How does one prove the following using natural deduction: $a ∧ ¬a \vdash b ∧ ¬b$ I am trying to prove the following using natural deduction and this is what I have so far. I am not sure, however, if this is entirely correct. 
If I could get some verification and be pointed to the right direction, that would be great.
    a ∧ ¬a |- b ∧ ¬b
1     a ∧ ¬a     assump 0
2     b |- a ∧ ¬a 
2.1   b          assump 2
2.2  ¬b         ¬I 2.1
3     b ∧ ¬b    ∧I 2.1, 2.2
4    ¬b |- a ∧ ¬a 
4.1  ¬b          assump 4
4.2   b         ¬E 4.1
5     b ∧ ¬b    ∧I 4.1, 4.2

This is not a duplicate as the symbols and solutions provided with the question it being marked against are different.
 A: Since $a ∧ ¬a$ is a contradiction we can derive anything from such a premise using explosion.
Here is a proof checker showing that we could use explosion in natural deduction to reach the goal:

The proof uses the following natural deduction rules:


*

*Conjunction elimination (∧E)

*Contradiction introduction (⊥I)

*Explosion (X)


See the text forallx for further details about these rules.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/
A: Negation introduction requires a contradiction to be derived from the assumption. Your premise is a contradiction, so just reiterate.
The negation elimination rule you are using appears to be Reductio Ab Absurdum, and operates similarly.  Different proof systems name negation rules differently, so it may be useful to note that you mean RAA.
Only after you have derived both conjuncts, may you finish with the conjunction introduction.
0  a ∧ ¬a |- b ∧ ¬b
1         a ∧ ¬a       assump 0
2         b |- a ∧ ¬a 
2.1         b          assump 2
2.2         a ∧ ¬a     reit 1
3        ¬b            ¬intro 2
4        ¬b |- a ∧ ¬a 
4.1         ¬b         assump 4
4.2         a ∧ ¬a     reit 1
5        b             ¬elim 4  (RAA)   
6        b ∧ ¬b        ∧intro 5,3

