Abstract
Logical consequence is typically construed as a metalinguistic relation between (sets of) sentences. Deflationism is an account of logic that challenges this orthodoxy. In Williamson’s recent presentation of deflationism, logic’s primary concern is with universal generalizations over absolutely everything. As well as an interesting account of logic in its own right, deflationism has also been recruited to decide between competing logics in resolving semantic paradoxes. This paper defends deflationism from its most important challenge to date, due to Ole Hjortland. It then presents two new problems for the view. Hjortland’s objection is that deflationism cannot discriminate between distinct logics. I show that his example of classical logic and supervaluationism depends on equivocating about whether the language includes a “definitely” operator. Moreover, I prove a result that blocks this line of objection no matter the choice of logics. I end by criticizing deflationism on two fronts. First, it cannot do the work it has been recruited to perform. That is, it cannot help adjudicate between competing logics. This is because a theory of logic cannot be as easily separated from a theory of truth as its proponents claim. Second, deflationism currently has no adequate answer to the following challenge: what does a sentence’s universal generalization have to do with its logical truth? I argue that the most promising, stipulative response on behalf of the deflationist amounts to an unwarranted change of subject.
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22 November 2019
The original version of this article unfortunately contains mistakes introduced by the publisher during the production process. The mistakes and corrections are described in the following list
Notes
 1.
This requires metaarguments to have a nonempty metapremise set (though this is compatible with a metapremise—an argument—having an empty premise set). But this seems unobjectionable. A “metaargument” with nothing above the line is just the assertion of some entailment. Even if we allow these degenerate kinds of metaargument, the result still holds. Proof. For reductio suppose some zerometapremise metaargument is Kmetavalid but Jmetainvalid. Let its conclusion be Δ ⊩ β. Then Δ ⊩_{K}β but \({\varDelta } \nvdash _{J} \beta \). So \(\beta \in Cn_{\vdash _{K}} ({\varDelta })\) and \(\beta \notin Cn_{\vdash _{J}} ({\varDelta })\). Hence the K and Jclosures of Δ are not the same set. Contradiction with the closure equivalence of K and J.
 2.
This requires that J is not the empty logic. The WLOG assumption a few lines above requires that K is not empty as well. Note that Γ could be the empty set, however.
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Acknowledgments
This paper has benefitted from discussion with Jared Henderson, Nathan Kellen, Gurpreet Rattan, and participants of the Society for Exact Philosophy annual conference at the University of Connecticut in 2018. I am also grateful to Joanna Lawson, Erik Stei, and Simon Varey for reading drafts. Special thanks are due to Gillian Russell. The idea for the paper came from participating in her Philosophy of Logic graduate class at UNC in the spring semester of 2017. She also provided extremely helpful feedback, and muchneeded encouragement, at multiple stages of the writing and rewriting process. Finally, thank you to two anonymous reviewers, whose generous comments improved the paper greatly.
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Appendix
Appendix
MetaArgumentative Equivalence (MAE)
If two logics are closure equivalent, then they are (globally) metaargumentatively equivalent.
Proof
Let K and J be closureequivalent logics. Let Γ_{1}…Γ_{n} and Δ be arbitrary (possibly empty) premise sets, and α_{1}…α_{n} and β be arbitrary conclusions. Consider the following arbitrary metaargument:
We show that Arg is Kmetavalid iff Arg is Jmetavalid. Consider the lefttoright direction.
Suppose for reductio that K metavalidates Arg, but that J does not. There are two cases to consider.
