The negation of $$\sim s \vee (\sim r \wedge s)$$ is equivalent to

We want the negation of the statement $$\sim s \vee (\sim r \wedge s)$$. In symbols we therefore start with

$$\neg\bigl(\,\sim s \;\vee\; (\sim r \wedge s)\bigr).$$

First recall De Morgan’s law for two statements: $$\neg(A\vee B)=\neg A\;\wedge\;\neg B.$$ Here we may identify

$$A=\sim s,\qquad B=(\sim r\wedge s).$$

Applying the law gives

$$\neg(\sim s\vee(\sim r\wedge s))=\neg(\sim s)\;\wedge\;\neg(\sim r\wedge s).$$

The first part simplifies directly because double negation cancels:

$$\neg(\sim s)=s.$$

So we now have

$$s\;\wedge\;\neg(\sim r\wedge s).$$

We still need to simplify the second negation. Again, by De Morgan’s law, but this time for a conjunction, $$\neg(A\wedge B)=\neg A\;\vee\;\neg B.$$ Taking

$$A=\sim r,\qquad B=s,$$

we get

$$\neg(\sim r\wedge s)=\neg(\sim r)\;\vee\;\neg s.$$

Double negation on the first term yields

$$\neg(\sim r)=r.$$

Thus

$$\neg(\sim r\wedge s)=r\;\vee\;\neg s.$$

Substituting this back, our whole expression is now

$$s\;\wedge\;(r\;\vee\;\neg s).$$

Next, we distribute the conjunction over the disjunction using the distributive law $$P\wedge(Q\vee R)=(P\wedge Q)\;\vee\;(P\wedge R).$$ Here $$P=s,\;Q=r,\;R=\neg s,$$ so we obtain

$$s\wedge(r\vee\neg s)=(s\wedge r)\;\vee\;(s\wedge\neg s).$$

However, $$s\wedge\neg s$$ is a contradiction and always evaluates to false, so that term can be dropped:

$$(s\wedge r)\;\vee\;\text{false}=s\wedge r.$$

We have therefore shown

$$\neg\bigl(\,\sim s \vee (\sim r \wedge s)\bigr)\equiv s\wedge r.$$

Hence, the correct answer is Option A.