Transmission Failure, AGM Style

Abstract

This article provides a discussion of the principle of transmission of evidential support across entailment from the perspective of belief revision theory in the AGM tradition. After outlining and briefly defending a small number of basic principles of belief change, which include a number of belief contraction analogues of the Darwiche–Pearl postulates for iterated revision, a proposal is made concerning the connection between evidential beliefs and belief change policies in rational agents. This proposal is found to be sufficient to establish the truth of a much-discussed intuition regarding transmission failure.

Appendix

Proof of Observation 1

Assume, for conditional proof, that $A\in K$. By (CL), (SUP) and (CON) it follows that $\neg A\notin K$. Now assume, again for conditional proof, that $B\in K$ and hence that $B \in{\rm Cn}(K\cup\{A\})$. It follows by (AGM*4) that $B\in K*A$. By the right-to-left half of (EV2), it then follows that $A\,\vartriangleright\,B\in K$. $\square$

Proof of Observation 2

Assume, for conditional proof, that $B \notin K$. By $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{3}})$, it follows that $K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B=K$. So $B\in K*A$ iff $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B)*A$, and by (EV3), iff $A\,\vartriangleright\,B\in K$. $\square$

Proof of Observation 3

Given $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{3}})$, (EV3) entails (EV1) (see Observation 2). Furthermore, (EV1) entails:

• (WM) If $K\subseteq K', A\vee C\notin K'$ and $C\in K*A$, then $C\in K'*A$

Indeed, assume that $K\subseteq K'$ and $A\vee C\notin K'$. It follows that $A\vee C\notin K$, and hence, by (CL), that $A,C\notin K$. Assume furthermore that $C\in K*A$. It follows by (EV1), and the fact that $K\subseteq K'$, that $A\,\vartriangleright\,C\in K$, that $A\,\vartriangleright\,C\in K'$, and therefore, by (EV1), that $C\in K'*A$.

Now assume for reductio that ${K\in\mathbb{K}}$ is such that, for pairwise inconsistent $A,B,C\in\mathcal{L}, \neg A, \neg B, \neg C\notin K$.

The general proof strategy is as follows. We first derive (i) $B\in (K*A)*B\vee C$ and (ii) $C\in (K*A)*B\vee C$. By (CL), this then entails that $B\wedge C\in (K*A)*B\vee C$, which violates (CON). Here, we simply prove (i). The proof for (ii) is analogous.

From (AGM*3+), (AGM*4+), (SUP) and (MON), it follows that $K*A\vee B\subseteq K*A$. Indeed consider an arbitrary $C\in K*A\vee B$. By (AGM*3+), it follows that $C\in {\rm Cn}(K\cup\{A\vee B\})$. By (MON), we have $C\in {\rm Cn}(K\cup\{A\vee B, A\})$. By (SUP), the latter is equal to Cn(K ∪ {A}). Now, by hypothesis, $\neg A\notin K$, so by (AGM*4+), it follows that $C\in K*A$.

We now show that $B\vee C\notin K*A$, and hence, by (SUP) and (CL), that $(B\vee C)\vee B\notin K*A$. Assume for reductio that $B\vee C\in K*A$. Then by (AGM*3), $B\vee C\in {\rm Cn}(K\cup\{A\})$. By (DT), $A\rightarrow(B\vee C)\in K$. Since A is inconsistent both with B and with C, it follows by (CL) that $\neg A\in K$, contrary to our assumptions.

Next, we prove that $B\in (K*A\vee B)*B\vee C$. We first show that $\neg(B\vee C)\notin K*A\vee B$. Assume for reductio that $\neg(B\vee C)\in K*A\vee B$. Then by (AGM*3), it follows that $\neg(B\vee C)\in {\rm Cn}(K\cup\{A\vee B\})$. By (DT), we have $(A\vee B)\rightarrow\neg(B\vee C)\in K$. By (SUP), we have $\neg B\in{\rm Cn}((A\vee B)\rightarrow\neg(B\vee C))$, so by (CL), it follows that $\neg B\in K$, contrary to our assumptions.

