Is $([P \wedge (\sim Q)] \Rightarrow Q) \Rightarrow P \vdash P$ a theorem in propositional logic?

Question

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?

Answer 1

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.

Answer 2

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} $

Answer 3

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.