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Consider the classical modal logic $\mathsf K$, given by the following axioms and rules over a language containing the standard propositional connectives and $\Box$:

  • A complete set of axiom schemes for classical propositional logic
  • $\Box(\phi\rightarrow\psi)\rightarrow(\Box\phi\rightarrow\Box\psi)$
  • From $\phi$ and $\phi\rightarrow\psi$, infer $\psi$
  • From $\vdash\phi$, infer $\vdash\Box\phi$

The formulation of the necessitation rule shall mean that it should only be applied to theorems (to allow derivations from premises) where $\vdash$ denotes provability in $\mathsf K$.

It is well known that $\mathsf K$ is complete with respect to its classical Kripke semantics.

What I am interested in is the following metatheorem:

$$\text{If }\vdash\Box\phi\text{, then }\vdash\phi$$

It is the converse to the necessitation rule and a weakening of the principle $\Box\phi\rightarrow\phi$ (of course only valid in reflexive models) to theorems of the logic.

By appealing to the completeness theorem, it is obvious that it is valid, by assuming $\not\vdash\phi$, i.e. there is a countermodel for $\phi$ and transforming this into a countermodel of $\Box\phi$.

This means that the rule is admissible in $\mathsf K$. My question is if there is a proof of the metatheorem which just appeal to the proof system $\mathsf K$ and not to any semantic issue.

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    Maybe this is overkill, but this follows from cut elimination in a sequent calculus presentation of $K$. – spaceisdarkgreen Jul 04 '19 at 17:09
  • @spaceisdarkgreen i was looking for something more simple, but this is very nice and at least something. –  Jul 04 '19 at 17:12
  • @spaceisdarkgreen Maybe you could elaborate a little more on how it follows from cut elimination. –  Jul 04 '19 at 20:51
  • You would devise a cut-free sequent calculus such that $\vdash\phi$ in the Hilbert system if and only if the sequent $ \Rightarrow\phi$ is provable in the sequent calculus. It being cut-free would guarantee that the only way one could deduce $\Rightarrow \square \phi$ as the endsequent would be to use the inference $$ \dfrac{\Rightarrow \phi}{\Rightarrow \square\phi}$$ (I hesitate to write down the general form of the right rule for $\square$ for fear of getting it wrong, since the kinds of side formulas that can appear would be restricted, in accord with some remarks in your question.) – spaceisdarkgreen Jul 04 '19 at 21:10
  • I wish I knew a good reference to point to here. The best I see offhand is that in Troelstra and Schwictenberg they sketch a cut elimination for a sequent calculus for S4. They give both the devising of the calculus and the proof of cut elimination for K as an exercise. – spaceisdarkgreen Jul 04 '19 at 21:20

1 Answers1

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A sequent calculus (of classical style, i.e., in contrast to hybrid proposals) for the system K (see Heinrich Wansing's "Sequent Systems for Modal Logics" for the Handbook of Philosophical Logic) is obtained with the addition of the following inference rule to the standard set, where $\Box\Gamma$ denotes the sequence of formulas $\Box A_{1},\dots,\Box A_{n}$ when $\Gamma$ is $A_{1},\dots,A_{n}\;$and $\Box\Gamma$ is the empty sequence when $\Gamma$ is empty:

$$ \frac{\Gamma\Rightarrow A}{\Box\Gamma\Rightarrow\Box A} $$

with the restriction that the succedent shall consist of exactly one formula. The restriction is in order to avoid fallacies; because in the "more than one" case, $\Box A\lor\Box\neg A$ would become a theorem:

$$ \cfrac{A\Rightarrow A}{\cfrac{\Rightarrow A,\neg A}{\Rightarrow\Box A,\Box\neg A}} $$

and in the "less than one" case $(\Box A\land \Box \neg A)\rightarrow\bot$ would become a theorem: $$ \cfrac{A\Rightarrow A}{\cfrac{A,\neg A\Rightarrow}{\Box A,\Box\neg A\Rightarrow}} $$

whereas neither of them is in K. It is worth remark that the restriction is in force only at the application of the inference rule within a derivation.

However, the mentioned metatheorem cannot be derived in this system. On the one side, a rule-wise inspection reveals that the second $\Box$ could not be eliminated, for instance, with a step further from a sequent

$\Box\Gamma\Rightarrow A,\Box A$, etc.

On the other side, the sequent $\Box A\Rightarrow A$ would entail

$\Rightarrow\Box A\rightarrow A$

which is not a theorem of K. Indeed, the required move can be provided by another inference rule

$$ \frac{A,\Gamma\Rightarrow\Delta}{\Box A,\Gamma\Rightarrow\Delta} $$

which is added to obtain a sequent calculus for the system T.

For a semantic demonstration of the metatheorem, see my answer to the question Is "If $\vDash\Box\phi$ then $\vDash\phi$" true in modal logic?.

Tankut Beygu
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