Abstract
I explore, from a prooftheoretic perspective, the hierarchy of classical and paraconsistent logics introduced by Barrio, Pailos and Szmuc in (Journal o f Philosophical Logic, 49, 93120, 2021). First, I provide sequent rules and axioms for all the logics in the hierarchy, for all inferential levels, and establish soundness and completeness results. Second, I show how to extend those systems with a corresponding hierarchy of validity predicates, each one of which is meant to capture “validity” at a different inferential level. Then, I point out two potential philosophical implications of these results. (i) Since the logics in the hierarchy differ from one another on the rules, I argue that each such logic maintains its own distinct identity (contrary to arguments like the one given by Dicher and Paoli in 2019). (ii) Each validity predicate need not capture “validity” at more than one metainferential level. Hence, there are reasons to deny the thesis (put forward in Barrio, E., Rosenblatt, L. & Tajer, D. (Synthese, 2016)) that the validity predicate introduced in by Beall and Murzi in (Journal o f Philosophy, 110(3), 143–165, 2013) has to express facts not only about what follows from what, but also about the metarules, etc.
Notes
Recall that R ∨ is a doubleline rule in the calculus given in Section 2.
Admissibility is defined for metainferences in the same way that it is defined for inferences: an admissible rule is one whose addition to the existing rules of the system doesn’t change the system’s set of theorems, or, in this case, the set of derivable metainferences. Thus, to prove that a rule is admissible, we need to show that if certain instances of the premise metainferences hold, the relevant instance of the conclusion metainference holds as well.
Nonetheless, here is a sketch of a proof that push is admissible: Assume that the (meta)sequent holds. So there is a proof tree whose leaves are ⇒ γ_{1},...,⇒ γ_{m},⇒ A and whose root is ⇒Δ. We show by induction on the proof height how to construct a proof tree of For a height of 1, if the sequent ⇒ A isn’t active in the proof, then there is a proof tree of Using \(Wea{k_{1}^{b}}\), we can weaken with A on the left of the bottom sequent and get the result. If ⇒ A is active in the proof, then there are three options: Δ is derived either by Ref_{1} or by ∨ R or by ∧ R. In the first case A ∈Δ and so A ⇒Δ is an identity axiom (and from there the proof is trivial). In the second case, there is some B such that A ∨ B ∈Δ. So and from there with the Weak_{1} rules the proof is trivial.
In the third case, there are two options: (i) there is some 1 ≤ i ≤ m such that γ_{i} ∧ A ∈Δ, and (ii) A is of the form B ∧ C and B (or C) ∈Δ. In the case of (i) we reason:
And once again, enough applications of the Weak_{1} rules will get the job done. As for (ii), B ∧ C ⇒ B(C)Δ is clearly derivable, and then we weaken m times with ⇒ γ_{1},...,⇒ γ_{m} respectively. The inductive step is analogous for, given Cutelimination, we can assume that the proof involves only sequents with an empty lefthand side.
References
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This paper was supported by a Minerva fellowship at Freie Universität Berlin. The paper benefited from fruitful discussions with Robert Brandom, Ulf Hlobil, Dan Kaplan, Shuhei Shimamura, Carl Posy, Ran Lanzet, Gil Sagi, Aviv Keren, David Kashtan, Balthasar Grabmayr, Michael Goldboim, Jonathan Fiat, and Filippos Papagiannopoulos. I would also like to thank two anonymous referees for this journal, for helpful comments and suggestions.
Appendix
Appendix
Lemma 2.8.
For all n ≥ 1, and for all \({\Gamma }\subseteq SEQ^{n1}({\mathscr{L}}),\ A\in SEQ^{n1}({\mathscr{L}})\):
⊧_{CL}Γ ⇒_{n}A iff ⊧_{CL}lower(Γ ⇒_{n}A).
Proof
By Theorem 2.6 ⊧_{CL}Γ ⇒_{n}A iff \(\models _{L_{n+1}}{\Gamma }\Rightarrow _{n}A\). By Theorem 2.7 the latter is the case iff \(\models _{L_{n}}lower({\Gamma }\Rightarrow _{n}A)\), and once again by Theorem 2.6 (since lower(Γ ⇒_{n}A) is a metainference of level n − 1), that’s the case iff ⊧_{CL}lower(Γ ⇒_{n}A). □
Proposition 3.3.
Here are some useful facts about these axioms:

(i)
\(Wea{k_{n}^{b}}\) for n ≥ 1 is redundant: given the wellknown Weakening rule (for inferences), it follows by induction for all n, via \(Tra{n_{n}^{c}}\).

