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The cut rule is given as follows by these two wikipedia articles 1, 2:

$$\Gamma \vdash \Delta, A \qquad \Sigma, A \vdash \Pi \over \Gamma, \Sigma \vdash \Delta, \Pi$$

I have the following proof which converts to and from implications.

Starting with the two sequents,

\begin{align} & \Gamma \vdash \Delta, A \qquad & \Sigma, A \vdash \Pi \\ & \Gamma \rightarrow \Delta \vee A \qquad & \Sigma \wedge A \rightarrow \Pi \\ & \neg \Gamma \vee (\Delta \vee A) \qquad & \neg (\Sigma \wedge A) \vee \Pi \\ & \neg \Gamma \vee \Delta \vee A \qquad & \neg \Sigma \vee \neg A \vee \Pi \\ \end{align}

IIUC, the original sequents ($\Gamma \vdash \Delta, A$ and $\Sigma, A \vdash \Pi$) must each be true, so we must have

$$(\neg \Gamma \vee \Delta \vee A) \wedge (\neg \Sigma \vee \neg A \vee \Pi)$$

Doing some case work, $A$ must be true or $A$ must be false (i.e. $A \vee \neg A$):

  • if $A$ is true, the above reduces to $\neg \Sigma \vee \Pi$
  • if $A$ is false, the above reduces to $\neg \Gamma \vee \Delta$

\begin{align} A & \vee \neg A \\ (A \rightarrow \neg \Sigma \vee \Pi) & \wedge (\neg A \rightarrow \neg \Gamma \vee \Delta) \\ (\neg \Sigma \vee \Pi) & \vee (\neg \Gamma \vee \Delta) \\ \neg \Sigma \vee \neg \Gamma & \vee \Delta \vee \Pi \\ \end{align}

Which can be converted back into a sequent:

\begin{align} & \vdash \neg \Sigma, \neg \Gamma, \Delta, \Pi \\ \Sigma, \Gamma & \vdash \Delta, \Pi \\ \Gamma, \Sigma & \vdash \Delta, \Pi \end{align}

Thus proving the cut rule.

Would there have been a way to prove the cut rule without converting the sequents into implications?

joseville
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    What exactly are you trying to prove here? What does it mean to "prove the cut rule"? – Mark Saving Nov 19 '21 at 19:13
  • To show that give "Γ⊢Δ," and "Σ,⊢Π", we can get to "Γ,Σ⊢Δ,Π". I may be using incorrect terminology, so curious to know how you would phrase it. – joseville Nov 19 '21 at 19:16
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    Your question doesn't make sense. The cut rule is one of the axiomatic ways you can come to new derivations. It is part of the definition of $\vdash$. So by definition, the cut rule is true. – Mark Saving Nov 19 '21 at 19:20
  • Oh, thanks for clarifying that - I guess my goal is then to get a sense of how the cut rule is consistent with the propositional logic I'm more familiar with. Kind of related question, when we start with two sequents as in $$\Gamma \vdash \Delta, A \qquad \Sigma, A \vdash \Pi$$, does that mean we're taking both to be true\hold? – joseville Nov 19 '21 at 19:25
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    No, it means that there is a derivation of each of the two sequents under the rules of sequent calculus. "Truth" is a separate question; the statement $\Gamma \vdash \Delta$ simply means that there is a formal derivation of this sequent using the rules of the LK sequent calculus. – Mark Saving Nov 19 '21 at 19:26
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    You're confusing the syntactic part of logic (eg the formal proof system of LK calculus) with the semantic part (which discusses "truth"). What you're doing is proving the "semantic cut rule" - that is, if $\Gamma \models \Delta, A$ and $\Sigma, A \models \Pi$, then $\Gamma, \Sigma \models \Delta, \Pi$. This would be a component of a proof that the LK system is sound - that is, if $\Theta \vdash \Omega$ then $\Theta \models \Omega$. – Mark Saving Nov 19 '21 at 19:45

1 Answers1

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What you tried to "prove" above semantically is actually called the admissibility of the cut rule in sequent calculus as referenced in this Wikipedia article:

In proof theory, admissibility is often considered in the context of sequent calculi, where the basic objects are sequents rather than formulas. For example, one can rephrase the cut-elimination theorem as saying that the cut-free sequent calculus admits the cut rule

However, cut rule as a basic rule is not derivable from other inference rules of the same system, otherwise Gentzen's cut-elimination theorem will be much easier to get. And your above "proof" is not acceptable to derive cut rule within $LK$ (not by translating to another Hilbert or mixed system), you need to try to use other clearly specified rules of $LK$ and you'll be certain to fail.

Your above effort makes sense in the context of the end of the linked Wikipedia article, which states:

(By abuse of language, it is also sometimes said that the (full) sequent calculus admits cut, meaning its cut-free version does.) However, admissibility in sequent calculi is usually only a notational variant for admissibility in the corresponding logic: any complete calculus for (say) intuitionistic logic admits a sequent rule if and only if IPC admits the formula rule which we obtain by translating each sequent to its characteristic formula.

In conclusion cut rule is an admissible but not derivable rule in most sequent calculi.

joseville
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cinch
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