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Brogaard and Salerno on antirealism and the conditional fallacy

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Abstract

Brogaard and Salerno (2005, Nous, 39, 123–139) have argued that antirealism resting on a counterfactual analysis of truth is flawed because it commits a conditional fallacy by entailing the absurdity that there is necessarily an epistemic agent. Brogaard and Salerno’s argument relies on a formal proof built upon the criticism of two parallel proofs given by Plantinga (1982, Proceedings and Addresses of the American Philosophical Association, 56, 47–70) and Rea (2000, Nous, 34, 291–301). If this argument were conclusive, antirealism resting on a counterfactual analysis of truth should probably be abandoned. I argue however that the antirealist is not committed to a controversial reading of counterfactuals presupposed in Brogaard and Salerno’s proof, and that the antirealist can in principle adopt an alternative reading that makes this proof invalid. My conclusion is that no reductio of antirealism resting on a counterfactual analysis of truth has yet been provided.

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Notes

  1. Plantinga 1982 and Rea 2000 take the conclusions of their proofs to show that the alethic antirealist is committed to some form of theism (i.e. to the thesis that, roughly, there is necessarily an omniscient epistemic agent/community). Brogaard and Salerno 2005 claim that this is a misguided interpretation of those alleged results, which should instead be seen as instances of the general problem of the conditional fallacy that plagues counterfactual analyses. (The same point is made in Wright 2000). I find this claim straightforward and I will not reconsider it in my paper.

  2. Namely, \(\square(\hbox{T}[P]\leftrightarrow P)\).

  3. By the necessitation of the Equivalence Schema, (Q) ‘epistemic conditions are ideal for determining whether [epistemic conditions are ideal for determining whether some statement is true]’ entails ‘epistemic conditions are ideal for determining whether it is true that [epistemic conditions are ideal for determining whether some statement is true]’. From the latter, it follows: (P) ‘epistemic conditions are ideal for determining whether some statement is true’.

  4. Indeed, Brogaard and Salerno 2005 complain that “Rea unapologetically embraced classical logic and S4, which is highly controversial in this context” (p. 137). This remark—which may be interpreted as questioning axiom 4—is very puzzling not only because Brogaard and Salerno do not raise any similar objection to Plantinga’s proof, which presupposes S4 too, but also because Brogaard and Salerno’s own proof is carried out in just an expansion of S4! Berit Brogaard has admitted (in personal communication) that ‘S4’ is a misprint here, which should be replaced with ‘S5’. To strengthen the conclusion of his proof, Rea has in fact appealed to S5 (cf. Brogaard and Salerno 2005, p. 130).

  5. This claim can be generalized, as the proofs in the literature that aim to show (or that can be read as aiming to show) that the alethic antirealist commits a conditional fallacy are typically carried out in S4—so they presuppose axiom 4. (I list these proofs in the conclusion of this paper). The only exception I am aware of is a weaker version of Plantinga’s proof due to Wright (2000, pp. 341–342), which aims to demonstrate that Peircean antirealism entails the actual (but not necessary) existence of an omniscient epistemic agent/community. Wright’s proof is made in the modal system T (expanded with resources for counterfactuals), which embeds axiom M but not axiom 4.

  6. A version of Salmon’s argument against axiom 4 runs as follows (cf. for instance Salmon 1989, pp. 4–5, and Hayaki 2005, p. 28). Axiom 4 is equivalent to \(\lozenge\lozenge P \rightarrow \lozenge P,\) where \(\lozenge\) is the possibility operator. Consider an artifact—for example a table, which we will call T. It seems intuitive that T, while retaining its numerical identity, could have originated from a piece of wood W1 very slightly different from the piece of wood W0 from which T has actually originated. Suppose, for instance, that W1 has the same shape and size as W0 but is taken from one millimeter further down the same tree trunk as W0. This intuition can be expressed by saying, shortly, that it is possible that T is made of W1. Consider now that if T had actually originated from W1, it would plausibly be true that T could have originated from W2, a piece of wood taken from one additional millimeter further down the same trunk. This intuition can be expressed by saying, shortly, that it is possible that it is possible that T is made of W2. Let us reiterate this reasoning by, say, one thousand times to reach the apparently correct conclusion that it is possible that...it is possible that...it is possible that T is made of W1,000, namely, a piece of wood that differs from T’s actual piece of wood by one meter. As \(\lozenge\lozenge P \rightarrow \lozenge P,\) entails (by reiterated applications of modus ponens) that any finite string of diamonds can be replaced by just one diamond, from the above claim that it is possible that...it is possible that...it is possible that T is made of W1,000, we should derive that it is just possible that T is made of W1,000. But this seems incorrect: if T had originated from a piece of wood that differs from T’s actual piece of wood by one meter, T would plausibly be a distinct individual! This apparent counterexample to axiom 4 might be used by an antirealist who preferred intuitionistic logic to classical logic. I describe systems of intuitionistic modal logic at the end of this section. As \(\square\) and \(\lozenge\) are intuitionistically non-interdefinable, \(\square P\rightarrow \square\square P\) does not entail \(\lozenge\lozenge P \rightarrow \lozenge P,\) and the intuitionistic version of axiom 4 comes in the terms of the conjunction \((\square P \rightarrow \square\square P) \wedge (\lozenge\lozenge P \rightarrow \lozenge P)\). One might perhaps worry that the sorites counterexamples hit only the conjunct \(\lozenge\lozenge P \rightarrow \lozenge P\) but not the conjunct \(\square P \rightarrow \square\square P\), which is actually used in Brogaard and Salerno’s proof. But Salmon’s argument can easily be formulated to hit directly \(\square P\rightarrow \square\square P\).

  7. The fact that when issues of vagueness come into play our intuitions become often confused may cast doubts on the soundness of Salmon’s argument. An appropriate evaluation of the latter would probably require a general solution to the problem of vagueness. As Hayaki (2005, p. 29) suggests, one might be no more convinced by Salmon’s argument than by a sorites argument purporting to show that there is no such thing as baldness!

  8. It seems reasonable that A1 should be accepted if M is accepted. This is why: for any possible world w, if Q \(\square\!\!\!\rightarrow\) R is true at w, intuitively, R must be true in all possible worlds closest to w in which Q is true (i.e. in all possible worlds that resemble w as much as Q’s truth permits it). Since M entails Q → \(\lozenge\) Q, if M is accepted, any possible world will count as possible to itself. So, if Q is true at a world w, w is one of the worlds closest to w at which Q is true. In conclusion, if M is accepted, for any world w, if both \(Q \square\!\!\!\rightarrow R\) and Q are true at w, R must also be true at w. This seems to validate A1. I do not see how the antirealist could reasonably question this simple and intuitive reasoning.

  9. In this paper, modus ponens is defined in a broadened sense to apply to both → and ↔.

  10. Rigorously speaking, Q → R follows from \(P\leftrightarrow (Q \square\!\!\!\rightarrow R)\), Q → P and axiom A1, used to derive line 5 from lines 3 and 4. The rule of closure correctly applies here because also A1, as an axiom schema, can be shown to be necessary by necessitation, i.e. the rule that if a statement is a theorem of a given modal logic, it can be necessitated in that logic. A1 represents a theorem of the logic \({{\mathbf{S4}}} \cup \{\hbox{A}1, \hbox{A}2\}\). As any axiom schema is necessary, for the sake of simplicity I will never mention any of the axioms implicitly appealed to in the applications of the rule of closure made this paper.

  11. These papers provide slightly different axiomatizations of the same intuitionistic modal logics.

  12. Following Simpson (1994, p. 52), IK can be axiomatized as follows (where \(\lozenge\) is the possibility operator, ∨ is logical disjunction, \(\lnot\) is logical negation, and \(\bot \) is any logical contradiction):

    Axioms

    1. (i)

      All substitution instances of theorems of intuitionistic propositional logic;

    2. (ii)

      \(\square(Q \rightarrow R) \rightarrow (\square Q\rightarrow \square R);\)

    3. (iii)

      \(\square(Q \rightarrow R) \rightarrow (\lozenge Q \rightarrow \lozenge R);\)

    4. (iv)

      \(\lnot \lozenge\bot \)

    5. (v)

      \(\lozenge\)(Q ∨ R) → (\(\lozenge\) Q ∨ \(\lozenge\) R);

    6. (vi)

      \((\lozenge Q \rightarrow \square R)\rightarrow \square(Q \rightarrow R).\)

    Rules.

    • (Modus Ponens) From Q → R and Q, deduce R;

    • (Necessitation) If Q is a theorem, then so is \(\square Q.\)

  13. The following two schemata hold valid in the logics of \({\tt I} \cup\{\mathbf{IK}\}: \lnot\lozenge P \leftrightarrow \square\lnot P\) and \(\lozenge\lnot P \rightarrow \lnot\square P\). Yet \(\lnot\square P \rightarrow \lozenge\lnot P\) does not hold valid in them.

