Key Concepts and Formulas

• Implication: $A \Rightarrow B \equiv \sim A \vee B$
• Tautology: A statement that is always true, regardless of the truth values of its components.
• Commutative and Associative Laws: $A \vee B \equiv B \vee A$, $(A \vee B) \vee C \equiv A \vee (B \vee C)$
• Law of Excluded Middle: $A \vee \sim A \equiv T$

Step-by-Step Solution

Step 1: Rewrite the implication using the equivalence $A \Rightarrow B \equiv \sim A \vee B$

We are given the expression $((\mathrm{p} \wedge \mathrm{q}) \Rightarrow(\mathrm{r} \vee \mathrm{q})) \wedge((\mathrm{p} \wedge \mathrm{r}) \Rightarrow \mathrm{q})$, which is a tautology. Using the implication rule, we can rewrite the expression as: $(\sim(p \wedge q) \vee (r \vee q)) \wedge (\sim(p \wedge r) \vee q)$

Step 2: Apply De Morgan's Law: $\sim(A \wedge B) \equiv \sim A \vee \sim B$

Applying De Morgan's Law to both negated conjunctions, we get: $((\sim p \vee \sim q) \vee (r \vee q)) \wedge ((\sim p \vee \sim r) \vee q)$

Step 3: Simplify the expression using associative and commutative laws

Rearranging the terms using the associative and commutative properties of disjunction, we have: $(\sim p \vee (\sim q \vee q) \vee r) \wedge (\sim p \vee \sim r \vee q)$

Step 4: Apply the Law of Excluded Middle: $\sim q \vee q \equiv T$

Since $\sim q \vee q$ is always true, we can replace it with $T$: $(\sim p \vee T \vee r) \wedge (\sim p \vee \sim r \vee q)$

Step 5: Simplify using the property $A \vee T \equiv T$

Since $\sim p \vee T \vee r$ contains $T$, the entire expression becomes $T$. Therefore, we have: $T \wedge (\sim p \vee \sim r \vee q)$ Which simplifies to: $\sim p \vee \sim r \vee q$

Step 6: Determine the condition for the expression to be a tautology

For the expression $\sim p \vee \sim r \vee q$ to be a tautology, it must always be true. We are given that $r \in \{p, q, \sim p, \sim q\}$. Let's analyze each case:

• Case 1: $r = p$. The expression becomes $\sim p \vee \sim p \vee q \equiv \sim p \vee q$. This is NOT a tautology. For example, if $p$ is true and $q$ is false, the expression is false.
• Case 2: $r = q$. The expression becomes $\sim p \vee \sim q \vee q \equiv \sim p \vee T \equiv T$. This IS a tautology.
• Case 3: $r = \sim p$. The expression becomes $\sim p \vee \sim (\sim p) \vee q \equiv \sim p \vee p \vee q \equiv T \vee q \equiv T$. This IS a tautology.
• Case 4: $r = \sim q$. The expression becomes $\sim p \vee \sim (\sim q) \vee q \equiv \sim p \vee q \vee q \equiv \sim p \vee q$. This is NOT a tautology (as shown in Case 1).

Therefore, only two values of $r$ (namely $q$ and $\sim p$) make the given expression a tautology.

Common Mistakes & Tips

• Be careful with De Morgan's Laws and implication rules. These are common sources of errors.
• Remember that any expression disjuncted with $T$ is always $T$.
• When checking for tautologies, consider cases where the expression might be false. If you find one such case, it's not a tautology.

Summary

We simplified the given expression using logical equivalences and De Morgan's laws to obtain $\sim p \vee \sim r \vee q$. We then checked each possible value of $r$ from the set $\{p, q, \sim p, \sim q\}$ to see which ones resulted in a tautology. We found that $r = q$ and $r = \sim p$ both resulted in tautologies, while $r = p$ and $r = \sim q$ did not. Therefore, there are two values of $r$ for which the given expression is a tautology.

Final Answer

The final answer is \boxed{2}, which corresponds to option (A).