Question:
Consider the following statements:
(i) \(\sim (p\wedge \sim q)\)
(ii) \((p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)\)
Result 1 : (i) and (ii) are logically equivalent
Result 2 : (ii) is a contradiction
then which of the following is correct?
Consider the following statements:
(i) \(\sim (p\wedge \sim q)\)
(ii) \((p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)\)
Result 1 : (i) and (ii) are logically equivalent
Result 2 : (ii) is a contradiction
then which of the following is correct?
(i) \(\sim (p\wedge \sim q)\)
(ii) \((p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)\)
Result 1 : (i) and (ii) are logically equivalent
Result 2 : (ii) is a contradiction
then which of the following is correct?
Show Hint
Simplify statement (i) to an implication and recall that a pattern with its contrapositive is always true.
Updated On: Nov 12, 2026
- Result 1 is true, Result 2 is true
- Result 1 is true, Result 2 is false
- Result 1 is false, Result 2 is true
- Result 1 is false, Result 2 is false
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Two statements are logically equivalent if they have the same truth table. A contradiction is always false.
Step 2: Simplify statement (i)
By De Morgan's law, \(\sim(p\wedge \sim q)\equiv \sim p\vee q\equiv p\rightarrow q\).
Step 3: Analyse statement (ii)
The implication \(p\rightarrow q\) is equivalent to its contrapositive \(\sim q\rightarrow \sim p\).
So \((p\rightarrow q)\leftrightarrow(\sim q\rightarrow \sim p)\) is always true, i.e. a tautology.
Step 4: Check the results
Statement (i) is not always true, while (ii) is a tautology, so they are not equivalent: Result 1 is false.
A tautology is not a contradiction: Result 2 is false.
Final Answer:
Both Result 1 and Result 2 are false. \[ \boxed{\text{(D) }\text{Result 1 is false, Result 2 is false}} \]
Two statements are logically equivalent if they have the same truth table. A contradiction is always false.
Step 2: Simplify statement (i)
By De Morgan's law, \(\sim(p\wedge \sim q)\equiv \sim p\vee q\equiv p\rightarrow q\).
Step 3: Analyse statement (ii)
The implication \(p\rightarrow q\) is equivalent to its contrapositive \(\sim q\rightarrow \sim p\).
So \((p\rightarrow q)\leftrightarrow(\sim q\rightarrow \sim p)\) is always true, i.e. a tautology.
Step 4: Check the results
Statement (i) is not always true, while (ii) is a tautology, so they are not equivalent: Result 1 is false.
A tautology is not a contradiction: Result 2 is false.
Final Answer:
Both Result 1 and Result 2 are false. \[ \boxed{\text{(D) }\text{Result 1 is false, Result 2 is false}} \]
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