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

We have to examine the statement $$\sim\!\left(p \leftrightarrow \sim q\right)$$ and decide which description among the four alternatives fits it.

First, recall the standard logical equivalence formula for a biconditional. We state it explicitly:

$$p \leftrightarrow r \;\equiv\; (p \wedge r)\;\vee\;(\sim p \wedge \sim r).$$

The symbol “$$\equiv$$” means “is logically equivalent to.” In our problem the role of $$r$$ is played by $$\sim q$$, so we substitute $$r = \sim q$$:

$$p \leftrightarrow \sim q \;\equiv\; \bigl(p \wedge \sim q\bigr)\;\vee\;\bigl(\sim p \wedge \sim(\sim q)\bigr).$$

Because a double negation cancels, $$\sim(\sim q)=q$$. Substituting this, we simplify the right-hand side:

$$p \leftrightarrow \sim q \;\equiv\; \bigl(p \wedge \sim q\bigr)\;\vee\;\bigl(\sim p \wedge q\bigr).$$

Now our original expression contains the negation of this whole biconditional, so we negate what we have just obtained:

$$\sim\!\left(p \leftrightarrow \sim q\right) \;\equiv\; \sim\!\Bigl[\, \bigl(p \wedge \sim q\bigr)\;\vee\;\bigl(\sim p \wedge q\bigr)\Bigr].$$

Next we invoke De Morgan’s law, which states

$$\sim(A \vee B) \;\equiv\; (\sim A) \wedge (\sim B).$$

Identifying $$A = (p \wedge \sim q)$$ and $$B = (\sim p \wedge q)$$, we apply the law:

$$\sim\!\left(p \leftrightarrow \sim q\right) \;\equiv\; \bigl(\sim(p \wedge \sim q)\bigr)\;\wedge\;\bigl(\sim(\sim p \wedge q)\bigr).$$

Each of the two negations inside the brackets can again be opened with De Morgan’s law, this time in its conjunctive form

$$\sim(X \wedge Y) \;\equiv\; (\sim X) \vee (\sim Y).$$

For the first bracket we have $$X=p$$ and $$Y=\sim q$$, giving

$$\sim(p \wedge \sim q) \;\equiv\; (\sim p) \vee \bigl(\sim(\sim q)\bigr).$$

The double negation $$\sim(\sim q)$$ collapses to $$q$$, so the first bracket becomes

$$\sim(p \wedge \sim q) \;\equiv\; (\sim p) \vee q.$$

For the second bracket we have $$X=\sim p$$ and $$Y=q$$, yielding

$$\sim(\sim p \wedge q) \;\equiv\; \bigl(\sim(\sim p)\bigr) \vee (\sim q).$$

Again the double negation $$\sim(\sim p)$$ reduces to $$p$$, giving

$$\sim(\sim p \wedge q) \;\equiv\; p \vee (\sim q).$$

Collecting both results, we have transformed the original statement into

$$\sim\!\left(p \leftrightarrow \sim q\right) \;\equiv\; \bigl((\sim p) \vee q\bigr)\;\wedge\;\bigl(p \vee (\sim q)\bigr).$$

Thus

$$\boxed{\; \sim\!\left(p \leftrightarrow \sim q\right) \;\equiv\; (\sim p \vee q)\;\wedge\;(p \vee \sim q) \;}$$

We now show that the right-hand side is exactly $$p \leftrightarrow q$$. To see this, we once more recall the biconditional formula, this time with $$r=q$$:

$$p \leftrightarrow q \;\equiv\; (p \wedge q)\;\vee\;(\sim p \wedge \sim q).$$

One way to verify equivalence of two compound statements is to check that their truth tables match. Below are the four possible truth-value assignments for $$p$$ and $$q$$, and the resulting truth values of $$(\sim p \vee q)\wedge(p \vee \sim q)$$:

$$ \begin{array}{c|c||c|c|c} p & q & \sim p \vee q & p \vee \sim q & \text{Conjunction} \\ \hline T & T & T & T & T\\ T & F & F & T & F\\ F & T & T & F & F\\ F & F & T & T & T \end{array} $$

The last column reads T, F, F, T—precisely the truth pattern of $$p \leftrightarrow q$$. Hence

$$ (\sim p \vee q)\;\wedge\;(p \vee \sim q) \;\equiv\; p \leftrightarrow q. $$

Combining this with the boxed equivalence obtained earlier, we finally deduce

$$\boxed{\; \sim\!\left(p \leftrightarrow \sim q\right) \;\equiv\; p \leftrightarrow q \;}$$

This matches Option C in the given list. It is therefore neither a tautology nor a fallacy in isolation; it is simply equivalent to another specific biconditional.

Hence, the correct answer is Option C.