Question about a basic logic problem The following problem is presented in "How to prove it":
$\neg(P\land \neg S )$
Where P stands for "I will buy the pants" and S for, "I will buy the shirt".
Here is how I would tackle this:
I first look at the statement in parantheses, which just says P but (and) not S. Then the negation of this statement would be, not P but (and) S.
Where am I wrong?
Thanks in advance.
 A: 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 $\lnot (P\land \lnot S)$.
A: $\lnot (A\land B) \ne \lnot A \land \lnot B$.
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}$
A: (Posted after answer was accepted.)
Assuming that $P$ and $S$ are both true. Using a form of natural deduction, we have:

