Difference-making grounds

Abstract

We define a notion of difference-making for partial grounds of a fact in rough analogy to existing notions of difference-making for causes of an event. Using orthodox assumptions about ground, we show that it induces a non-trivial division with examples of partial grounds on both sides. We then demonstrate the theoretical fruitfulness of the notion by applying it to the analysis of a certain kind of putative counter-example to the transitivity of ground recently described by Jonathan Schaffer. First, we show that our conceptual apparatus of difference-making enables us to give a much clearer description than Schaffer does of what makes the relevant instances of transitivity appear problematic. Second, we suggest that difference-making is best seen as a mark of good grounding-based explanations rather than a necessary condition on grounding, and argue that this enables us to deal with the counter-example in a satisfactory way. Along the way, we show that Schaffer’s own proposal for salvaging a form of transitivity by moving to a contrastive conception of ground is unsuccessful. We conclude by sketching some natural strategies for extending our proposal to a more comprehensive account of grounding-based explanations.

Appendix 1: Logical cases of non-difference-making

We have claimed above that plausible assumptions concerning the logic of ground allow us to identify a systematic way of producing instances of non-difference-making ground given a pair of suitably independent facts to start with. This appendix develops our above, rough and informal argument for this claim in more detail.

The required logical assumptions concern the pure and the propositional logic of ground. Unfortunately, there is currently no standard, fully worked out propositional logic of ground. Rather, what we have is: (I) a system of natural inference rules that are plausible relative to a very fine-grained conception of ground, proposed by Kit Fine, but with as yet no adequate semantics, and acknowledged as incomplete with respect to any plausible understanding of ground (cf. Fine 2012a, 67); (II) a natural truthmaker-semantics, also proposed in Fine (2012a), which yields a logic adequate to a much more coarse-grained conception of ground; (III) a logical system proposed by Correia (2010), addressing again a coarse-grained conception of ground, proven sound and complete for a corresponding algebraic semantics.³⁵ An additional difficulty is that the languages of these logical systems are expressively too weak to state anything like the quantified claim that for every $\Gamma$, if $\Gamma , P < Q$, then $\Gamma ^0 < Q$ for some $\Gamma ^0 \subseteq \Gamma$ with $P \notin \Gamma ^0$.

Nevertheless, we believe a useful case for our claim can be made. To this end, we shall do two things. Firstly, we discuss a sharpening of our above informal argument in the context of the truthmaker-semantics for ground. We suggest that in this context, it is best to work with a slightly more refined conception of what it is for $\Gamma$ to contain a full ground of some fact than the simple set-theoretic conception employed above. It can then be shown that for suitably independent, true P and Q, the informal argument given above goes through. It would take a lot of work to reconstruct the argument within the logic of (III), so we choose not to do so here. However, it is known that the logic obtained on the semantics in (II) is very close to that in (III) (cf. Fine ms, 11), and it is clear that a version of our argument can be given for Correia’s system, too. Secondly, we discuss our informal argument in the context of the system (I). We show that given an additional rule that turns out valid on both of the only two semantics known to us that validate the others of Fine’s rules, our argument goes through given very modest assumptions of independence of P and Q. Without the additional rule, a move parallel to that made before of refining the relevant notion of containment will secure our result.

1.1 Appendix 1.1: The coarse-grained framework

In his truthmaker-semantics for ground, Fine associates with each sentence A a set of states that verify A and a set of states that falsify A. States are here thought of as obtaining. Since no sentence is both true and false and no sentence is neither, exactly one of the two sets of states associated with a given sentence is empty.³⁶ For our purposes, it is easiest to reason directly about the sets of states and forget about the sentences to which they are assigned. We may then think of a non-empty set of states as a true proposition which is verified by exactly its members. Following Fine, we assume that (i) whenever there are states $s, t, u, \ldots$, there is also their fusion $\bigsqcup \{s, t, u, \ldots \} = s \sqcup t \sqcup u \sqcup \ldots$,³⁷ and that (ii) any proposition P is closed under non-empty fusions, so that $\bigsqcup P^\prime \in P$ whenever $\emptyset \subset P^\prime \subseteq P$.

The full and partial notions of (strict) ground are defined by Fine in terms of prior notions of full and partial weak ground. The definitions are as follows:³⁸

(Df. $\le$):

$P_1, P_2, \ldots \le Q =_{\text {df.}}s_1 \sqcup s_2 \sqcup \ldots \in Q$ whenever $s_1 \in P_1, s_1 \in P_2, \ldots$

(Df. $\preceq$):

$P \preceq Q =_{\text {df.}}\Gamma , P \le Q$ for some set of propositions $\Gamma$

(Df. ${<}$):

$P_1, P_2, \ldots < Q =_{\text {df.}}P_1, P_2, \ldots \le Q$ and $Q \not \preceq P_i$ for all i

(Df. $\prec$):

$P \prec Q =_{\text {df.}}\Gamma , P < Q$ for some set of propositions $\Gamma$

Now say that a state s is part of a state t iff for some state $s^\prime$, $t = s \sqcup s^\prime$, and say that two states s and t overlap iff they share some (non-null³⁹) part.