Case 1. K metavalidates Arg because both its metapremises and metaconclusion are Kvalid, and J fails to metavalidate Arg. More precisely:
 (i)
Γ_{1} ⊩_{K}α_{1}⋯Γ_{n} ⊩_{K}α_{n}
 (ii)
Δ ⊩_{K}β
 (iii)
Γ_{1} ⊩_{J}α_{1}⋯Γ_{n} ⊩_{J}α_{n}
 (iv)
\({\varDelta } \nvdash _{J} \beta \)
By (ii), \(\beta \in Cn_{\vdash _{K}} ({\varDelta })\). But by (iv), \(\beta \notin Cn_{\vdash _{J}} ({\varDelta })\). Hence \(Cn_{\vdash _{K}} ({\varDelta }) \neq Cn_{\vdash _{J}} ({\varDelta })\). But, since K and J are closure equivalent, \(Cn_{\vdash _{K}} ({\varDelta }) = Cn_{\vdash _{J}} ({\varDelta })\). Contradiction.
Case 2. K metavalidates Arg because at least one of its metapremises is invalid (thus the conditional “if all metapremises, then metaconclusion” obtains), and J fails to metavalidate Arg. More precisely, and (iii) and (iv) hold as in Case 1, and:
 (v)
For some Γ_{i} and α_{i}, \({\varGamma }_{i} \nvdash _{K} \alpha _{i}\).^{Footnote 1}
By (v), \(\alpha _{i} \notin Cn_{\vdash _{K}} ({\varGamma }_{i})\). But by (iii) \(\alpha _{i} \in Cn_{\vdash _{J}} ({\varGamma }_{i})\). Hence \(Cn_{\vdash _{K}} ({\varGamma }_{i}) \neq Cn_{\vdash _{J}} ({\varGamma }_{i})\). But since K and J are closure equivalent, \(Cn_{\vdash _{K}} ({\varGamma }_{i}) = Cn_{\vdash _{J}} ({\varGamma }_{i})\). Contradiction.
Either way, we get a contradiction. It follows by reductio that if Arg is Kmetavalid, then it is Jmetavalid.
The proof for the righttoleft direction of the biconditional is almost exactly the same.
Hence, Arg is Kmetavalid iff Arg is Jmetavalid.
Since this was proved on the assumption that K and J are closure equivalent, and since the logics and metaarguments were arbitrary, it follows that if two logics are closure equivalent, then they are metaargumentatively equivalent. □
Converse of MAE
If two logics are (globally) metaargumentatively equivalent, then they are closure equivalent.
Proof
Suppose for reductio that K and J are metaargumentatively equivalent, but they are not closure equivalent: for some Δ, \(Cn_{\vdash _{K}} ({\varDelta }) \neq Cn_{\vdash _{J}} ({\varDelta })\). Without loss of generality assume that:
 (i)
\(\delta \in Cn_{\vdash _{K}} ({\varDelta })\)
 (ii)
\(\delta \notin Cn_{\vdash _{J}} ({\varDelta })\)
By (i), Δ ⊩_{K}δ. Hence:
 (iii)
Any metaargument to the metaconclusion that Δ ⊩ δ is Kmetavalid.
Consider in particular the following metaargument, where:^{Footnote 2}

(iv)
Γ ⊩_{J}α
By (iii), Arg1 is Kmetavalid. So, by the assumption of metaargumentative equivalence, Arg1 is Jmetavalid too. Either (a) the metapremise is Jinvalid, or (b) the metaconclusion is Jvalid. But (a) contradicts (iv), because then both \({\varGamma } \nvdash _{J} \alpha \) and Γ ⊩_{J}α. And (b) contradicts (ii), because then Δ ⊩_{J}δ. Hence \(\delta \in Cn_{\vdash _{J}} ({\varDelta })\) and \(\delta \notin Cn_{\vdash _{J}} ({\varDelta })\). Either way, contradiction.
It follows by reductio that if two logics are metaargumentatively equivalent, then they are closure equivalent. □
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BlakeTurner, C. Deflationism About Logic. J Philos Logic (2019). https://doi.org/10.1007/s10992019095295
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Keywords
 Philosophy of logic
 Deflationism
 Deflationism about logic
 Williamson
 Hjortland
 Antiexceptionalism
 Metavalidity
 Metaargument
 Metaargumentative equivalence
 Abduction
 Truth