Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Simplify $(p \vee q) \Rightarrow q$

We aim to simplify the expression $(p \vee q) \Rightarrow q$ using the implication rule. $(p \vee q) \Rightarrow q \equiv \sim(p \vee q) \vee q$

Step 2: Apply De Morgan's Law

Apply De Morgan's Law to simplify $\sim(p \vee q)$. $\sim(p \vee q) \vee q \equiv (\sim p \wedge \sim q) \vee q$

Step 3: Apply the Distributive Law

Apply the distributive law to expand $(\sim p \wedge \sim q) \vee q$. $(\sim p \wedge \sim q) \vee q \equiv (\sim p \vee q) \wedge (\sim q \vee q)$

Step 4: Simplify $(\sim q \vee q)$

Since $\sim q \vee q$ is always true (either $q$ is true or its negation is true), it is a tautology and can be represented as $T$. $(\sim p \vee q) \wedge (\sim q \vee q) \equiv (\sim p \vee q) \wedge T$

Step 5: Simplify $(\sim p \vee q) \wedge T$

Since anything ANDed with true is itself, we have $(\sim p \vee q) \wedge T \equiv \sim p \vee q$ Therefore, $(p \vee q) \Rightarrow q$ simplifies to $\sim p \vee q$.

Step 6: Analyze the truth table for $(p \wedge q) \Delta (\sim p \vee q)$

Now, we are given that $(p \wedge q) \Delta (\sim p \vee q)$ is a tautology. We need to find which logical connective $\Delta$ makes this statement always true. Let's construct a truth table:

Step 7: Test the options for $\Delta$ using the truth table.

We need $(p \wedge q) \Delta (\sim p \vee q)$ to be true for all combinations of $p$ and $q$.

• If $\Delta$ is $\wedge$, then $(p \wedge q) \wedge (\sim p \vee q)$ is T only when both $(p \wedge q)$ and $(\sim p \vee q)$ are T. This is only true when $p$ and $q$ are T. So, $\Delta$ cannot be $\wedge$.
• If $\Delta$ is $\vee$, then $(p \wedge q) \vee (\sim p \vee q)$ is T if either $(p \wedge q)$ or $(\sim p \vee q)$ is T. This is true in all cases except when both $(p \wedge q)$ and $(\sim p \vee q)$ are F. This occurs when $p$ is T and $q$ is F. So, $\Delta$ cannot be $\vee$.
• If $\Delta$ is $\Rightarrow$, then $(p \wedge q) \Rightarrow (\sim p \vee q)$ is equivalent to $\sim(p \wedge q) \vee (\sim p \vee q)$, which is $(\sim p \vee \sim q) \vee (\sim p \vee q) = \sim p \vee (\sim q \vee q) = \sim p \vee T$, which is always T. Thus, $(p \wedge q) \Rightarrow (\sim p \vee q)$ is a tautology.
• If $\Delta$ is $\Leftrightarrow$, then $(p \wedge q) \Leftrightarrow (\sim p \vee q)$ is T only when both $(p \wedge q)$ and $(\sim p \vee q)$ have the same truth value. From the truth table, they both have the same truth value for all rows. Thus, $(p \wedge q) \Leftrightarrow (\sim p \vee q)$ is a tautology. However, the given answer is $\wedge$, so we made an error.

Let's re-examine the options based on the truth table in Step 6, and see what the result of each connective would be:

The question states that $(p \wedge q) \Delta (\sim p \vee q)$ is a Tautology. From the table above, $(p \wedge q) \Rightarrow (\sim p \vee q)$ is a tautology, but that is not option (A). Also, $(p \wedge q) \Leftrightarrow (\sim p \vee q)$ is NOT a tautology, because it is not always true.

The correct answer is (A), so let us re-examine the question. We are told the answer is (A), which corresponds to $\wedge$. This means that $(p \wedge q) \wedge ((\sim p) \vee q)$ is a tautology. But from the truth table, we see that this is not a tautology.

Let us reconsider the simplification. We want $(p \wedge q) \Delta ((p \vee q) \Rightarrow q)$ to be a tautology. We have shown that $(p \vee q) \Rightarrow q \equiv \sim p \vee q$. So we want $(p \wedge q) \Delta (\sim p \vee q)$ to be a tautology.

If $\Delta$ is $\Leftrightarrow$, then we need $(p \wedge q)$ and $(\sim p \vee q)$ to always have the same truth value. $p \wedge q = T$ only when $p = T$ and $q = T$. Then $\sim p \vee q = \sim T \vee T = F \vee T = T$. $p \wedge q = F$ in all other cases. If $p = T$ and $q = F$, then $\sim p \vee q = \sim T \vee F = F \vee F = F$. If $p = F$ and $q = T$, then $\sim p \vee q = \sim F \vee T = T \vee T = T$. If $p = F$ and $q = F$, then $\sim p \vee q = \sim F \vee F = T \vee F = T$. Thus the truth values are the same when $p=T, q=T$ and $p=T, q=F$. The truth values are different when $p=F, q=T$ and $p=F, q=F$. Thus, $(p \wedge q) \Leftrightarrow (\sim p \vee q)$ is NOT a tautology.

Consider $(p \wedge q) \wedge (\sim p \vee q)$. This is only true when both are true. $p \wedge q = T$ only when $p=T$ and $q=T$, and $\sim p \vee q = \sim T \vee T = F \vee T = T$. If $p=T$ and $q=F$, then $p \wedge q = F$, so the conjunction is $F$. If $p=F$ and $q=T$, then $p \wedge q = F$, so the conjunction is $F$. If $p=F$ and $q=F$, then $p \wedge q = F$, so the conjunction is $F$. Thus $(p \wedge q) \wedge (\sim p \vee q)$ is equivalent to $p \wedge q$. This is not a tautology. There must be an error with the question or the given answer.

Looking back at the truth table, we need $(p \wedge q) \Delta (\sim p \vee q)$ to be a tautology. The only connective that yields a tautology is $\Rightarrow$. However, we are told that the answer is (A) $\wedge$. There appears to be an error.

Common Mistakes & Tips

• Carefully apply De Morgan's Laws and Distributive Laws.
• Remember the implication rule: $p \Rightarrow q \equiv \sim p \vee q$.
• Double-check truth tables to avoid errors.
• When simplifying, look for opportunities to use known logical equivalences.

Summary

We simplified $(p \vee q) \Rightarrow q$ to $\sim p \vee q$. Then, we analyzed the truth table for $(p \wedge q) \Delta (\sim p \vee q)$ to determine which logical connective $\Delta$ would make the statement a tautology. We found that $\Rightarrow$ results in a tautology. However, the correct answer is given as $\wedge$, which is incorrect. There might be an error in the question or the given answer.

Final Answer The question or given answer appears to be flawed as the correct connective based on the derivation is $\Rightarrow$, not $\wedge$. The given correct answer corresponds to option (A), which is incorrect.