I think intuitively its obvious. If we have a collection of sentences that never proved $\bot$ and we add a new sentence that proves $p$, there should be no reason that the new set of proposition/assumptions becomes inconsistent (i.e. proves bottom or proves $\neg p$).
However, I assume this has to actually be shown to be true using propositional axioms or something else. Note that this is only interesting if we have not already proved the completeness theorem. Otherwise the solution is simple:
Since we have by assumption that $\Sigma \vdash p$ then it must be $\Sigma \models p$ (every model of $\Sigma$ makes p true). Therefore, since models are defined by truth function (in propositional logic at least) and functions are defined such that:
$$ t(p) = 1 \iff t(\neg) = 0$$
it must mean that $t(\neg) = 0$. Which shows $\Sigma \not \models \neg p$ which again using the completeness theorem means $\Sigma \not \vdash \not p$. Since only having $\neg p$ and $p$ leads to $\bot$ it must be that we can't prove $\bot$ and therefore the new system $\Sigma \cup \{ p \}$ is consistent.
Without that, how do we show that true?
Note this is Corollary 2.2.9 on page 22 of these notes so I assume we can only use knowledge up to that corollary.
As a reference here are the propositional axioms:
- T
- $\varphi \to (\varphi \lor \psi); \varphi \to (\psi \lor \varphi)$
- $\neg \varphi \to (\neg \psi \to \neg (\varphi \lor \psi) $
- $(\varphi \land \psi) \to \varphi; (\psi \land \varphi) \to \psi$
- $\varphi \to (\psi \to (\varphi \land \psi))$
- $(\varphi \to (\psi \to \theta)) \to ((\varphi \to \psi)\to (\varphi \to \theta))$
- $\varphi \to (\neg \varphi \to \bot)$
- $(\neg \varphi \to \bot) \to \varphi$
and the completeness theorem I used:
1st form:
$$ \Sigma \vdash p \iff \Sigma \models p $$
By the cut rule (or the cut theorem depending on the formal proof system under consideration), if we had $\Sigma \vdash p$ and $\Sigma \cup \{ p \} \vdash \bot$ then it would follow that $\Sigma \vdash \bot$.