Step 1: Understanding the Concept:
A statement pattern is a tautology if it is always true ($T$) regardless of the truth values assigned to its constituent variables. To solve this, we should first simplify the premise (antecedent) of the implication. Then, we determine what the consequent ($r$) must be to ensure the entire implication never evaluates to false ($F$).
Step 2: Key Formula or Approach:
1. Implication Equivalence: $A \rightarrow B \equiv \sim A \lor B$.
2. De Morgan's Laws and Distributive Laws.
3. An implication $A \rightarrow B$ is a tautology if $A$ logically implies $B$, meaning whenever $A$ is True, $B$ must also be True.
Step 3: Detailed Explanation:
Let's simplify the antecedent part of the given expression: $A \equiv (p \rightarrow q) \land \sim q$.
Using the equivalence $p \rightarrow q \equiv \sim p \lor q$, substitute into $A$:
\[ A \equiv (\sim p \lor q) \land \sim q \]
Apply the distributive law:
\[ A \equiv (\sim p \land \sim q) \lor (q \land \sim q) \]
Since $(q \land \sim q)$ is a logical contradiction (always $F$):
\[ A \equiv (\sim p \land \sim q) \lor F \]
\[ A \equiv \sim p \land \sim q \]
So, the entire statement pattern becomes:
\[ (\sim p \land \sim q) \rightarrow r \]
For an implication $A \rightarrow r$ to be a tautology, it must be True in all cases. The only way an implication can be False is if the antecedent ($A$) is True and the consequent ($r$) is False. Therefore, to prevent it from ever being False, whenever $A$ is True, $r$ must also be True.
The antecedent $\sim p \land \sim q$ is True only in one specific case: when both $p$ is False and $q$ is False.
So, when $p = F$ and $q = F$, we absolutely must have $r$ evaluate to True.
Let's check the given options under the condition $p = F, q = F$:
(A) $p \land \sim q = F \land T = F$ (Fails)
(B) $q \lor p = F \lor F = F$ (Fails)
(C) $p \land q = F \land F = F$ (Fails)
(D) $\sim q = \sim F = T$ (Works!)
Since option D is the only one that evaluates to True when the antecedent is True, it guarantees the implication is a tautology.
Let's verify analytically. If $r \equiv \sim q$:
\[ (\sim p \land \sim q) \rightarrow \sim q \]
Rewrite using $\sim A \lor B$:
\[ \sim(\sim p \land \sim q) \lor \sim q \]
Apply De Morgan's law:
\[ (p \lor q) \lor \sim q \]
Associativity:
\[ p \lor (q \lor \sim q) \]
Since $q \lor \sim q \equiv T$:
\[ p \lor T \equiv T \]
It is indeed a tautology.
Step 4: Final Answer:
The pattern is a tautology when $r$ is equivalent to $\sim q$.