I'm asked if adding the following rule to natural deduction would maintain the soundness and completeness of natural deduction. I think with the first one, natural deduction would maintain its completeness, because it doesn't change or take away any rules, so everything true can be proved true with the logic system, but it wouldn't be sound, but I'm not exactly sure why, it seems like maybe because one assumption could be false, then from that you're concluding true with the OR, which seems like a contradiction.
ϕ ψ
.............. (∨I')
ϕ ∨ ψ
With the second one, I feel it's the same that it would be complete still, but not sound because you can't just prove true from nothing.
................ (¬⊥I)
¬⊥
Both ND+$\lor I'$ and ND+$\neg\bot I$ are sound:
ND+$\lor I'$-Soundness: Suppose $\{\phi,\psi\}\vdash(\phi\lor\psi)$ by $\lor I'$, then we have to show $\{\phi,\psi\}\models(\phi\lor\psi)$, but this is of course true by the usual semantics of ND.
ND+$\neg\bot I$-Soundness: Similarly, we have $\models\neg\bot$ since every model $A$ has $A(\bot)=F$ and hence has $A(\neg\bot)=T$.
And ND+$R$ where $R$ is any rule is complete:
ND+$R$-Completeness: Suppose $\Gamma\models\phi$, then $\Gamma\vdash_{\mathrm{ND}}\phi$ by ND-completeness, which is a still a proof of $\Gamma\vdash_{\mathrm{ND}+R}\phi$ (without any uses of $R$).
I'm not an expert in logic so if there are any mistakes, please me know!