Is $([P \wedge (\sim Q)] \Rightarrow Q) \Rightarrow P \vdash P$ a theorem in propositional logic? By constructing truth tables, I have found that $([P \wedge (\sim Q)] \Rightarrow Q) \Rightarrow P \vdash P$.
In attempting to prove it, so far I have:
$1 \: (1) \: ([P \wedge (\sim Q)] \Rightarrow Q) \Rightarrow P$ [Assumption]
$2 \: (2) \: [P \wedge (\sim Q)] \Rightarrow Q$ [Assumption]
$3 \: (3) \: P \wedge (\sim Q)$ [Assumption]
$3 \: (4) \: P$ [$\wedge E$, 3]
$3 \: (5) \: \sim Q$ [$\wedge E$, 3]
$2, 3 \: (6) \: Q $ [MP 2, 3]
However, it seems to me that it is cannot be a theorem since $P \wedge (\sim Q)$ cannot imply $Q$.
Is this a theorem, and if so, how can it be proven?
 A: Using standard rules of propositional logic, we have:
$$\begin{equation} \begin{aligned}
(P \land (\sim Q)) \Rightarrow Q
&\equiv \ \sim (P \land (\sim Q) \land (\sim Q)) \\[6pt]
&\equiv \ \sim (P \land (\sim Q)) \\[6pt]
&\equiv \ (\sim P) \lor Q. \\[6pt]
\end{aligned} \end{equation}$$
Hence, we have:
$$\begin{equation} \begin{aligned}
((P \wedge (\sim Q)) \Rightarrow Q) \Rightarrow P
&\equiv \ ((\sim P) \lor Q) \Rightarrow P \\[6pt]
&\equiv \ \sim (((\sim P) \lor Q) \lor (\sim P)) \\[6pt]
&\equiv \ \sim ((\sim P) \lor Q) \\[6pt]
&\equiv \ P \land (\sim Q). \\[6pt]
\end{aligned} \end{equation}$$
Thus, your statement boils down to $(P \land (\sim Q)) \vdash P$, which is indeed a theorem.
A: 
However, it seems to me that it is cannot be a theorem since $P∧(∼Q)$cannot imply $Q$.

Hint:  It can when you have already assumed $\lnot P$ for a proof by reduction to absurdity.
Start here:
$1. ~((P\land\lnot Q)\to Q)\to P\hspace{10ex}\text{Premise}
\\\quad 2. ~\lnot P\hspace{26ex}\text{Assumption}
\\\qquad 3. ~(P\land\lnot Q)\hspace{17.5ex}\text{Assumption}
$
A: You know you can do truth tables in propositional logic right?  To establish $Y(P, Q) \vdash X$ :
$$\begin{array} {ll}
Y \\
\quad P & \text{assumption} \\
\quad \quad Q & \text{assumption} \\
\quad \quad \vdots  \\
\quad \quad X  \\ \\
\quad \quad \lnot Q & \text{assumption} \\
\quad \quad \vdots  \\
\quad \quad X  \\ \\
\quad Q \lor \lnot Q & \text{LEM} \\
\quad X & \text{or elimination} \\\\
\quad \lnot P & \text{assumption} \\
\quad \quad Q & \text{assumption} \\
\quad \quad \vdots  \\
\quad \quad X  \\ \\
\quad \quad \lnot Q & \text{assumption} \\
\quad \quad \vdots  \\
\quad \quad X  \\ \\
\quad Q \lor \lnot Q & \text{LEM} \\
\quad X & \text{or elimination} \\ \\
P \lor \lnot P & \text{LEM} \\
X & \text{or elimination} \\
\end{array}$$
Every $\vdots$ is just a row from the truth table.  Your $Y(P, Q)$ is $((P \land \lnot Q)  \to Q) \to P$ and your $X$ is $P$.  You don't need all these tricky tricks.
