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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)

    ¬⊥

Jamie Whitaker
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  • What is a conclusion in the first rule? – Cohesion Dec 29 '18 at 06:27
  • ϕ ∨ ψ Sorry I'm not sure what happened to it. – Jamie Whitaker Dec 29 '18 at 06:28
  • To prove soundness you have to show that the new rules are truth-preserving. It's quite easy. The second rule is obviously truth-preserving, because there is nothing above the line and below there is a tautology. – Cohesion Dec 29 '18 at 06:34
  • Is the rule $\lor_I'$ added to the usual rules $\lor_{I_1}$ and $\lor_{I_2}$ in natural deduction, or does it replace them? – Taroccoesbrocco Dec 29 '18 at 07:10
  • I believe this is true since the new system is no more powerful than the old system. The fact that all proofs in the new system could be re-formulated in the old system means that IF a contradiction could be found in the new system, THEN a contradiction could be found in the old system. By contrapositive, soundness is implied. Not as sure about completeness. – Quelklef Dec 29 '18 at 22:00

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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!

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    For completeness in both cases, shouldn't it be sufficient just to say any valid sequent $\Gamma \vdash \phi$ has a proof in ND which fairly trivially lifts to a proof in ND+$\vee I'$ and to a proof in ND+$\lnot\bot I$? – Daniel Schepler Jan 09 '19 at 18:21
  • Wow I completely missed that, yeah it sounds like completeness is trivial then. Let me edit that in. –  Jan 09 '19 at 23:57