  14. Although this simple and powerful systematisation of intuitionistic modal logic is today accepted by most logicians and computer scientists, weaker and interesting logics have also been studied, often called constructive rather than intuitionistic. See, for instance, De Paiva and Mendler 2005 and Alechina et al. 2001.

  15. The parallelism between the semantics of intuitionistic and classical modal logics is rigorous. Each logic of \({\tt I}\) is sound and complete with respect to classes of frames < W, ≤, R > in which R has properties that coincide with those of R in the classes of frames < W, R > with respect to which each parallel logic of \({\tt C}\) is sound and complete. Thus, T and IT are sound and complete with respect to frames in which R is reflexive, S4 and IS4 are sound and complete with respect to frames in which R is a preorder, and S5 and IS5 are sound and complete with respect to frames where R is an equivalence relation.

  16. As the intuitionistic \(\square\) and \(\lozenge\) are not dual, the intuitionistic versions of axiom schema M and axiom schema 4 are, respectively, \((\square P \rightarrow P)\wedge (P \rightarrow \lozenge P)\) and (\(\square\) P \(\rightarrow \square\square P) \wedge (\lozenge\lozenge P \rightarrow \lozenge P)\) (cf. Gabbay et al. 2003, p. 190, and Simpson 1994, pp. 55–56). For the sake of simplicity, in the logical proofs of this paper I will often use the expressions ‘axiom M’ and ‘axiom 4’ to refer to only one of the two conjuncts of, respectively, axiom M and axiom 4.

  17. To obtain the logic \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}2\}\), we first expand the alphabet of intuitionistic modal logic with the connective \(\square\!\!\!\rightarrow\), and add to the formation rules of the language of intuitionistic modal logic the one such that, if P and Q are well-formed formulae, so is P \(\square\!\!\!\rightarrow\) Q. Then, we extend the set of axioms of IS4 with A1 and A2. It would be easy to show that the logics of \({\tt I} \cup\{{\mathbf{IK}}\}\) are closed under all elementary inference rules additional to modus ponens used in the proof of CF, and under all those I use in this paper. These elementary rules are: →introduction, \(\wedge\)introduction, \(\wedge\)elimination, and closure. If IS4 is extended to \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}2\}\), the latter logic will prove closed under the same rules. Since IS4 is logically consistent, \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}2\}\) is also logically consistent. Proof Let us reinterpret \(\square\!\!\!\rightarrow\) as →. As a result, A1 and A2 are turned into, respectively, (Q \(\wedge\) (Q → R)) → R and \(\square\) (Q → R) → (Q → R), while there is no change in the inference rules. As both the latter schemata are valid in IS4 (this can easily be shown by using →introduction, \(\wedge\)introduction, axiom M and modus ponens) and no formula with the form \(P \wedge \lnot P\) can be deduced from IS4, no such a formula can be deduced from \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}2\}\) either. QED.

  18. To demonstrate that \({\mathbf{IK}} \cup \{\hbox{A}2\}\) validates \(\lnot\lozenge Q \rightarrow (Q \square\!\!\!\rightarrow R)\), let us prove first that \(\lnot\lozenge Q \rightarrow \square(Q\rightarrow R)\) is valid in IK. Proof As IK is conservative over intuitionistic propositional logic, IK validates the law of contradiction, i.e. \(\lnot Q \rightarrow (Q \rightarrow R)\). By necessitation (see above, note 12), we turn \(\lnot Q \rightarrow (Q \rightarrow R)\) into \(\square(\lnot Q \rightarrow (Q\rightarrow R))\). This schema and axiom (ii), i.e. \(\square(Q \rightarrow R) \rightarrow (\square Q \rightarrow \square R)\) (see ibid.), entail \(\square\lnot Q \rightarrow \square(Q \rightarrow R)\) by modus ponens. IK validates the schema \(\lnot\lozenge Q \rightarrow \square\lnot Q\) (see above, note 13). Assume \(\lnot\lozenge Q\) for →introduction. \(\lnot\lozenge Q\) and \(\lnot\lozenge Q \rightarrow \square\lnot Q\) entail \(\square\lnot Q\) by modus ponens. \(\square\lnot Q\) and \(\square\lnot Q \rightarrow \square(Q \rightarrow R)\), derived before, entail \(\square(Q \rightarrow R)\) by modus ponens. Finally, \(\lnot\lozenge Q\), assumed above, and \(\square(Q \rightarrow R)\) entail \(\lnot\lozenge Q \rightarrow \square(Q \rightarrow R)\) by →introduction. QED. Let us now demonstrate that \({\mathbf{IK}} \cup \{\hbox{A}2\}\) validates \(\lnot\lozenge Q \rightarrow (Q \square\!\!\!\rightarrow R)\). Proof Assume \(\lnot\lozenge Q\) for →introduction. This assumption and the schema \(\lnot\lozenge Q \rightarrow \square(Q \rightarrow R)\), which has been shown to hold valid in IK, entail \(\square(Q \rightarrow R)\) by modus ponens. \(\square(Q \rightarrow R)\) and A2—i.e. \(\square(Q \rightarrow R) \rightarrow (Q \square\!\!\!\rightarrow R)\)—entail \(Q \square\!\!\!\rightarrow R\) by modus ponens. Finally, \(\lnot\lozenge Q\), assumed initially, and Q \(\square\!\!\!\rightarrow\) R entail \(\lnot\lozenge Q \rightarrow (Q \square\!\!\!\rightarrow R)\) by →introduction. QED.

