I am trying to improve my understanding of the negation of a quantified statement where the statement is an implication. I am doing a practice problem which I dont have the answer to from the textbook and would like to check if mine is right. The question is as follows:
Let $P(x)$ and $Q(x)$ be open sentences where the domain of the variable $x$ is $T$. Which of the following implies that $(\thicksim P(x)) \implies Q(x)$ is false for some $x \in S$.
a) $P(x) \wedge Q(x)$ is false for all $x \in S$
b) $P(x)$ is true for all $x \in S$
c) $Q(x)$ is true for all $x \in S$
d) $P(x) \vee Q(x)$ is false for some $x \in S$
e) $P(x) \wedge (\thicksim Q(x))$ is false for all $x \in S$
How I approached the question is: I want to show that the statement $\exists x \in S$ s.t $(\thicksim P(x)) \implies Q(x)$ is false. Then, $$\thicksim(\exists x \in S \text{ s.t } (\thicksim P(x)) \implies Q(x)) \text{ is true }$$ $$\implies \forall x \in S \text{ s.t } (\thicksim P(x) \wedge \thicksim Q(x)) \text{ is true }$$ This is because $$\thicksim (\thicksim P(x) \implies Q(x))\ \ \text{is} \ \ \thicksim(\thicksim(\thicksim P(x)) \vee Q(x)$$ $$\implies \thicksim(P(x) \vee Q(x))$$ $$\implies \thicksim P(x) \wedge \thicksim Q(x)$$
Hence, the answer that I got is D. Is there any other options or am I correct here ? I would appreciate some opinions on this. Thank you