\documentclass{article}
\input linear
\input theorem
%\usepackage{enumerate}
\title{Classical Truth as a Modal Operator}
\author{Jason Reed}
\def\says{\Longrightarrow}
\def\limp{\supset}
\def\D{{\cal D}}
\def\E{{\cal E}}
\def\true{\mathop{\textsf t}\nolimits}
\def\false{\mathop\textsf f}
\def\cent{{\textbf{\textsf{C}}}}
\begin{document}
\maketitle
We have a judgment $\Gamma \prov \gamma$ where contexts are defined by
$$\Gamma ::= \cdot \celse \Gamma, A \celse \Gamma,  A\true \celse \Gamma, A\false$$
and propositions by
$$A,B,C ::= P \celse \bot \celse \cent A \celse A\limp B$$
and the optionally empty conclusion $\gamma$ is
$$\gamma ::= \cdot \celse A$$

Hypotheses $A\true$ and $A\false$ in $\Gamma$ are assumptions of
classical truth and falsehood. Bare hypotheses $A$ in $\Gamma$ are
intuitionistic assumptions. 
The right-hand side of the judgment $\Gamma\prov C$
is an intuitionistic conclusion. $\Gamma\prov \cdot$ means that
the assumptions in $\Gamma$ are contradictory.

Propositions are either propositional variables $P$ or falsehood $\bot$
or constructed from other propositions by 
 the `classically true' modality $\cent$ and implication $\limp$.

The sequent calculus is as follows. The initial sequents are

$$\begin{prooftree}
\using init_i \justifies
\Gamma, P \prov P
\end{prooftree}
\qquad
\begin{prooftree}
\using init_c \justifies
\Gamma, P\true, P\false \prov \gamma
\end{prooftree}
\qquad
\begin{prooftree}
\using init_{ic} \justifies
\Gamma, P, P\false \prov \gamma
\end{prooftree}$$

The rules for $\bot$ are

$$\begin{prooftree}
\using \bot L \justifies
\Gamma, \bot  \prov \gamma
\end{prooftree}$$
$$\begin{prooftree}
\using \bot T \justifies
\Gamma, \bot  \true \prov \gamma
\end{prooftree}
\qquad
\begin{prooftree}
\Gamma \prov \gamma
\using \bot F \justifies
\Gamma, \bot \false \prov \gamma
\end{prooftree}$$


The rules for $\cent$ are

$$\begin{prooftree}
\Gamma,  A \true \prov \gamma
\using \cent L \justifies
\Gamma, \cent A  \prov \gamma
\end{prooftree}
\qquad
\begin{prooftree}
\Gamma, A \false \prov \cdot
\using \cent R \justifies
\Gamma \prov \cent A
\end{prooftree}$$
$$\begin{prooftree}
\Gamma,  A \true \prov \gamma
\using \cent T \justifies
\Gamma, \cent A \true \prov \gamma
\end{prooftree}
\qquad
\begin{prooftree}
\Gamma,  A \false \prov \gamma
\using \cent F \justifies
\Gamma, \cent A \false \prov \gamma
\end{prooftree}$$

The rules for $\limp$ are
$$\begin{prooftree}
\Gamma, A\limp B \prov A \qquad \Gamma, B \prov \gamma
\using \limp L \justifies
\Gamma,  A \limp B  \prov \gamma
\end{prooftree}
\qquad
\begin{prooftree}
\Gamma, A  \prov B
\using \limp R \justifies
\Gamma \prov A\limp B
\end{prooftree}$$
$$\begin{prooftree}
\Gamma,  A \false \prov \gamma_1 \qquad \Gamma,B\true \prov \gamma_2
\using \limp T \justifies
\Gamma, A\limp B \true \prov \gamma_1 + \gamma_2
\end{prooftree}
\qquad
\begin{prooftree}
\Gamma,  A\true, B\false  \prov \gamma
\using \limp F \justifies
\Gamma, A\limp B \false \prov \gamma
\end{prooftree}$$
Where $+$ is defined by
$$\cdot + \cdot = \cdot$$
$$C + \cdot = \cdot$$
$$\cdot + C = \cdot$$
where $C_1 + C_2$ is undefined even if $C_1 = C_2$.
That is, $\limp T$ is really three rules where one or both (but not neither)
of the premises have an empty right-hand side.

We may ask, does cut elimination hold for this calculus?
Is the rule
$$\begin{prooftree}
\Gamma\prov A \qquad \Gamma, A \prov \gamma
\using cut \justifies
\Gamma \prov \gamma
\end{prooftree}$$
admissible? In the event that $A$ is of the form $\cent B$ and
the derivations of $\Gamma\prov \cent B$ and $\Gamma, \cent B \prov \gamma$
both end with a rule that introduces the connective in the cut formula,
the derivation looks like
$$\begin{prooftree}
\[\Gamma, B\false \prov \cdot
\using \cent R\justifies
\Gamma\prov \cent B\]
\[\Gamma, B\true \prov \gamma
\using \cent L\justifies
\Gamma, \cent B \prov \gamma\]
\using cut \justifies
\Gamma \prov \gamma
\end{prooftree}$$
In order to proceed, it seems necessary that we also
 show the admissibility of a classical
cut principle
$$\begin{prooftree}
\Gamma, B\false \prov \cdot
\qquad
\Gamma, B\true \prov \gamma
\using sync \justifies
\Gamma \prov \gamma
\end{prooftree}$$
where false assumptions $A\false$ (thought of as continuations, or `$A$ readers') synchronize with true assumptions $A\true$ (thought of as `$A$ writers').
But proving this rule admissible is problematic. Consider the left-derivation
true-commutative case for $\limp$:

$$\begin{prooftree}
\[
\Gamma, A_1 \false, B\false \prov \cdot
\qquad
\Gamma, A_2\true, B\false \prov \cdot
\justifies
\Gamma, A_1, \limp A_2\true, B\false \prov \cdot
\]
\qquad
\Gamma, A_1\limp A_2\true, B\true \prov \gamma
\using sync \justifies
\Gamma, A_1\limp A_2\true \prov \gamma
\end{prooftree}$$
We can sync $\Gamma, A_2\true, B\false \prov \cdot$ with
$\Gamma, A_1\limp A_2\true, B\true \prov \gamma$ (on $B \true,B\false$)
to obtain $\Gamma, A_2\true, A_1\limp A_2\true\prov \gamma$.
We can sync $\Gamma, A_1\false, B\false \prov \cdot$ with
$\Gamma, A_1\limp A_2\true, B\true \prov \gamma$ (on $B \true,B\false$)
to obtain $\Gamma, A_1\false, A_1\limp A_2\true\prov \gamma$.

We would like to derive
$$\begin{prooftree}
\Gamma, A_1 \false , A_1\limp A_2 \true \prov \gamma
\qquad
\Gamma, A_2 \true  A_1\limp A_2 \true \prov \gamma
\using \limp T\justifies
\Gamma, A_1\limp A_2\true, A_1\limp A_2 \true \prov \gamma
\end{prooftree}$$
and apply contraction, but it may not be that $\gamma + \gamma = \gamma$!
I still haven't found a concrete counterexample, however. So the question
is still open:

\begin{question}
Are the rules cut and sync admissible?
\end{question}
\end{document}