  19. Consider any possible world w. If \(\lozenge Q\) is true at w, there are worlds that are possible to w and in which Q is true. If \(\square(Q \rightarrow R)\) is also true at w, in all these worlds—and thus in the closest to w —R is true. Given Lewis’ only clause (b), \(Q \square\!\!\!\rightarrow R\) is also true at w. This seems to validate A3. There is no apparent reason why the antirealist should reject this intuitive reasoning.

  20. More precisely, as the intuitionistic \(\square\) and \(\lozenge\) are not dual, axiom 5 is to be formulated as follows: (\(\lozenge Q\rightarrow \square\lozenge Q) \wedge (\lozenge\square Q\rightarrow \square Q)\) (cf. Gabbay et al. 2003, p. 190, and Simpson 1994, pp. 55–56).

  21. Although I will not pursue this possible line of defence of antirealism here, it is worth emphasizing that the difficulties that the antirealist might raise against the adequacy of axiom 4—as I have suggested in Sect 2—would count against IS5 too, as the latter validates the former.

  22. The following sketched argument, based on the version of combinatorialism presented in Divers (2002, pp. 174–176 and pp. 207–208) appears to strike axiom 5 independently of whether classical or intuitionistic logic is presupposed. For the combinatorialist, a state of affairs is possible if and only if it can be obtained by recombining actually existing simple individuals and actually instantiated simple properties. Simple individuals are those that lack proper parts, and simple properties are those that do not have other properties as constituents. Simple individuals and simple properties exist only contingently. (To simplify, I assume that the actual states of affairs represent a limiting case of recombination). Consider any simple property P actually instantiated by some individual and any existing simple individual a. The state of affairs that Pa can be obtained by recombining a and P. Therefore, it holds true that \(\lozenge\)Pa. Consider now all recombinations of the actually instantiated property P with all existing simple individuals: one recombination is to the effect that P is instantiated by no individual at all. This constitutes a genuine possibility for the combinatorialist—something that could have happened. But, in that case, if P had been instantiated by no individual at all, it would not have been possible that Pa, for there would have been no existing P to be recombined with the individual a in the state of affairs that Pa. Therefore, although it is true that \(\lozenge\)Pa, it is also possible that \(\lnot\lozenge\)Pa—briefly, it is also true that \(\lozenge\lnot\lozenge\)Pa. As both K and IK validate the schema \(\lozenge\lnot P\rightarrow \lnot\square P\), \(\lozenge\lnot\lozenge\)Pa entails \(\lnot\square\lozenge\)Pa in both classical and intuitionistic modal logic. In conclusion, for any simple property P actually instantiated and any existing simple individual a, it turns out that \(\lozenge\hbox{Pa} \wedge \lnot\square\lozenge\hbox{Pa}\). This is in plain contradiction with axiom 5. The combinatorialist cannot accept this axiom.

  23. For a critical overview of the principal theories of possibility including combinatorialism, see for instance Divers 2002.

  24. It can easily be shown that, in \(\mathbf {IS4} \cup \{\hbox{A}3\}\) (and so in \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}3\}\)), A3 is equivalent to (A3*) \(\lozenge Q \rightarrow (\square(Q \rightarrow R) \rightarrow (Q \square\!\!\!\rightarrow R))\). Appealing to A3* rather than A3 in the derivation of \(Q \square\!\!\!\rightarrow R\), on line 20, produces no improvement. In this case, \(\square(Q \square\!\!\!\rightarrow R)\) should be derived from \(\square\square(Q \rightarrow R)\) and \(\square \hbox{(}Q \rightarrow R \hbox{)} \rightarrow Q\) \( (Q\square\!\!\!\rightarrow R)\), by closure. The problem is that only the first schema but not the second is necessary.

