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


*

*Premise 

*Premise

*Inference from( step #(s)) Justification


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