The conjunction $P \wedge Q$ of two propositions is the set $\{s \sqcup t: s \in P$ and $t \in Q\}$, and the disjunction $P \vee Q$ is the set $P \cup Q \cup (P \wedge Q)$. Now consider any truths P, Q which are independent in the sense that $\bigsqcup P$ and $\bigsqcup Q$ do not overlap. We may now show that

• (A) $P \prec Q \vee (P \wedge Q)$

This is immediate by definition given that $P, Q < Q \vee (P \wedge Q)$, which may be established as follows. Suppose $s \in P$ and $t \in Q$. Then $s \sqcup t \in P \wedge Q$, and hence $s \sqcup t \in Q \vee (P \wedge Q)$. So $P, Q \le Q \vee (P \wedge Q)$. Now consider $\bigsqcup (Q \vee (P \wedge Q)) = \bigsqcup P \sqcup \bigsqcup Q$. Since $\bigsqcup P$ and $\bigsqcup Q$ do not overlap, $\bigsqcup P \sqcup \bigsqcup Q$ is not a part of $\bigsqcup P$ or $\bigsqcup Q$. But it is easy to verify that $R \preceq P$ only if $\bigsqcup R$ is part of $\bigsqcup P$, so it follows that $Q \vee (P \wedge Q) \not \preceq P$ and that $Q \vee (P \wedge Q) \not \preceq Q$. Hence $P, Q < Q \vee (P \wedge Q)$.

Now suppose that some scenario S which includes the fact that P contains a scenario which is a strict full ground of $Q \vee (P \wedge Q)$. To show that P is not a difference-maker, we need to show that $S{\setminus} \{$the fact that $P\}$ still contains a strict full ground of P. If this fails for any scenario, it fails for a scenario which is a strict full ground of $Q \vee (P \wedge Q)$, so we may restrict attention to such scenarios. Suppose, therefore, that

• (H) $\Gamma , P < Q \vee (P \wedge Q)$

Writing $\bigwedge \Gamma$ for the conjunction of all the members of $\Gamma$,⁴⁰ it follows that every verifier of $\bigwedge \Gamma \wedge P$ is a verifier of $Q \vee (P \wedge Q)$. Since any such verifier has a part that verifies P, by the assumption of independence, no such verifier is a verifier of Q, and hence it must be a verifier of $P \wedge Q$. But since no verifier of P is part of a verifier of Q, it follows that every verifier of $\bigwedge (\Gamma {\setminus} \{P\})$ must have a part that verifies Q.

But does it follow that $\Gamma {\setminus} \{P\}$ contains a strict full ground of $Q \vee (P \wedge Q)$? If a scenario contains another just in case the latter is a subset of the former, it does not.⁴¹ In a rough approximation, the problem is that although some subset of $\Gamma {\setminus} \{P\}$ must be strong enough to establish Q, it may be that every such subset is, as it were, strictly between Q and $P \wedge Q$, and thereby fail to ground $Q \vee (P \wedge Q)$.⁴²

However, intuitively, what this shows is not that P may after all be a difference-maker with respect to $Q \vee (P \wedge Q)$, but that the set-theoretic interpretation of the containment of one scenario in another is unsatisfactory. For in the intuitively relevant sense, since every verifier of $\bigwedge (\Gamma {\setminus} \{P\})$ contains a verifier of Q, $\Gamma {\setminus} \{P\}$ contains $\{Q\}$, and thereby a strict full ground of $Q \vee (P \wedge Q)$. So what we should do is refine the conception of containment appealed to in the definition of difference-making. We shall therefore say that a scenario $\Gamma$ contains a scenario $\Gamma ^0$ just in case every verifier of $\bigwedge \Gamma$ has a part that verifies $\bigwedge \Gamma ^0$. Then since $\Gamma {\setminus} \{P\}$ contains $\{Q\}$ whenever $\Gamma , P < Q \vee (P \wedge Q)$, P comes out a non-difference-making partial ground of $Q \vee (P \wedge Q)$.

1.2 Appendix 1.2: The fine-grained framework

We use the following rules:⁴³

Sub(${<}/\prec$):

From $\Gamma , A < B$ infer $A \prec B$

Sub(${<}$/$\le$):

From $\Gamma < B$ infer $\Gamma \le B$

Sub($\le$/$\preceq$):

From $\Gamma , A \le B$ infer $A \preceq B$

Sub($\prec$/$\preceq$):

From $A \prec B$ infer $A \preceq B$

Trans(${<}$/${<}$):

From $A < B$ and $B < C$ infer $A < C$

Trans($\le$/${<}$):

From $\Gamma \le A$ and $A < B$, infer $\Gamma < B$

Trans($\prec ^{sp}$/$\preceq$):

From $A \prec ^{sp} B$ and $B \preceq C$ infer $A \prec ^{sp} C$

Irr($\prec ^{sp}$):

From $A \prec ^{sp} A$, infer $\bot$

Rev Sub($\le$/${<}$):

From $A_1, A_2, \ldots \le B$ and $A_1 \prec ^{sp} B$, $A_2 \prec ^{sp} B$, $\ldots$, infer $A_1, A_2, \ldots < B$

($\wedge$I):

From A and B, infer $A, B < A \wedge B$

($\vee$I):

From A, infer $A < A \vee B$ or $B < A \vee B$

($\vee$E):

From $\Gamma < A \vee B$, infer that either $\Gamma \le A$, or $\Gamma \le B$, or for some $\Gamma _A, \Gamma _B$ with $\Gamma = \Gamma _A \cup \Gamma _B$: $\Gamma _A \le A$ and $\Gamma _B \le B$

($\wedge$E):

From $\Gamma < A \wedge B$ infer that for some $\Gamma _A, \Gamma _B$ with $\Gamma = \Gamma _A \cup \Gamma _B$: $\Gamma _A \le A$ and $\Gamma _B \le B$

We make the following assumptions.