(ii)
For each instance of such axioms/rules r of level n (except for \(Wea{k_{n}^{b}}\) which is redundant, given (i)), lower(r), higher(r) instantiate that same axiom/rule at levels n − 1,n + 1, respectively.

(iii)
Given Cut, meta_{n}Cut (for n ≥ 1) is derivable via \(Tra{n_{n}^{c}}\).
Proof
(i) By induction. Assuming \(Wea{k_{n}^{b}}\) is derivable, we can derive \(Weak_{n+1}^{b}\) from \(Wea{k_{n}^{b}}\) and \(Tran_{n+1}^{c}\) in the following way:
Part (ii) is easy to prove, and so it is left for the reader as an exercise.
Part (iii) is proven by induction on n. The base case, Cut, is given. Assume that the claim holds for n, and let \({\Gamma }=\{\gamma _{1},...,\gamma _{m}\}\subseteq SEQ^{n}({\mathscr{L}}),A\in SEQ^{n}({\mathscr{L}})\). Then meta_{n+ 1} − Cut is derivable from meta_{n} − Cut (guaranteed by the inductive hypothesis) and \(Tran_{n+1}^{c}\):
□
Proposition 3.5.
For all \(n\in \mathbb {N}\cup \{0\}\), and for every metainference Γ ⇒_{n}A: \(\vdash _{CL^{\infty }}{\Gamma }\Rightarrow _{n}A\) iff \(\vdash _{CL^{\infty }}lower({\Gamma }\Rightarrow _{n}A)\).
Proof
By induction. I shall prove the claim for n = 0, 1, 2, and then turn to the inductive step.
The n = 0 case is the deduction theorem, which is known to hold in CL, so \(\vdash _{CL^{\infty }}{\Gamma }\Rightarrow {\Delta }\) iff \(\vdash _{CL^{\infty }}(\wedge {\Gamma })\supset (\vee {\Delta })\), which is \(\vdash _{CL^{\infty }}lower({\Gamma }\Rightarrow {\Delta })\).
The n = 1 case follows from the deduction theorem along with the fact that CL coincides with its external logic. Let be some metainference of level 1. By the deduction theorem, for all 1 ≤ i ≤ m: \(\vdash _{CL^{\infty }}{\Gamma }_{i}\Rightarrow {\Delta }_{i}\) iff \(\vdash _{CL^{\infty }}lower({\Gamma }_{i}\Rightarrow {\Delta }_{i})\), and \(\vdash _{CL^{\infty }}{\Sigma }\Rightarrow {\Pi }\) iff \(\vdash _{CL^{\infty }}lower({\Sigma }\Rightarrow {\Pi })\). Now, by m + 1 applications of the transitivity rules, we can replace each of these sequents with its equivalent one, and get that Since CL coincides with its external logic, the latter holds iff
which is
We also need to prove the n = 2 case. It is enough to show that for each instance r of a rule/axiom of level 2, lower(r) is derivable in \(CL^{\infty }\), and that for each instance q of a rule/axiom of level 1, higher(q) is derivable in \(CL^{\infty }\). If so, then:

1.
For each proof tree of metainferences of level 2 whose leaves are r_{1},...,r_{n} and whose root is b there is a proof tree of metainferences of level 1 (a “regular” proof tree) whose leaves are lower(r_{1}),...,lower(r_{n}) and whose root is lower(b).

