Question

Consider the following statements:
Statement I : \[\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)\] is a fallacy.
Statement II : \[\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)\] is a tautology.
A. statement I is true; statement II is true; statement II is not a correct explanation for statement I.
B. statement I is true; statement II is false.
C. statement I is false; statement II is true.
D. statement I is true; statement II is true; statement II is a correct explanation for statement I.

Answer 1

Hint: To solve this question, we will use the concept of algebra of propositions. A proposition function which is true in every possible case (in final result) is called tautology. A proposition function which is false in every possible case (in final result) is called fallacy.

Complete step-by-step answer:
As we know that,
\[
   \Rightarrow {\text{ }} \wedge {\text{ }} = {\text{ conjunction (AND)}} \\
   \Rightarrow {\text{ }} \sim {\text{ = negation (NOT)}} \\
   \Rightarrow {\text{ }} \to {\text{ = conditional statement (if p, then q)}} \\
   \Rightarrow {\text{ }} \leftrightarrow {\text{ = biconditional statement (p iff q)}} \\
\]
Now, we will make the truth table for both statements.
Truth table of statement II:

p	q	$ \sim p$	$ \sim q$	$\left( {p \wedge \sim q} \right)$	$\left( { \sim p \wedge q} \right)$	\[\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)\]
T	T	F	F	F	F	F
T	F	F	T	T	F	F
F	T	T	F	F	T	F
F	F	T	T	F	F	F

As we know that,
A proposition function which is false in every possible case (in final result) is called fallacy.
Therefore, from the above table we can see that statement I: \[\left( {p \wedge \sim q} \right) \wedge \left( { \sim p \wedge q} \right)\] is false in every statement that is why it is a fallacy.
Truth table of statement II:

p	q	$ \sim p$	$ \sim q$	$\left( {p \to q} \right)$	$\left( { \sim q \to \sim p} \right)$	\[\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)\]
T	T	F	F	T	T	T
T	F	F	T	F	F	T
F	T	T	F	T	T	T
F	F	T	T	T	T	T

As we know that,
A proposition function which is true in every possible case (in final result) is called tautology.
Therefore, from the above truth table of statement II, we can see that \[\left( {p \to q} \right) \leftrightarrow \left( { \sim q \to \sim p} \right)\] is true in every statement.
Hence, statement II is a tautology.
Now, we can say that statement I and statement II both are true, but they are independent of each other and any of the statements cannot explain the other statement.

So, the correct answer is “Option A”.

Note: Whenever we ask such types of questions, we have to remember some basic propositional logic. The proposition function which is true sometimes and false sometimes, is called contingency or we can say that the proposition which is neither a tautology nor a fallacy is called contingency. The fallacy proposition is also known as contradiction.