I’ve been working through Shamkanov’s paper on cyclic proofs for Gödel-Löb Logic (https://arxiv.org/pdf/1401.4002). While it’s a landmark paper in the field of (cyclic) proof theory, I’m finding that many details are left implicit, especially from the perspective of someone still learning the ropes. Right now, I’m stuck on Lemma 3.7.
Definition: For a circular derivation $\pi = (\kappa, d)$, the set of assumption leaves of $\pi$ is the set of non-axiomatic leaves of $\kappa$ that are not connected by the back-link function $d$. We say that an assumption leaf is boxed if there is an application of the rule $\Box$ on the path from the root to the leaf. Let $BH(\pi)$ be the set of boxed assumption leaves of $\pi$, and $H(\pi)$ be the set of all other assumption leaves. For a subderivation $\tau$ of $\kappa$, we define the back-link function $d_\tau$ as the set of all links from $d$ with images inside $\tau$.
Recall that $\boxed{\cdot} A$ is an abbreviation for $A \land \Box A$.
Lemma 3.7. $\textsf{GL}_{\textsf{circ}} \vdash \Gamma \implies \textsf{GL}_{\textsf{Seq}} \vdash \Gamma$.
Proof. Given a circular derivation $\pi = (\kappa, d)$ of $\Gamma$, we claim that $$ \textsf{GL} \vdash \bigwedge \{\boxed{\cdot} \Delta_a^\# : a \in H(\pi)\} \land \bigwedge \{\Box \Delta_a^\# : a \in BH(\pi)\} \rightarrow \Gamma^\#, $$ where $\Delta_a$ is the sequent of $a$. The proof is by induction on the structure of $\kappa$.
If $\kappa$ consists of a single sequent, then our claim is obvious. Otherwise, consider the last application of an inference rule in $\kappa$. We have three cases: $$ \frac{\Delta, A \quad \Delta, B}{\Delta, A \land B} \quad \land \quad \frac{\Delta, A, B}{\Delta, A \lor B} \quad \lor \quad \frac{\Delta, \Diamond \Delta, A}{\Diamond \Delta, \Box A, \Sigma} \quad \Box $$
Suppose the conclusion of $\kappa$ is not connected by the back-link function. Recall that: $$ \textsf{GL} \vdash (\Delta, A)^\# \land (\Delta, B)^\# \rightarrow (\Delta, A \land B)^\#, $$ $$ \textsf{GL} \vdash (\Delta, A, B)^\# \rightarrow (\Delta, A \lor B)^\#, $$ $$ \textsf{GL} \vdash \Box(\Delta, \Diamond \Delta, A)^\# \rightarrow (\Diamond \Delta, \Box A, \Sigma)^\#. $$
We see that the claim follows from the induction hypothesis for $(\tau, d_\tau)$ (or for $(\tau_1, d_{\tau_1})$ and $(\tau_2, d_{\tau_2})$) immediately.
The only remaining case is that the conclusion of $\kappa$ is connected by the back-link function. Recall that there is always an application of the rule $\Box$ on the path between two nodes connected by the back-link function. By the induction hypothesis and from the previous cases, we have: $$ \textsf{GL} \vdash \bigwedge \{\boxed{\cdot} \Delta_a^\# : a \in H(\pi)\} \land \bigwedge \{\Box \Delta_a^\# : a \in BH(\pi)\} \land \Box \Gamma^\# \rightarrow \Gamma^\#. $$
Since the Löb rule is admissible for $\textsf{GL}$, the assumption $\Box \Gamma^\#$ can be dropped. Hence, $$ \textsf{GL} \vdash \bigwedge \{\boxed{\cdot} \Delta_a^\# : a \in H(\pi)\} \land \bigwedge \{\Box \Delta_a^\# : a \in BH(\pi)\} \rightarrow \Gamma^\#. $$
Now if $\pi$ is a circular proof of $\Gamma$, then $\textsf{GL} \vdash \Gamma^\#$. By Proposition 2.2, $\textsf{GL}_{\textsf{Seq}} \vdash \Gamma$.
Proposition 2.2 is:$ \textsf{ GL}_{\textsf{Seq}} \vdash \Gamma \iff \textsf{GL} \vdash \Gamma^\#.$
Where: $\Gamma^\# := \begin{cases} \bot& \text{if } n = 0, \\ A_1 \lor \dots \lor A_n & \text{otherwise.} \end{cases}$
And: $\Diamond\Gamma := \Diamond A_1, \dots, \Diamond A_n$
The Hilbert-style axiomatization of GL is as follows:
Axioms:
- Boolean tautologies;
- $\Box(A \rightarrow B) \rightarrow (\Box A \rightarrow \Box B)$;
- $\Box(\Box A \rightarrow A) \rightarrow \Box A$.
Rules:
- Modus ponens,
- $A / \Box A$.
And $GL_{circ}$ is obtained from $K4_{Seq}$ by admitting circular proofs, it contains the modal-rule $\Box$ shown in the proof above.
Now, to my questions:
Why are we splitting the assumptions of $\Gamma$ in the way described above and why are they boxed differently in GL?
Why is $ \textsf{GL} \vdash \Box(\Delta, \Diamond \Delta, A)^\# \rightarrow (\Diamond \Delta, \Box A, \Sigma)^\# $ true? I assume by combining of the necessitation rule and Löb's Axiom, but I am unable to derive it after multiple attempts
And for the case that the conclusion of $\kappa$ is connected by the back-link function, I cannot follow the reasoning behind the proof fully. I get that the last rule before reaching the conclusion-sequent again (that is linked back to the conclusion) is the boxed version of it, so I guess the induction hypothesis might be that the proof holds up to this point and for all sequents in between we proved the statement holds already? But I might be totally wrong.
Hope somebody can help me here, I would really appreciate it!