  25. It would be easy to show that, in \({{{\mathbf{IS4}}}} \cup \{\hbox{A}4\}\) (and so in \({{{\mathbf{IS4}}}} \cup \{\hbox{A}1, \hbox{A}4\}\)), A4 is equivalent to \((\square(Q \rightarrow R)\wedge \lozenge Q)\rightarrow \square(Q\square\rightarrow R\)).

  26. Proof Suppose CF2 is demonstrable in \({\mathbf{IS4}} \cup \{\hbox{A}1\}\). Let us substitute, in CF2, P for Q, a tautology T for R, and ↔ for \(\square\!\!\!\rightarrow\). The resulting meta-statement is:

    $$(1) \{\square(P \leftrightarrow (P\leftrightarrow T)), \square(P \rightarrow P), \square(P \rightarrow T)\rightarrow (P \leftrightarrow T)\}\enspace\hbox{entails}\ \square P.$$

    If CF2 were provable in \({\mathbf{IS4}} \cup \{\hbox{A}1\}\), (1) should be provable in IS4 (since ↔ obeys modus ponens, which is a rule of IS4, and modus ponens is also the only principle that \({\mathbf{IS4}} \cup \{\hbox{A}1\}\) specifies for \(\square\!\!\!\rightarrow\)). As IK is conservative over intuitionistic propositional logic, IK, and so IS4, validates both \(P \leftrightarrow (P \leftrightarrow T)\) and P → P and, by necessitation, both \(\square (P \leftrightarrow (P \leftrightarrow T))\) and \(\square (P \rightarrow P)\). Consequently, (1) can be simplified into:

    $$(2) \square(P \rightarrow T) \rightarrow (P \leftrightarrow T) \enspace\hbox{entails}\ \square P.$$

    If CF2 were provable in \({\mathbf{IS4}} \cup \{\hbox{A}1\}\), (2) should be provable in IS4. Yet, as IS4 is conservative over intuitionistic propositional logic, IS4 validates \(P \rightarrow (P \leftrightarrow T)\). By this schema, modus ponens and →introduction, it is possible to show that the following meta-statement is provable in IS4:

    $$(3)P\,\,\hbox{entails}\,\square(P \rightarrow T) \rightarrow (P \leftrightarrow T).$$

    As a result, if (2) were provable in IS4, through (3) this would yield a proof of \(\square P\) from P, which is known to be invalid in IS4. Since (2) is not provable in IS4, (1) is not provable in \({\mathbf{IS4}} \cup \{\hbox{A}1\}\). QED. I am grateful to a referee of this journal for suggesting this proof.

  27. It would be easy to show that, in \({\mathbf{IS4}} \cup \{\hbox{A}1\}\), the schema \(\square(\square(Q \rightarrow R)\rightarrow (Q \square\!\!\!\rightarrow R))\) is equivalent to the schema \(\square(Q \rightarrow R)\rightarrow \square(Q \square\!\!\!\rightarrow R)\).

  28. For instance, \({\mathbf{IS4}} \cup \{\hbox{A}1, \hbox{A}3\}\) could be supplemented with some of the rules and principles for counterfactuals assumed by Lewis (cf. 1973, p. 132), provided that the logics so obtained do not validate the questionable axiom schema A2.

  29. Indeed, the other proofs available in the literature aiming to show (or that can be read as aiming to show) that the alethic antirealist commits a conditional fallacy (see the conclusion of this article) assume no axiom schema specific for counterfactuals other than A1 and the problematic A2 (or the corresponding inference rules).

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Acknowledgements

I am very grateful to Nicola Ciprotti, John Divers, Michael Gabbay, Uriah Kriegel, Jonathan Kvanvig, Julien Murzi, Valeria de Paiva, Tommaso Piazza, Alex K. Simpson, Nicholas J.J. Smith and a referee of this Journal for valuable discussions and important criticisms upon previous versions of this paper. A special thank to Arif Ahmed and Berit Brogaard.

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    Luca Moretti

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Moretti, L. Brogaard and Salerno on antirealism and the conditional fallacy. Philos Stud 140, 229–246 (2008). https://doi.org/10.1007/s11098-007-9139-3

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