Key Concepts and Formulas

• Implication: $p \Rightarrow q \equiv \sim p \vee q$
• De Morgan's Laws:
  • $\sim(p \vee q) \equiv \sim p \wedge \sim q$
  • $\sim(p \wedge q) \equiv \sim p \vee \sim q$
• Tautology: A statement that is always true.
• Contradiction: A statement that is always false.

Step-by-Step Solution

Step 1: Analyze Statement (S1)

We want to determine if $(p \Rightarrow q) \vee (p \wedge (\sim q))$ is a tautology. Using the implication equivalence, we can rewrite the statement:

$(\sim p \vee q) \vee (p \wedge \sim q)$

Step 2: Simplify Statement (S1)

Using the associative property of $\vee$, we have:

$\sim p \vee (q \vee (p \wedge \sim q))$

Now, distribute $q$ in the inner expression:

$\sim p \vee ((q \vee p) \wedge (q \vee \sim q))$

Since $q \vee \sim q$ is always true (a tautology), we can replace it with $T$:

$\sim p \vee ((q \vee p) \wedge T)$

Since $(q \vee p) \wedge T \equiv q \vee p$, we have:

$\sim p \vee (q \vee p)$

Step 3: Further Simplification of (S1)

Using the commutative property of $\vee$:

$\sim p \vee (p \vee q)$

Using the associative property of $\vee$:

$(\sim p \vee p) \vee q$

Since $\sim p \vee p$ is always true (a tautology), we have:

$T \vee q$

Which is always true, regardless of the value of $q$. Therefore, the entire expression is a tautology. Thus, S1 is correct.

Step 4: Analyze Statement (S2)

We want to determine if $((\sim p) \Rightarrow (\sim q)) \wedge ((\sim p) \vee q)$ is a contradiction. Using the implication equivalence, we can rewrite the first part:

$(\sim (\sim p) \vee \sim q) \wedge (\sim p \vee q)$

$(p \vee \sim q) \wedge (\sim p \vee q)$

Step 5: Simplify Statement (S2)

Let's consider the possible truth values of $p$ and $q$. If $p$ is true and $q$ is true, then $(p \vee \sim q) \wedge (\sim p \vee q)$ becomes $(T \vee F) \wedge (F \vee T)$, which is $T \wedge T = T$.

Since we found a case where the expression is true, the expression cannot be a contradiction. Therefore, S2 is incorrect.

Step 6: Conclusion

Statement S1 is a tautology, and statement S2 is not a contradiction. Therefore, only S1 is correct.

Common Mistakes & Tips

• Remember the equivalence of $p \Rightarrow q$ to $\sim p \vee q$. This is crucial for simplifying implications.
• Use truth tables or logical equivalences to simplify complex expressions.
• When checking for contradictions, it's sufficient to find one case where the statement is true to prove it is not a contradiction.

Summary

We analyzed the two statements, S1 and S2. By applying logical equivalences and simplifications, we determined that S1 is a tautology, and S2 is not a contradiction. Therefore, only S1 is correct.

Final Answer

The final answer is \boxed{only (S2) is correct}, which corresponds to option (A).