On the Ramsey Test Analysis of ‘Because’

Abstract

The well-known formal semantics of conditionals due to Stalnaker (in: Rescher (ed) Studies in logical theory, Blackwell, Oxford, 1968), Lewis (Counterfactuals, Blackwell, Oxford, 1973a), and Gärdenfors (in: Niiniluoto, Tuomela (eds) The logic and 1140 epistemology of scientific change, North-Holland, Amsterdam, 1978, Knowledge in flux, MIT Press, Cambridge, 1988) all fail to distinguish between trivially and nontrivially true indicative conditionals. This problem has been addressed by Rott (Erkenntnis 25(3):345–370, 1986) in terms of a strengthened Ramsey Test. In this paper, we refine Rott’s strengthened Ramsey Test and the corresponding analysis of explanatory relations. We show that our final analysis captures the presumed asymmetry between explanans and explanandum much better than Rott’s original analysis.

Appendix: Proofs

Proposition 2

LetKbe a non-absurd belief set and$\gamma$a non-tautology. Then (UPC) and$\alpha , \gamma \in K$implies ($\hbox {Because}_R$).

Proof

The proof presents the simplification of (UPC) to ($\hbox {Because}_R$). Assume $K\ne K_\bot$ (K is not the absurd belief set) and $\gamma$ is not a tautology. Further, suppose $\alpha , \gamma \in K$. Then, by Gärdenfors’s contraction postulate $(K^{-}4)$, $\gamma \not \in K - \gamma$.¹⁹ Moreover, since $\gamma \in K$ and $K\ne K_\bot$, $\lnot \gamma \not \in K$. By ($K^-2)$ this implies that $\lnot \gamma \not \in (K - \gamma )$. By Proposition 1, $\gamma \not \in (K - \gamma )$ implies $\gamma \not \in (K - \gamma ) * \alpha$ or $\gamma \not \in (K - \gamma ) * \lnot \alpha$, and $\lnot \gamma \not \in (K - \gamma )$ implies $\lnot \gamma \not \in (K - \gamma ) * \alpha$ or $\lnot \gamma \not \in (K - \gamma ) * \lnot \alpha$. Hence, (i) $\gamma \not \in (K - \gamma ) * \alpha$ or $\gamma \not \in (K - \gamma ) * \lnot \alpha$ and (ii) $\lnot \gamma \not \in (K - \gamma ) * \alpha$ or $\lnot \gamma \not \in (K - \gamma ) * \lnot \alpha$. (i) implies that (iii), if $\gamma \in (K - \gamma ) * \alpha$, then $\gamma \not \in (K - \gamma ) * \lnot \alpha$. (ii) implies that (iv) if $\lnot \gamma \in (K - \gamma ) *\lnot \alpha$, then $\lnot \gamma \not \in (K - \gamma ) * \alpha$. From (iii), (iv), and (UPC), we can infer ($\hbox {Because}_R$). $\square$

Proposition 3

Let$\alpha$and$\gamma$be literals or conjunctions of literals. Further, $\alpha , \gamma \in K$. Suppose that$\alpha \rightarrow \gamma$is a non-trivial implication inKand Assumption1holds for$\alpha \rightarrow \gamma$. Then, $\alpha {\,{}^{a} {\Rightarrow } \,} \gamma \in K$and$\gamma {\,{}^{a} {\Rightarrow } \,} \alpha \in K$.

Proof

($\alpha {\,{}^{a} {\Rightarrow } \,} \gamma \in K$) Suppose $\alpha \rightarrow \gamma$ is a non-trivial implication in K. Then, by Definition 3 and Assumption 1, $\alpha \rightarrow \gamma$ is a non-trivial implication in $K - \gamma$. By Definition 3, $(K - \gamma ) - \lnot \alpha \vdash \alpha \rightarrow \gamma$. We obtain, by the deduction theorem, $(K - \gamma ) - \lnot \alpha , \alpha \vdash \gamma$, which we can rewrite as $((K - \gamma ) - \lnot \alpha ) + \alpha \vdash \gamma$. By the Levi identity, we obtain $\gamma \in (K - \gamma ) * \alpha$. Using $\alpha , \gamma \in K$, we can infer therefrom that $\alpha {\,{}^{a} {\Rightarrow } \,} \gamma \in K$.

($\gamma {\,{}^{a} {\Rightarrow } \,} \alpha \in K$) Suppose $\alpha \rightarrow \gamma$ is a non-trivial implication in K. Then, by Assumption 1, $\alpha \rightarrow \gamma$ is a non-trivial implication in $K - \alpha$. By contraposition, $\lnot \gamma \rightarrow \lnot \alpha$ is a non-trivial implication in $K - \alpha$. By Definition 3, $\lnot \gamma \rightarrow \lnot \alpha$ is a non-trivial implication in $(K - \alpha ) - \gamma$. Via the deduction theorem, we obtain $(K - \alpha ) - \gamma , \lnot \gamma \vdash \lnot \alpha$. By the Levi identity, we obtain $\lnot \alpha \in (K - \alpha ) * \lnot \gamma$. Using $\alpha , \gamma \in K$, we can infer therefrom that $\gamma {\,{}^{a} {\Rightarrow } \,} \alpha \in K$. $\square$

Proposition 4

Assume a ($\hbox {Because}_R$) agent accepts all facts and the generalisation of the tower-shadow scenario, i. e. $t, s, sh, t \wedge s \rightarrow sh \in K$, where the order of epistemic entrenchment is$t, s, sh < t \wedge s \rightarrow sh$. Then, t ${}^{a} {\Rightarrow }$  $sh \in K$if$t \le sh$andsh ${}^{a} {\Rightarrow }$  $t \in K$if$sh \le t$.

Proof

By ($\hbox {Because}_R$):

$$\begin{aligned} \begin{aligned} t {\,{}^{a} {\Rightarrow } \,} sh \in K&\;{\text{ iff }}\;sh \in (K - sh) * t {\text{ or }} \lnot sh \in (K - sh) * \lnot t \\& {\text{ and }}\;t, sh \in K. \end{aligned} \end{aligned}$$

$t, s, sh \in K$ holds by assumption. We show that (a) t ${}^{a} {\Rightarrow }$  $sh \in K$ if $t \le sh$. Let us assume $t \le sh$. By (G-), (EE2), (EE1), this implies (i) $t \not \in K - sh$. By the recovery postulate

$$\begin{aligned} {\text{If }} \alpha \in K,\;{\text{ then }}\;K \subseteq (K-\alpha )+\alpha \qquad \qquad \qquad (K^-5) \end{aligned}$$

$t \in (K - sh) *sh$. By the Levi identity, the deduction theorem, and (i), this implies that $sh \rightarrow t \in (K - sh)$. Hence, $\lnot t \rightarrow \lnot sh \in (K-sh)$. Using $t\notin K -sh$ and the Levi identity, we can infer therefrom that $\lnot sh \in (K - sh) *\lnot t$. This entails (a) in light of ($\hbox {Because}_R$).

The proof of (b) $sh {\,{}^{a} {\Rightarrow } \,} t$ if $sh \le t$ is completely analogous to that of (a). $\square$

Proposition 5

Assume a ($\hbox {Because}_R$) agent accepts all the formulas in$K(H, <) = K(S)$for$H=\{ t, s, sh, t \wedge s \rightarrow sh \}$, where the order of epistemic priority is$t \sim s \sim sh < t \wedge s \rightarrow sh$. Thent ${}^{a} {\Rightarrow }$$sh \not \in K_>(S)$andsh ${}^{a} {\Rightarrow }$$t \in K_>(S)$.

Proof

t ${}^{a} {\Rightarrow }$sh and sh ${}^{a} {\Rightarrow }$t remain well-defined, when replacing K with K(S):

$$\begin{aligned} \begin{aligned}&t {\,{}^{a} {\Rightarrow } } sh \in K_>(S) {\text{ iff }} sh \in K((S - sh) * t)\;{\text{ or }}\;\lnot sh \in K((S - sh) * \lnot t) \\&\qquad {\text{ and }}\;t, sh \in K(S)\\&sh {\,{}^{a} {\Rightarrow } \,} t \in K_>(S) {\text{ iff }} t \in K((S - t) * sh) {\text{ or }} \lnot t \in K((S - t) * \lnot sh) \\&\qquad {\text{ and }} sh, t \in K(S). \end{aligned} \end{aligned}$$

t ${}^{a} {\Rightarrow }$$sh \not \in K_>(S)$, where $S = (H, <)$: $t, sh \in K(S)$ is satisfied by assumption. By Definition 1, the remainder set $H \bot sh$ contains three sets, $H' = \{ t \wedge s \rightarrow sh, s \}$, $H'' = \{ t \wedge s \rightarrow sh, t \}$, and $H''' = \{ s, t \}$. By Definition 2, $H' \le H''$ and $H'' \le H'$, but $H' \not \le H'''$. Hence, by (Def $\sigma$), both $H'$ and $H''$ are selected for the partial meet base contraction (PMBC) which yields $H - sh = \bigcap \sigma (H \bot sh) = \{ t \wedge s \rightarrow sh \}$.

By (PMBR), $(H - sh) * t = \bigcap \sigma ((H - sh) \bot \lnot t) + t$. Since $\lnot t \not \in Cn(H - sh)$, by Definition 1, $H-sh$ is the unique member of $(H - sh) \bot \lnot t$. By Definition 2, $(H-sh) \le (H-sh)$ and by (Def $\sigma$), the partial meet base contraction (PMBC) yields $(H-sh) - \lnot t = \bigcap \sigma ((H - sh) \bot \lnot t) = \{ t \wedge s \rightarrow sh \}$. Notice that when $\lnot t \not \in H-sh$, then $(H-sh) - \lnot t = H-sh$. By (H + $\alpha$), $(H - sh) * t = \{ t \wedge s \rightarrow sh \} \cup \{ t \}$. Hence, $(S - sh) * t = ((H-sh)*t, <')$, where $<'$ is such that generalisations have strict priority over literals. Then $sh \not \in K((S-sh)*t)$.

By (PMBR), $(H - sh) * \lnot t = \bigcap \sigma ((H - sh) \bot t) + \lnot t$. By similar reasoning as above, $t \not \in H - sh$ and thus $(H - sh) - t = H - sh$. By (H + $\alpha$), $(H - sh) * \lnot t = \{ t \wedge s \rightarrow sh \} \cup \{ \lnot t \}$. Hence, $(S - sh) * \lnot t = ((H-sh)* \lnot t, <')$, where $<'$ is such that generalisations have strict priority over literals. Then $\lnot sh \not \in K((S-sh)* \lnot t)$.

$sh {\,{}^{a} {\Rightarrow } \,} t \in K_>(S)$, where $S = (H, <)$: $t, sh \in K(S)$ is satisfied by assumption. By Definition 1, the remainder set $H \bot t$ contains only $H' = \{ t \wedge s \rightarrow sh, s, sh \}$. By Definition 2, $H' \le H'$ and by (Def $\sigma$), the partial meet base contraction (PMBC) yields $(H-t) = \bigcap \sigma (H \bot t) = \{ t \wedge s \rightarrow sh, s, sh \}$. By similar reasoning, $(H-t)-sh = \{ t \wedge s \rightarrow sh, s \}$. By (H + $\alpha$), $((H-t)-sh) + \lnot sh = \{ t \wedge s \rightarrow sh, s, \lnot sh \}$, which is by (PMBR) $(H-t)* \lnot sh$. Hence, $(S - t) * \lnot sh = ((H-t)* \lnot sh, <')$, where $<'$ is such that generalisations have strict priority over literals. Then $\lnot t \in K((S-t)* \lnot sh)$. $\square$

Proposition 6

Assume a ($\hbox {Because}_{P}$) agent accepts all facts and the single, more entrenched generalisation of the tower-shadow scenario, i. e. $t, s, sh, t \wedge s \rightarrow sh \in K$, where the order of epistemic entrenchment is$t \sim s \sim sh < t \wedge s \rightarrow sh$. Then$t {\,{}^{P} {\Rightarrow }\,} sh \in K_>$and$sh {\,{}^{P} {\Rightarrow }\,} t \in K_>$.

Proof

$t {\,{}^{P} {\Rightarrow }\,} sh \in K_>$: The agnostic belief set is $K' = K - (t \vee sh)$. By (G-), $t, s, sh \not \in K'$, but by recovery (i) $(t \vee sh) \rightarrow sh \in K'$. By assumption, $\lnot t \not \in K$ and so by $(K^-2)$, (ii) $\lnot t \not \in K^\prime$. Using $t \vdash t \vee sh$ and the Levi identity, we can infer from (i) and (ii) that $sh \in K' * t$.

$sh {\,{}^{P} {\Rightarrow }\,} t \in K_>$: The agnostic belief set is, again, $K' = K - (t \vee sh)$ such that $t, s, sh \not \in K'$, but by recovery$(t \vee sh) \rightarrow t \in K'$. Using $sh \vdash t \vee sh$, we infer that $t \in K' * sh$. $\square$

Proposition 7

Assume a ($\hbox {Because}_{P}$) agent accepts all the formulas in$K(H, <) = K(S)$ for $H=\{ t, s, sh, t \wedge s \rightarrow sh \}$, where$t, s, sh < t \wedge s \rightarrow sh$and the order of epistemic priority is$t \sim s \sim sh < t \wedge s \rightarrow sh$. Then$t {\,{}^{P} {\Rightarrow }\,} sh \in K_>(S)$, but$sh {\,{}^{P} {\Rightarrow }\,} t \not \in K_>(S)$.

Proof

$t {\,{}^{P} {\Rightarrow }\,} sh \in K_>(S)$, where $S=(H, <)$: $t, sh \in K(S)$ is satisfied by assumption. By Definition 1, the remainder set $H \bot (t \vee sh)$ contains only $H'' = \{ t \wedge s \rightarrow sh, s \}$. By Definition 2, $H'' \le H''$ and by (Def $\sigma$), the partial meet base contraction (PMBC) yields the agnostic belief base $H' = H - (t \vee sh) = \bigcap \sigma (H \bot (t \vee sh)) = \{ t \wedge s \rightarrow sh, s \}$.

By (PMBR), $H' * t = \bigcap \sigma (H' \bot \lnot t) + t$. Since $\lnot t \not \in Cn(H')$, by Definition 1, $H'$ is the only member of $H' \bot \lnot t$. By Definition 2 and (Def $\sigma$), $\bigcap \sigma (H' \bot \lnot t) = H'$. By (H + $\alpha$), $H' + t = \{ t \wedge s \rightarrow sh, s \} \cup \{ t \}$. Hence, $S' * t = (H' + t, <')$, where $<'$ is such that generalisations have strict priority over literals. Then $sh \in K(S' * t)$ so that $t \gg sh \in K_> (S)$.

$sh {\,{}^{P} {\Rightarrow }\,} t \not \in K_>(S)$, where $S=(H, <)$: $t, sh \in K(S)$ is satisfied by assumption. The agnostic belief base is again $H' = H - (t \vee sh) = \{ t \wedge s \rightarrow sh, s \}$. By (PMBR), $H' * sh = \bigcap \sigma (H' \bot \lnot sh) + sh$. Since $\lnot sh \not \in Cn(H')$, by Definition 1, $H'$ is the only member of $H' \bot \lnot sh$. By Definition 2 and (Def $\sigma$), $\bigcap \sigma (H' \bot \lnot t) = H'$. By (H + $\alpha$), $H' + sh = \{ t \wedge s \rightarrow sh, s \} \cup \{ sh \}$. Hence, $S' * sh = (H' + sh, <')$, where $<'$ is such that generalisations have strict priority over literals. Then $t \not \in K(S' * sh)$ so that $sh \gg t \not \in K_> (S)$. $\square$

Proposition 8

Let$\alpha$and$\gamma$be literals. Epistemic states are represented by prioritised belief bases with two levels: an upper levelGof generalisations and a lower levelLof literals, as explained in Sect.2.5. A ($\hbox {Because}_{P'}$) agent accepts ‘$\gamma$because of$\alpha$’ with respect to$(H,<)$iff$\alpha$inferentially explains$\gamma$ – in the sense of Definition6 – in the eyes of the agent accepting all members ofH.

Proof

Suppose (i) $\gamma$ because of $\alpha$ is verified by an epistemic state $(H,<)$ (in the sense of ($\hbox {Because}_{P'}$)). Let G be the set of generalisations of H, while L is the set of literals of H. Hence, (ii) there are $(H'', <'')$ and $L^-$ such that ${(H',<')}= {( H, <)} - \bigvee L^-$ and $\alpha \gg \gamma \in K_>(H', <')$. Therefore, (iii) there is $(H'',<'')= {( H', <')} - \alpha \vee \gamma$ such that $H''=G'' \cup L''$, $G''= G \cap H''$, and $L''= L \cap H''$. Hence, the pair $(G'', L'')$ satisfies conditions (1), (2), and (4) for an agent who accepts all members of H. Moreover, (ii) and (iii) imply that $G'' \cup L'', \alpha \vdash \gamma$. Hence, Condition (5) is satisfied as well for $(G'', L'')$. Finally, Condition (6) holds for $(G'', L'')$ because of (ii) and (iii). (i) implies that Condition (3) of Definition 6 is satisfied for an agent who accepts all members of H. Hence, all conditions of this definition are satisfied for such an agent. Thus, $\alpha$ inferentially explains $\gamma$—in the sense of Definition 6—in the eyes of an agent who accepts all members of H.

For the other direction, suppose (i) $\alpha$ inferentially explains $\gamma$ in the eyes of an agent a, in the sense of Definition 6. Hence, there is a set G of generalisations and a set L of literals such that conditions (1)–(6) of Definition 6 are satisfied for a. (ii) $H:=G \cup L \cup \{\alpha \}$. < is such that generalisations are prioritised over literals. Obviously, (iii) $\alpha \in K(H)$. By Condition (5) of Definition 6, (iv) $\gamma \in K(H, <)$. We show that $\alpha \vee \gamma \notin Cn(G \cup L)$. Suppose, for contradiction, $\alpha \vee \gamma \in Cn(G \cup L)$. This implies that (v) $\lnot \alpha \rightarrow \gamma \in Cn(G \cup L).$ By Condition (5) of Definition 6, we know that (vi) $\alpha \rightarrow \gamma \in Cn(G \cup L)$. Since $\alpha \vee \lnot \alpha \in Cn(G \cup L)$, (v) and (vi) imply that $\gamma \in Cn(G \cup L)$. This contradicts Condition (6) of Definition 6. Hence, $\alpha \vee \gamma \notin Cn(G \cup L)$. Therefore, $\sigma ((H, <)\bot \alpha \vee \gamma ):=\{G \cup L\}$, where $\sigma$ is defined by (Def $\sigma$) and Definition 2. Using Condition (5) of Definition 6, we can infer therefrom that $\alpha \gg \gamma \in K_>(H,<)$. Using (iii) and (iv), we can infer therefrom that ‘$\gamma$ because of $\alpha$’ is verified by the epistemic state $(H,<)$ (in the sense of ($\hbox {Because}_{P'}$)). $\square$