Natural deduction prove that A$\vee$B, $\neg$ A, $\neg$B $\vdash$ A $\wedge \neg$A

Question

I want to prove the following deduction in the natural deduction system:

A$\vee$B, $\neg$ A, $\neg$B $\vdash$ A $\wedge$$\neg$A

Specifically the difficult part:

A$\vee$B, $\neg$ A, $\neg$B $\vdash$ A

How can the result be proven by using the common inference rules?

Answer 1

You have the premise $A\lor B$, as well as the premise $\lnot B$

$\rightarrow (A\lor B) \land (\lnot B)\quad \land$-Introduction

Then

$\big((A\lor B) \land \lnot B\big) \rightarrow A$ by disjunction elimination, also known as disjunction syllogism.

So you can derive $A$ as above. And you're given $\lnot A$. Then through $\land$-Introduction with A, your have $A \land \lnot A$.

$\bot$

I'll let you set up the actual proof with steps numbered, and steps involved, and justifications given

1. Premise
2. Premise
3. Inference from( step #(s)) Justification

$\quad \vdots$

$\quad \vdots$

$A\land \lnot A$

$\bot$

Answer 2

$$\begin{array} {r|ll} (1) & A \lor B & \text{Given} \\ (2) & \lnot A & \text{Given} \\ (3) & \lnot B & \text{Given} \\ & & \\ (4) & \quad \quad A & \text{Premise} \\ (5) & \quad \quad \bot & \text{Contradiction of 2 and 4} \\ & & \\ (6) & \quad \quad B & \text{Premise} \\ (7) & \quad \quad \bot & \text{Contradiction of 3 and 6} \\ & & \\ (8) & \bot & \text{Or Eliminination of 1, 4 to 5, 6 to 7} \\ (9) & \text{Anything} & \text{Vaccuous Implication} \\ \end{array}$$