2.
For each proof tree of metainferences of level 1 whose leaves are l_{1},...l_{m} and whose root is q, there is a proof tree of level 2 whose leaves are higher(l_{1}),...,higher(l_{m}) and whose root is higher(q).
Let us begin with the first claim. Consider an instance r of some rules/axiom of level 2. It is necessarily one of the structural rules, and so by Proposition 3.3 lower(r) also instantiates that same axiom/rule at level 1.
Consider then the axioms and rules of level 1, namely, the regular sequent rules and axioms of CL. We have to show that for each instance of some such axiom/rule q, higher(q) is derivable in \(CL^{\infty }\). I shall discuss two cases, and leave the rest as an exercise.
can be derived in the following way:^{Footnote 1}
(ii) The proof of the higher(R¬) case is more complex. I shall discuss the toptobottom direction, but the proof of the other direction is analogous. What we need to prove is that for all Γ,Δ,A:
What we really need to show is that the metarule push: is admissible in \(CL^{\infty }\).^{Footnote 2} If so, then whatever is derivable with push will be derivable without it. And so, for all Γ,Δ,A:
But actually, push is admissible simply because CL coincides with its external logic.^{Footnote 3} That completes the proof of the n = 2 case.
Now, suppose that the claim holds for some n ≥ 2. We want to show that it holds for n + 1. Since all the rules and axioms of levels n,n + 1 are structural, this is an immediate result of Proposition 3.3: for each instance r of some axiom/rule of level n, higher(r) instantiates that same axiom/rule at level n + 1, and for each instance q of some axiom/rule of level n + 1, lower(q) instantiates that same axiom/rule at level n. This completes the entire proof. □
Theorem 3.11.
For all m ≥ 2, for all \(n\in \mathbb {N}\cup \{0\}\), and for every metainference Γ ⇒_{n}A: \(\vdash _{L_{m}^{\infty }}{\Gamma }\Rightarrow _{n}A\) iff \(\vdash _{L_{m1}^{\infty }}lower({\Gamma }\Rightarrow _{n}A)\).
Proof
By nested induction.
Base case: m = 2, n = 0. The relevant rules governing inference (metainference of level 0) are those of CL^{0}. But it is known that ⊩_{CL}Γ ⇒Δ iff ⊩_{ST}lower(Γ ⇒Δ). I shall also discuss the n = 1 case. Consider a proof tree of a metainference (of the first level) Γ ⇒_{1}A. Now, by Proposition 3.3: \({L_{2}^{1}}=CL^{0}\cup Weak_{1}\cup Ref_{1}\cup Aux_{1}\cup Sub_{1}\) and so for each instance of axiom/rule \(r\in {L_{2}^{1}}\): \(\vdash _{ST^{0}}lower(r)\). Hence, \(\vdash _{ST^{0}}lower({\Gamma }\Rightarrow _{1}A)\). We also need to show that for each instance of axiom/rule q ∈ ST^{0}, higher(q) is derivable in \({L_{2}^{1}}\). Once again, the proof is similar to that of Proposition 3.5: it relies on the fact that all the operational rules of ST^{0}, which are also the operational rules of \({L_{2}^{1}}\), are doubleline rules. Therefore, it is easy to prove that their higher versions are derivable, using instances of Sub_{1}. For example, both directions of higher(R∨) are derivable as follows:
Hence, \(\vdash _{{L_{2}^{1}}}{\Gamma }\Rightarrow _{1}A\) iff \(\vdash _{ST^{0}}lower({\Gamma }\Rightarrow _{1}A)\).
For n > 1, however, observe that \({L_{2}^{n}}\) and ST^{n− 1} have only structural rules at these levels, and so the proof is trivial, given Proposition 3.3.
Inductive step: Assume that the claim holds for some m; then we have to show that it holds for m + 1, for all n. For 0 ≤ n ≤ m − 1, the relevant rules are those of CL^{m− 1}, and we know from Proposition 3.5 that \(\vdash _{CL^{m1}}{\Gamma }\Rightarrow _{n}A\) iff \(\vdash _{CL^{m2}}lower({\Gamma }\Rightarrow _{n}A)\). For n = m, note that \(L_{m+1}^{m}\) and \(L_{m}^{m1}\) have the same rules (modulo degrees) which are all structural (at level m of \(L_{m+1}^{m}\) and at level m − 1 of \(L_{m}^{m1}\) ), and so, once again, the claim holds, by Proposition 3.3. The same method of proof also applies to all n > m. This completes the proof. □
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Golan, R. Metainferences from a ProofTheoretic Perspective, and a Hierarchy of Validity Predicates. J Philos Logic 51, 1295–1325 (2022). https://doi.org/10.1007/s10992021096166
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DOI: https://doi.org/10.1007/s10992021096166
Keywords
 Metainferences
 Validity predicate
 Proof theory