Key Concepts and Formulas

• Logical Equivalence: Two statements are logically equivalent if they have the same truth values in all possible cases.
• Biconditional (p ⇔ q): True when both p and q have the same truth value (both true or both false).
• Conditional (p ⇒ q): False only when p is true and q is false; otherwise, it's true. Equivalent to ¬p ∨ q.
• Negation (¬p): Reverses the truth value of p.
• Conjunction (p ∧ q): True only when both p and q are true.
• De Morgan's Laws:
  • ¬(p ∧ q) ≡ ¬p ∨ ¬q
  • ¬(p ∨ q) ≡ ¬p ∧ ¬q

Step-by-Step Solution

Step 1: Simplify the expression using the equivalence of the biconditional

We start with the given expression: $(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q}$. We know that $p \Leftrightarrow q \equiv (p \rightarrow q) \wedge (q \rightarrow p)$. Therefore, $p \Leftrightarrow \sim q \equiv (p \rightarrow \sim q) \wedge (\sim q \rightarrow p)$. Then, $\sim(p \Leftrightarrow \sim q) \equiv \sim((p \rightarrow \sim q) \wedge (\sim q \rightarrow p))$.

Applying De Morgan's Law: $\sim((p \rightarrow \sim q) \wedge (\sim q \rightarrow p)) \equiv \sim(p \rightarrow \sim q) \vee \sim(\sim q \rightarrow p)$

Step 2: Convert conditionals to disjunctions

We know that $p \rightarrow q \equiv \sim p \vee q$. So, $p \rightarrow \sim q \equiv \sim p \vee \sim q$ and $\sim q \rightarrow p \equiv \sim(\sim q) \vee p \equiv q \vee p$.

Substituting these into the expression from Step 1:

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

Applying De Morgan's Law again:

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

Step 3: Combine with the conjunction

Now we combine this with the 'q' from the original expression using conjunction:

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

Distributing the 'q':

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

Since $q \wedge q \equiv q$ and $q \wedge \sim q \equiv F$ (False), we have:

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

Since anything AND False is False:

$(p \wedge q) \vee F$

Which simplifies to:

$p \wedge q$

Step 4: Show equivalence to (p ⇒ q) ∧ p

We want to show that $p \wedge q$ is equivalent to $(p \Rightarrow q) \wedge p$.

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

Distributing the 'p':

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

Since $\sim p \wedge p \equiv F$:

$F \vee (q \wedge p)$

Which simplifies to:

$p \wedge q$

Therefore, $(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q}$ is equivalent to $(p \Rightarrow q) \wedge p$.

Step 5: Construct the truth table for (p ⇒ q) ∧ p

Step 6: Re-examine the options

Since the expression simplifies to $p \wedge q$, let's re-evaluate the options. The truth table above matches the truth table of $p \wedge q$.

• (A) a tautology: A tautology is always true, but $p \wedge q$ is not always true. So, it's not a tautology.
• (B) a contradiction: A contradiction is always false, but $p \wedge q$ is not always false. So, it's not a contradiction.
• (C) equivalent to $(p \Rightarrow q) \wedge q$: $(p \Rightarrow q) \wedge q \equiv (\neg p \vee q) \wedge q \equiv (\neg p \wedge q) \vee (q \wedge q) \equiv (\neg p \wedge q) \vee q \equiv q$. This is not equivalent to $p \wedge q$.
• (D) equivalent to $(p \Rightarrow q) \wedge p$: As shown in step 4, $(p \Rightarrow q) \wedge p \equiv p \wedge q$. This is equivalent to $p \wedge q$.

Therefore the correct answer is (D).

Common Mistakes & Tips

• Carefully apply De Morgan's laws. Double-check the negations.
• Remember the equivalence of $p \Rightarrow q$ with $\sim p \vee q$. This is crucial for simplification.
• When simplifying, always look for opportunities to use the properties of logical connectives (e.g., $p \wedge p \equiv p$, $p \vee p \equiv p$, $p \wedge F \equiv F$, $p \vee T \equiv T$).

Summary

We started with the expression $(\sim(\mathrm{p} \Leftrightarrow \,\sim \mathrm{q})) \wedge \mathrm{q}$, used the equivalence of biconditional, De Morgan's Laws, and the conditional statement to simplify the expression to $p \wedge q$. We then showed that $p \wedge q$ is equivalent to $(p \Rightarrow q) \wedge p$. This corresponds to option (D).

Final Answer

The final answer is \boxed{(p \Rightarrow q) \wedge p}, which corresponds to option (D).