Understanding the Logical Statement: ~(p ↔ ~q)

Let's analyze the given logical statement step by step:

Step 1: Understand the Components

The statement is $\sim (p\leftrightarrow \sim q)$.

• $\leftrightarrow$ represents the biconditional (if and only if) operator.
• $\sim$ represents the negation operator.

Step 2: Construct a Truth Table

We'll evaluate the truth values for all combinations of p and q:

Step 3: Compare with Given Options

Now let's compare the truth values of $\sim (p\leftrightarrow \sim q)$ with each option:

Option 1: p ↔ q

This does NOT match our original statement.

Option 2: Tautology (always true)

Our statement is not always true (see truth table), so it's not a tautology.

Option 3: Fallacy (always false)

Our statement is not always false, so it's not a fallacy.

Option 4: ~p ↔ q

This does NOT match our original statement either.

Final Answer

None of the given options exactly match the truth values of $\sim (p\leftrightarrow \sim q)$. The correct equivalent is actually $p\leftrightarrow q$, which matches option 1, but our truth table shows they are different. Let me re-examine:

Looking more carefully, $\sim (p\leftrightarrow \sim q)$ is actually equivalent to $p\leftrightarrow q$! Let me verify this equivalence using logical identities:

$\sim (p\leftrightarrow \sim q)\equiv \sim \left[\right(p\to \sim q)\wedge (\sim q\to p\left)\right]\equiv \sim (p\to \sim q)\vee \sim (\sim q\to p)\equiv (p\wedge q)\vee (q\wedge \sim p)\equiv p\leftrightarrow q$

So the correct answer is: equivalent to p ↔ q (option 1)

Related Topics

• Propositional Logic
• Logical Equivalences
• Truth Tables
• Biconditional Statements
• Negation Laws

Key Formulae

$p\leftrightarrow q\equiv (p\to q)\wedge (q\to p)$

$\sim (p\leftrightarrow q)\equiv p\leftrightarrow \sim q\equiv \sim p\leftrightarrow q$

$\sim (p\to q)\equiv p\wedge \sim q$