• (A1) P

• (A2) Q

• (A3) $P \not \preceq Q$

Using the rules ($\wedge$I), ($\vee$I), Trans(<), and Sub(</$\prec$) we obtain

• (1) $P \prec Q \vee (P \wedge Q)$

Now suppose

• (S) $\Gamma , P < Q \vee (P \wedge Q)$

By ($\vee$E), we may infer from (S) that one of the following three claims holds:

• (i) $\Gamma , P \le Q$

• (ii) $\Gamma , P \le P \wedge Q$

• (iii) For some $\varDelta _1, \varDelta _2$ with $\varDelta = \varDelta _1 \cup \varDelta _2$: $\varDelta _1 \le Q$ and $\varDelta _2 \le P \wedge Q$

But from (i), it follows by Sub($\le$/$\preceq$) that $P \preceq Q$, contrary to our assumption.

Now suppose (iii), and let $\varDelta _1 \le Q$. Note that $P \notin \varDelta _1$, for otherwise again $P \preceq Q$, contrary to our assumption. So $\varDelta _1$ is a subset of $\Gamma \cup \{P\}$, not including P, with $\varDelta _1 \le Q$. Using that $Q < Q \vee (P \wedge Q)$ as well as Trans($\le$/<), we obtain that $\varDelta _1 < Q \vee (P \wedge Q)$. So $\varDelta _1$ is the required witness for the claim that P is a non-difference-making partial ground of $Q \vee (P \wedge Q)$ for the case that (iii) holds.

Suppose finally that (ii) holds. Then by Sub($\le$/$\preceq$), we have $R \preceq P \wedge Q$ for all $R \in \Gamma , P$. Since any relationship of weak partial ground is either mutual or a relationship of strict partial ground, it follows by Rev Sub($\le$/<) that one of the following two conditions holds:

• (a) $\Gamma , P < P \wedge Q$

• (b) $P \wedge Q \preceq R$ for some $R \in \Gamma , P$

If (a), then by ($\wedge$E) we obtain that for some subset $\varDelta _1$ of $\Gamma , P$, we have $\varDelta _1 \le Q$. Note that $P \notin \varDelta _1$ for otherwise $P \preceq Q$, contrary to our assumption. So $\varDelta _1$ is a subset of $\Gamma \cup \{P\}$, not including P, with $\varDelta _1 \le Q$, and thereby as before, $\varDelta _1 < Q \vee (P \wedge Q)$. So $\varDelta _1$ is the required witness for the claim that P is a non-difference-making partial ground of $Q \vee (P \wedge Q)$ for the case that (a) holds.

If (b) holds, then we have for some $R \in \Gamma \cup \{P\}$ both $P \wedge Q \preceq R$ and $R \preceq P \wedge Q$. It is tempting to infer from this that $P \wedge Q$ and R are mutual weak full grounds, as per the rule:

Rev Sub($\preceq$/$\le$):

From $A \preceq B$ and $B \preceq A$, infer $A \le B$

And on the only proposals for a semantics for a logic of ground incorporating Fine’s rules—in as yet unpublished work by Krämer (ms) and Correia (ms)—this inference indeed comes out valid. But then $R \le P \wedge Q$, and since $P \wedge Q < Q \vee (P \wedge Q)$ we obtain $R < Q \vee (P \wedge Q)$. Now it can be shown that $R \ne P$. For otherwise we obtain $P \le P \wedge Q$. But $P \wedge Q \not \preceq P$, for otherwise we obtain by Trans($\prec ^{sp}$/$\preceq$) that $P \prec ^{sp} P$, and thus $\bot$ by Irr($\prec ^{sp}$). It follows by Rev Sub($\le$/<) that $P < P \wedge Q$. But then by the elimination rule ($\wedge$E), $P \le Q$ and hence $P \preceq Q$, contrary to our assumption. So $P \notin \{R\}$, and hence $\{R\}$ is our required witness for the claim that P is a non-difference-making partial ground of $Q \vee (P \wedge Q)$ for the case that (iii) holds.

Now it may be that there could also be a plausible semantics for the kind of logic of ground Fine proposes which delivers natural counter-models to Rev Sub($\preceq$/$\le$). In that case R would again, roughly speaking, have to lie between Q and $P \wedge Q$ in strength. But then similar means to those proposed above would give a natural way to salvage our case by refining the conception of containment to fit the relevant sense of ‘between’.