Now, since $\neg A, \neg B\notin K$, by (CL), $\neg A, \neg B\notin {\rm Cn}(\varnothing)$, and hence, by (SUP), it follows that $\neg(A\vee B)\notin {\rm Cn}(\varnothing)$. By (AGM*2), therefore, $A\vee B\in K*A\vee B$. Since A, B and C were assumed to be pairwise inconsistent, it follows from this and (SUP) that $B\in {\rm Cn}(K*A\vee B\cup\{B\vee C\})$. By (AGM*4), and the fact that $\neg(B\vee C)\notin K*A\vee B$, it then follows that $B\in (K*A\vee B)*B\vee C$.

From the fact that $K*A\vee B\subseteq K*A, (B\vee C)\vee B\notin K*A$, and $B\in (K*A\vee B)*B\vee C$, which have all now been derived, it then follows by (WM) that (i) $B\in (K*A)*B\vee C$. $\square$

Proof of Observation 4

Assume, for conditional proof, that (i) $C\in {\rm Cn}(B)$ and (ii) $A\,\vartriangleright\,B\in K$. By the left-to-right half of (EV2) and (ii), it follows that $B\in K*A$. From this, by (i) and (CL), we have $C\in K*A$. By the right-to-left half of (EV2), it then follows that $A\,\vartriangleright\,C\in K$. $\square$

Proof of Observation 5

We divide the proof into three cases: (a) the case in which $C\in{\rm Cn}(\varnothing)$, (b) the case in which $\neg A\in {\rm Cn}(\varnothing)$, and (c) the case in which neither $C\in{\rm Cn}(\varnothing)$, nor $\neg A\in {\rm Cn}(\varnothing).$

Regarding (a), the proof is trivial. Assume that $C\in{\rm Cn}(\varnothing)$. It follows from this, by $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} }\! {{3}})$, that (i) $K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C = K$. Assume $A\,\vartriangleright\,B\in K$. From (i), it then follows that $A\,\vartriangleright\,B\in K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$. It also follows from $C\in{\rm Cn}(\varnothing)$, this time by (CL), that (ii) $C\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C )*A$. From (ii), by the right-to-left direction of (EV3), it follows that $A\,\vartriangleright\,C\in K$. So we have $A\,\vartriangleright\,C\in K$ and $A\,\vartriangleright\,B\in K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$, and therefore $A\,\vartriangleright\,C\notin K$ iff $A\,\vartriangleright\,B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$.

Case (b) is equally easy to handle. There are two subcases to consider, namely $B\notin{\rm Cn}(\varnothing)$ and $B\in{\rm Cn}(\varnothing):$

• Assume $B\notin{\rm Cn}(\varnothing)$. By $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{4}})$ we have (i) $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B)$. Assume $\neg A\in {\rm Cn}(\varnothing)$. By (AGM * V), we then have (ii) $(K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! B)*A=K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B$. It follows from (i) and (ii) that $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B)*A$ and hence, by the left-to-right direction of (EV3), that $A\,\vartriangleright\,B\notin K$. It is then vacuously true that if $C\in {\rm Cn}(B)$ and $A\,\vartriangleright\,B\in K$, then $A\,\vartriangleright\,C\notin K$ iff $A\,\vartriangleright\,B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$.

• Assume $B\in{\rm Cn}(\varnothing)$. It then follows, by (CL), that $B\in((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B)*A$, and hence, by the right-to-left direction of (EV3), $A\,\vartriangleright\,B\in K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$. Furthermore, since $B\in{\rm Cn}(\varnothing)$ and $C\in {\rm Cn}(B)$, we have $C\in{\rm Cn}(\varnothing)$. But then, by (CL), we have $C\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$ and therefore, by the right-to-left direction of (EV3), $A\,\vartriangleright\,C\in K$. So $A\,\vartriangleright\,C\in K$ and $A\,\vartriangleright\,B\in K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$, and therefore $A\,\vartriangleright\,C\notin K$ iff $A\,\vartriangleright\,B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$.

Regarding (c), we consider each direction of the biconditional separately.

