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Deflationism About Logic


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


  1. 1.

    This requires meta-arguments to have a non-empty metapremise set (though this is compatible with a metapremise—an argument—having an empty premise set). But this seems unobjectionable. A “meta-argument” with nothing above the line is just the assertion of some entailment. Even if we allow these degenerate kinds of meta-argument, the result still holds. Proof. For reductio suppose some zero-metapremise meta-argument is K-metavalid but J-meta-invalid. Let its conclusion be Δβ. Then ΔKβ but \({\varDelta } \nvdash _{J} \beta \). So \(\beta \in Cn_{\vdash _{K}} ({\varDelta })\) and \(\beta \notin Cn_{\vdash _{J}} ({\varDelta })\). Hence the K- and J-closures of Δ are not the same set. Contradiction with the closure equivalence of K and J.

  2. 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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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 much-needed 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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Correspondence to Christopher Blake-Turner.

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Meta-Argumentative Equivalence (MAE)

If two logics are closure equivalent, then they are (globally) meta-argumentatively equivalent.


Let K and J be closure-equivalent logics. Let Γ1Γn and Δ be arbitrary (possibly empty) premise sets, and α1αn and β be arbitrary conclusions. Consider the following arbitrary meta-argument:


We show that Arg is K-metavalid iff Arg is J-metavalid. Consider the left-to-right 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 K-valid, and J fails to metavalidate Arg. More precisely:

  1. (i)


  2. (ii)


  3. (iii)


  4. (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:

  1. (v)

    For some Γi and αi, \({\varGamma }_{i} \nvdash _{K} \alpha _{i}\).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 K-metavalid, then it is J-metavalid.

The proof for the right-to-left direction of the biconditional is almost exactly the same.

Hence, Arg is K-metavalid iff Arg is J-metavalid.

Since this was proved on the assumption that K and J are closure equivalent, and since the logics and meta-arguments were arbitrary, it follows that if two logics are closure equivalent, then they are meta-argumentatively equivalent. □

Converse of MAE

If two logics are (globally) meta-argumentatively equivalent, then they are closure equivalent.


Suppose for reductio that K and J are meta-argumentatively 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:

  1. (i)

    \(\delta \in Cn_{\vdash _{K}} ({\varDelta })\)

  2. (ii)

    \(\delta \notin Cn_{\vdash _{J}} ({\varDelta })\)

By (i), ΔKδ. Hence:

  1. (iii)

    Any meta-argument to the metaconclusion that Δδ is K-metavalid.

Consider in particular the following meta-argument, where:Footnote 2

  1. (iv)



By (iii), Arg1 is K-metavalid. So, by the assumption of meta-argumentative equivalence, Arg1 is J-metavalid too. Either (a) the metapremise is J-invalid, or (b) the metaconclusion is J-valid. But (a) contradicts (iv), because then both \({\varGamma } \nvdash _{J} \alpha \) and ΓJα. And (b) contradicts (ii), because then ΔJδ. 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 meta-argumentatively equivalent, then they are closure equivalent. □

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Blake-Turner, C. Deflationism About Logic. J Philos Logic (2019). https://doi.org/10.1007/s10992-019-09529-5

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  • Philosophy of logic
  • Deflationism
  • Deflationism about logic
  • Williamson
  • Hjortland
  • Anti-exceptionalism
  • Metavalidity
  • Meta-argument
  • Meta-argumentative equivalence
  • Abduction
  • Truth