Question:
The statement $[(p\rightarrow q)\wedge\sim q]\rightarrow r$ is tautology, when $r$ is equivalent to
The statement $[(p\rightarrow q)\wedge\sim q]\rightarrow r$ is tautology, when $r$ is equivalent to
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Logic Tip:Modus Tollens rule states that $[(p\rightarrow q)\wedge\sim q] \rightarrow \sim p$ is a tautology. Similarly, it implies $\sim q$.Constructing a quick truth table is a foolproof alternative to algebraic simplification.
Updated On: Apr 23, 2026
- $p\wedge\sim q$
- $q\vee p$
- $p\wedge q$
- $\sim q$
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The Correct Option is D
Solution and Explanation
Concept:
Mathematical Logic - Tautology and Truth Tables.
Step 1: Understand the condition for a tautology.
A compound statement is a tautology if it is true (T) for all possible truth values of its component propositions. The statement given is an implication: $A \rightarrow r$, where $A = [(p\rightarrow q)\wedge\sim q]$.
Step 2: Analyze the antecedent part of the statement.
We simplify the antecedent $[(p\rightarrow q)\wedge\sim q]$ using logical equivalences. The implication $(p\rightarrow q)$ is equivalent to $(\sim p \vee q)$.
Step 3: Apply logical equivalences to simplify.
Substitute this back: $(\sim p \vee q) \wedge \sim q$. Applying the distributive law, this becomes $(\sim p \wedge \sim q) \vee (q \wedge \sim q)$. Since $(q \wedge \sim q)$ is always false (a contradiction), the expression simplifies strictly to $(\sim p \wedge \sim q)$.
Step 4: Evaluate the requirement for the overall implication to be a tautology.
We now need the implication $(\sim p \wedge \sim q) \rightarrow r$ to always be true.
This implication is only in danger of being false if the antecedent is true and the consequent $r$ is false. The antecedent is true only when both $p$ is false and $q$ is false.
Step 5: Verify which option prevents the implication from being false.
When $p=F$ and $q=F$, $r$ must be True for the statement to be a tautology. Let's test the options for $p=F, q=F$: (A) $F\wedge T = F$; (B) $F\vee F = F$; (C) $F\wedge F = F$; (D) $\sim F = T$. Since only $\sim q$ evaluates to True under these specific conditions, $r$ must be equivalent to $\sim q$. $$ \therefore \text{The correct answer is Option D: } \sim q $$
Mathematical Logic - Tautology and Truth Tables.
Step 1: Understand the condition for a tautology.
A compound statement is a tautology if it is true (T) for all possible truth values of its component propositions. The statement given is an implication: $A \rightarrow r$, where $A = [(p\rightarrow q)\wedge\sim q]$.
Step 2: Analyze the antecedent part of the statement.
We simplify the antecedent $[(p\rightarrow q)\wedge\sim q]$ using logical equivalences. The implication $(p\rightarrow q)$ is equivalent to $(\sim p \vee q)$.
Step 3: Apply logical equivalences to simplify.
Substitute this back: $(\sim p \vee q) \wedge \sim q$. Applying the distributive law, this becomes $(\sim p \wedge \sim q) \vee (q \wedge \sim q)$. Since $(q \wedge \sim q)$ is always false (a contradiction), the expression simplifies strictly to $(\sim p \wedge \sim q)$.
Step 4: Evaluate the requirement for the overall implication to be a tautology.
We now need the implication $(\sim p \wedge \sim q) \rightarrow r$ to always be true.
This implication is only in danger of being false if the antecedent is true and the consequent $r$ is false. The antecedent is true only when both $p$ is false and $q$ is false.
Step 5: Verify which option prevents the implication from being false.
When $p=F$ and $q=F$, $r$ must be True for the statement to be a tautology. Let's test the options for $p=F, q=F$: (A) $F\wedge T = F$; (B) $F\vee F = F$; (C) $F\wedge F = F$; (D) $\sim F = T$. Since only $\sim q$ evaluates to True under these specific conditions, $r$ must be equivalent to $\sim q$. $$ \therefore \text{The correct answer is Option D: } \sim q $$
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