The left-to-right direction of this biconditional, from (1) to (2), is easy enough to establish. Indeed, assume that $A\,\vartriangleright\,C\notin K$. By the right-to-left direction of (EV3), it follows that $C\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$. Now assume $C\in {\rm Cn}(B)$. It then follows from $C\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A$, by (CL), that (i) $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A$. Assume $C\notin{\rm Cn}(\varnothing)$. By $({\text{AGM}}\!\buildrel\textstyle\!\!\!\centerdot\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{4}})$, it then follows that $C\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$. From this, (CL) and the fact that $C\in {\rm Cn}(B)$, we obtain $B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C$. Therefore, by $({\text{AGM}}\!\buildrel\textstyle\!\!\!\centerdot\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{3}})$, $(K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } B=(K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)$. Given this equality, (i) is equivalent to $B\notin ((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! B)*A$, from which it follows, by the left-to-right direction of (EV3), that $A\,\vartriangleright\,B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C$.

Regarding the other direction of the biconditional, from (2) to (1), things are a little trickier. We first derive

Lemma 1

If $A\,\vartriangleright\,B\in K$, then $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A\wedge C$.

Assume for conditional proof that $A\,\vartriangleright\,B\in K$. It follows, by the left-to-right direction of (EV3), that (i) $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! B) * A$. From this, and the fact that $C\in{\rm Cn}(B)$, it follows by (CL) that (ii) $C\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! B) * A$. From (i), (ii) and (CM), it then follows that $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B) * A\wedge C$. Now assume for reductio that $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C) * A\wedge C$. Since $C\in{\rm Cn}(A\wedge C)$, it follows by $({\text{I}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} }\! {{2}})$ that $B\notin K* A\wedge C$. By $({\text{I}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} }\! {{4}})$ it then follows that $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! B) * A\wedge C:$ contradiction. Hence $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A\wedge C$.

We now prove

Lemma 2

If $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A\wedge C$ and $A\,\vartriangleright\,B\notin K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } C$, then $A\,\vartriangleright\,C\notin K$.

Assume for conditional proof both that $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A\wedge C$ and that $A\,\vartriangleright\,B\notin K\!\!\! \buildrel\textstyle\centerdot\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\!C$. From the second assumption, by the right-to-left direction of (EV3), it follows that $B\notin ((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C) \!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!B)*A$. As noted above, in the proof of the left-to-right direction of the biconditional, given $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} }\! {{4}})$, $({\text{AGM}}\!\buildrel\textstyle{\!\!\centerdot}\over{\hbox{\vrule height3pt depth0pt width0pt}{-} } \!{{3}})$ and (CL), this is equivalent to $B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A$. Given that $\neg A\notin {\rm Cn}(\varnothing)$, by (AGM*2), it follows that $A\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$. But by (SUP), we have $A\rightarrow B\in {\rm Cn}(\{(A\wedge C)\rightarrow B, A \rightarrow C\})$. So it follows by (CL) that either:

• (i) $(A\wedge C)\rightarrow B\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$, or

• (ii) $A \rightarrow C\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$

Assume (i) for reductio. But note that we have already assumed that $B\in (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A\wedge C$. Since $A\in {\rm Cn}(A\wedge C)$, it follows from this, by (I*1), that $B\in ((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A)*A\wedge C$. By (AGM*3), we then have $B\in {\rm Cn}((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\!C)*A\cup\{A\wedge C\})$. Finally, by (DT), it follows that $(A\wedge C)\rightarrow B\in {\rm Cn}((K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A)$. Contradiction.

It therefore follows that (ii) $A\rightarrow C\notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} }\! C)*A$. But since, by (SUP), $A \rightarrow C\in {\rm Cn}(C)$, it follows by (CL) that $C \notin (K\!\!\!\buildrel\textstyle{\!\centerdot}\over{\hbox{ \vrule height3pt depth0pt width0pt}{-} } \!C)*A$. Finally, by the left-to-right direction of (EV3), it follows that $A\,\vartriangleright\,C\notin K$.

The desired result then follows immediately from the conjunction of the two lemmas. $\square$