Question about a basic logic problem

The negation outside the parentheses will change the 'and' connective inside the parentheses to an 'or' connective. Thus, the equivalent statement would be

"I will not buy the pant or I will buy the shirt"

You can verify this by comparing the truth tables for $\lnot P\lor S$ and $\lnot (P\land \lnot S)$.

$\lnot (A\land B) \ne \lnot A \land \lnot B$.

Instead $\lnot(A\land B) = \lnot A \lor \lnot B$.

So $\lnot (P\land \lnot Q) = \lnot P \lor Q$.

It is not the case that I will buy the pants but not the shirt.

So either I wont buy the pants OR I will buy the shirt.

Now it's possible that $\lnot P \land Q$. It's possible that I might not but the pants but buy the shirt, but I don't have to. I could simple not buy the pants; Then $\lnot (P\land \lnot Q)$, because $\lnot P$, whether I buy the shirt or not. Or I might simply buy the shirt; then $\lnot(P \land \lnot Q)$, because $Q$, whether I buy the pants or not.

Look at the truth tables.

$\begin{matrix} P & Q & \lnot (P\land \lnot Q) & \lnot P \lor Q& \lnot P\land Q\\T&T&\color{blue}{T}\text{(bcs $Q$ is not false)}&\color{blue}T\text{(bcs $Q$ is true)}& \color{red} F\text{(bcs $P$ is not false)}\\ T&F&F&F&F\\F&T&T&T&T\\F&F&\color{blue}{T}\text{(bcs $P$ is false)}&\color{blue}T\text{(bcs $P$ is not true)}& \color{red} F\text{(bcs $Q$ is false)}\\ \end{matrix}$

(Posted after answer was accepted.)

Assuming that $P$ and $S$ are both true. Using a form of natural deduction, we have: