth table for $S_1$

• Columns 1 and 2 represent all possible combinations of truth values for $p$ and $q$.
• Column 3, $\sim q$, is the negation of column 2.
• Column 4, $q \vee p$, is true if either $q$ or $p$ (or both) are true, and false otherwise.
• Column 5, $p \leftrightarrow \sim q$, is true if $p$ and $\sim q$ have the same truth value, and false otherwise.
• Column 6, $(q \vee p) \to (p \leftrightarrow \sim q)$, is true if the implication holds. It's only false when $q \vee p$ is true and $p \leftrightarrow \sim q$ is false.

Step 3: Determine if $S_1$ is a tautology

• From the truth table, we see that the statement $(q \vee p) \to (p \leftrightarrow \sim q)$ is not always true (it's false in the first row). Therefore, $S_1$ is not a tautology.

Step 4: Analyze Statement $S_2$: $\sim q \wedge (\sim p \leftrightarrow q)$

• We need to determine if $\sim q \wedge (\sim p \leftrightarrow q)$ is a fallacy. We'll construct a truth table.

Step 5: Explain the truth table for $S_2$

• Columns 1 and 2 represent all possible combinations of truth values for $p$ and $q$.
• Column 3, $\sim q$, is the negation of column 2.
• Column 4, $\sim p$, is the negation of column 1.
• Column 5, $\sim p \leftrightarrow q$, is true if $\sim p$ and $q$ have the same truth value, and false otherwise.
• Column 6, $\sim q \wedge (\sim p \leftrightarrow q)$, is true only if both $\sim q$ and $(\sim p \leftrightarrow q)$ are true.

Step 6: Determine if $S_2$ is a fallacy

• From the truth table, we see that the statement $\sim q \wedge (\sim p \leftrightarrow q)$ is not always false (it's true in the second row). Therefore, $S_2$ is not a fallacy.

Step 7: Conclusion

• $S_1$ is not a tautology, and $S_2$ is not a fallacy. Thus, both statements are incorrect.

Common Mistakes & Tips

• Carefully construct truth tables. Double-check each entry to avoid errors.
• Remember the truth conditions for conditional and biconditional statements. These are often a source of mistakes.
• Understand the difference between a tautology (always true) and a fallacy (always false).

Summary

We analyzed the two given statements, $S_1$ and $S_2$, by constructing truth tables. We found that $S_1$, $(q \vee p) \to (p \leftrightarrow \sim q)$, is not a tautology because it is not always true. We also found that $S_2$, $\sim q \wedge (\sim p \leftrightarrow q)$, is not a fallacy because it is not always false. Therefore, both statements are incorrect.

Final Answer

The final answer is \boxed{both (S 1 ) and (S 2 ) are not correct}, which corresponds to option (A).