The negation of the Boolean expression $$x \leftrightarrow \sim y$$ is equivalent to:

We have to find the negation (logical NOT) of the Boolean statement $$x \leftrightarrow \sim y$$ and then express the result in a simplified disjunctive form so that it can be matched with one of the given options.

First of all, let us recall the standard equivalence for a biconditional. The formula for any two propositions $$p$$ and $$q$$ is:

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

Now we apply this rule to the concrete biconditional $$x \leftrightarrow \sim y$$ by replacing $$p$$ with $$x$$ and $$q$$ with $$\sim y$$. Doing so gives:

$$x \leftrightarrow \sim y \;=\; (x \wedge \sim y)\;\vee\;(\sim x \wedge \sim(\sim y)).$$

The double negation inside the second conjunct can be removed with the rule $$\sim(\sim y)=y$$. Hence we get:

$$x \leftrightarrow \sim y \;=\; (x \wedge \sim y)\;\vee\;(\sim x \wedge y).$$

We must now take the negation of the whole expression. So we write:

$$\sim\bigl(x \leftrightarrow \sim y\bigr) \;=\; \sim\Bigl[(x \wedge \sim y)\;\vee\;(\sim x \wedge y)\Bigr].$$

To push the negation inside the parentheses, we invoke De Morgan’s law, which states:

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

Here $$A$$ is $$x \wedge \sim y$$ and $$B$$ is $$\sim x \wedge y$$. Therefore:

$$\sim\Bigl[(x \wedge \sim y)\;\vee\;(\sim x \wedge y)\Bigr] \;=\; \bigl[\sim(x \wedge \sim y)\bigr] \;\wedge\; \bigl[\sim(\sim x \wedge y)\bigr].$$

The next step is to remove each inner negation. Again, we apply De Morgan’s law, this time in the form $$\sim(A \wedge B) = (\sim A) \vee (\sim B).$$ For the first bracket we have $$A = x$$ and $$B = \sim y$$:

$$\sim(x \wedge \sim y) \;=\; (\sim x) \;\vee\; (\sim(\sim y)) \;=\; (\sim x) \vee y.$$

For the second bracket we have $$A = \sim x$$ and $$B = y$$:

$$\sim(\sim x \wedge y) \;=\; (\sim(\sim x)) \;\vee\; (\sim y) \;=\; x \vee (\sim y).$$

So the entire expression is now

$$\bigl[(\sim x) \vee y\bigr] \;\wedge\; \bigl[x \vee (\sim y)\bigr].$$

To simplify, we distribute the conjunction across the disjunctions. The distributive law tells us:

$$(A \vee B)\wedge(C \vee D) \;=\; (A \wedge C)\;\vee\;(A \wedge D)\;\vee\;(B \wedge C)\;\vee\;(B \wedge D).$$

Letting $$A = \sim x,\; B = y,\; C = x,\; D = \sim y,$$ we compute each product term one by one:

1. $$(\sim x) \wedge x = \text{False},$$ because a proposition and its negation cannot both be true.

2. $$(\sim x) \wedge (\sim y) = \sim x \wedge \sim y.$$

3. $$y \wedge x = x \wedge y.$$

4. $$y \wedge (\sim y) = \text{False},$$ again because a variable and its negation cannot both be true.

Combining the surviving non-false terms under disjunction gives:

$$(\sim x \wedge \sim y)\;\vee\;(x \wedge y).$$

Conventionally we write the positive term first, so we obtain the final simplified negation as:

$$ (x \wedge y)\;\vee\;(\sim x \wedge \sim y). $$

Now we match this with the options provided. Option B is exactly

$$(x \wedge y) \vee (\sim x \wedge \sim y).$$

Hence, the correct answer is Option 2.