Reductio ad absurdum when there are no premises and just a supposition $S$

Question

I was reading An Introduction to Formal Logic by Peter Smith - https://www.logicmatters.net/resources/pdfs/IFL2_LM.pdf , and on page 36 he explains Reductio ad absurdum as:

If $A_1, A_2, \dots, A_n$ (as background premisses) plus the temporary supposition $S$ entail a contradiction, then $A_1, A_2, \dots, A_n$ by themselves entail $not$-$S$

He "proves" this by showing that $A$s plus $S$ cannot all be true together. Now he asks the question - "What happens when $n = 0$ and there are no background premisses?"

From the statement of RAA, since there are no premisses, $S$ itself must be a contradiction. So we are supposing a contradiction in our proof. So can we now just directly conclude $not$-$S$? In Formal Logic, is this the same as $\vdash\top$? If yes, what exactly does it mean?

Answer 1

In general, as you report, we have

If $A_1, A_2, \ldots, A_n$ (as background premisses) plus the temporary supposition $S$ entail a contradiction, then $A_1, A_2, \ldots, A_n$ by themselves entail $not$-$S$.

Hence, when there are no background premisses $A_i$, given what is said about "entails" on p. 9,

If the temporary supposition $S$ entails a contradiction, then there is no possible situation in which $not$-$S$ is false, i.e. necessarily $not$-$S$ is true, so $S$ is necessarily false.

A formal correlate of that is if $S \vdash \bot$ then $\vdash \neg S$. But $S$ can be arbitrarily complex, and $\vdash \neg S$ doesn't say the same as the trivial $\vdash \top$. Though, still assuming $S \vdash \bot$, we can derive the non-trivial $\vdash \neg S \equiv \top$.