Ytukyg

The answer mentions it is a tautology but does not explain the procedure.

So, not every logic does this! (There are many different logics based on the rules that you are allowed to use to infer things.)

In particular, this is called the principle of explosion and the logics which reject it are called paraconsistent logics. Usually to make normal logic into a paraconsistent system you delete one of the following rules:

1. If we have concluded $X$ then we can conclude, for any $Y$, $X vee Y$, which you can read as "either $X$ or $Y$ (or both)."




2. If we have concluded both $X vee Y$ and $neg X$, then we can conclude $Y$.

If you have both of these rules in your arsenal then it is easy to see why knowing both $P$ and $neg P$ would entail any $Q$, just first use $P$ to conclude $P vee Q$ and then use that plus $neg P$ to entail $Q$.

Since these are standard rules, dropping them from your logic results in a weaker logic: it doesn't prove any new things, it just fails to prove some existing things that the stronger logic has no problem with.

One motivation why some mathematicians (and philosophers etc.) wish to do this sort of thing has to do with the very shape of this claim "for all $Q$, $(P wedge neg P)to Q.$" Classical logic is very happy to rewrite $X to Y$ as $neg X vee Y$ in a way that enables vacuous consistency, for example I might say "if it is raining in Ithaca right now then the President of the United States is a gerbil!" and that is a completely true statement because it happens to be snowing, not raining, in Ithaca right now. But for a lot of folks this doesn't "smell right" because these things are not related to each other, and I want to say that I should only be able to phrase $X to Y$ when there is some sort of meaningful connection called "relevance" between $X$ and $Y$. Well the problem with "for all $Q$, $(P wedge neg P)to Q$" is precisely that the "for all $Q$" must include statements which are not relevant to either $P$ or $neg P$ and therefore cannot be relevant to $P wedge neg P$.

Think about it this way:

If we have {$p,neg p,neg q$} $vdash p$ and we also have {$p,neg p,neg q$} $vdash neg p$ (we know we have these because we can assume anything that's inside the premise set)
then we can move the negation of one of the things inside our premise set onto the right side (since we had a contradiction). It doesn't matter which thing we move outside the premise set, so we can choose to move the $neg q$ outside. This means that

{$p,neg p$} $vdash neg neg q$

{$p,neg p$} $vdash q$

Notice that we could have actually used just $q$ in our premise set instead of $neg q$, so this means that

{$p,neg p$} $vdash neg q$

In fact, {$p,neg p$} can imply anything.

To prove $Aimplies [neg A implies B]$ you need only the following elementary methods and rules of natural deduction:

1. 
   $implies$-Introduction (Conlusion, direct proof)


2. 
   $neg$-Introduction (Conclusion, proof by contradiction)


3. 
   $land$-Introduction (Join)


4. 
   $negneg$-Elimination (Rem DNeg)

In DC Proof, for example (user comments in blue font):