Formal proof for $q \land \neg q \vdash r \land \neg r$

Question

Having some issue with some logic. The question is to formally prove;

$$q \land \neg q \vdash r \land \neg r$$

I've never done this before so would appreciate some help with it. No idea really where to start as $r \land \neg r$ doesn't exist on the left hand side of the equation. I guess I break down the left hand side first but then I don't know where to go.

Would really appreciate any help anyone can provide.

Thanks.

Update

Have attempted it myself but still not sure what to do. With the first part of the formula $q \land \neg q$ ... do I declare this as false? Really have no idea where to go with this, other than deducting the $q \land \neg q$ apart.

Would really appreciate some help on this.

Thanks.

Image of solution so far

Answer 1

From the contradiction $q \land \lnot q$, by $\lnot$-elimination, we have to derive $\bot$, and then apply $\bot$-elimination :

$\dfrac \bot \varphi \ $, with $\varphi$ whatever,

to conclude.

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Proof

1) $q \land \lnot q$ --- premise

2) $q$ --- from 1) by $(\land E)$

3) $\lnot q$ --- from 1) by $(\land E)$

4) $\bot$ --- from 1) and 2) by $(\lnot E)$

5) $r \land \lnot r$ --- from 4) by $(\bot E)$.

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See Natural Deduction: Rules for the rules.

Answer 2

What you probably find useful is to use RAA in some kind of form. From $q\land\neg q$ you can deduce both $q$ and $\neg q$. Assuming $\neg(r \land \neg r)$ would then lead to a contradiction (both $q$ and $\neg q$) and from that we could conclude that $r\land\neg r$.