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Two Concepts of Plausibility in Default Reasoning

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Abstract

In their unifying theory to model uncertainty, Friedman and Halpern (1995–2003) applied plausibility measures to default reasoning satisfying certain sets of axioms. They proposed a distinctive condition for plausibility measures that characterizes “qualitative” reasoning (as contrasted with probabilistic reasoning). A similar and similarly fundamental, but more general and thus stronger condition was independently suggested in the context of “basic” entrenchment-based belief revision by Rott (1996–2003). The present paper analyzes the relation between the two approaches to formalizing basic notions of plausibility as used in qualitative default reasoning. While neither approach is a special case of the other, translations can be found that elucidate their relationship. I argue that Rott’s notion of plausibility allows for a more modular set-up and has a better philosophical motivation than that of Friedman and Halpern.

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Notes

  1. That A ∪ B < C implies A < B ∪ C follows from other conditions of Friedman and Halpern. See the dominance conditions in Sect. 4.1 below.

  2. The concept of dispositional coherence is briefly discussed in Rott (2003, pp. 262–263).

  3. However, somewhat confusingly, Halpern (1997, p. 5) uses the term ‘qualitative’ for relations that satisfy the unrestricted condition which is the left-to-right direction of Choice. Since this usage was not maintained in (Friedman and) Halpern’s later work, I will ignore it from now on.

  4. Halpern (2003, p. 289).

  5. These are the words of Friedman and Halpern (1996, p. 1301, 2001, p. 658) and Halpern (2003, p. 303).

  6. See the principles of Left logical equivalence and Right weakening below.

  7. On the account of plausibility that I will advocate in this paper (and many related accounts), even all propositions that are consistent with one’s expectations enjoy maximal plausibility. But it needs to be pointed out that this is not true for Friedman and Halpern’s account.

  8. Compare Rott (2009) with references to the seminal work produced within the frameworks of AGM-style belief revision and Dubois and Prade’s possibility theory. Both entrenchment and plausibility relations may be considered as representing degrees of belief. It appears, however, that this duality between plausibility and entrenchment cannot be fruitfully brought to bear upon Friedman and Halpern’s work on plausibilities.

  9. Alchourrón et al. (1985, pp. 525–526) discuss a postulate called Ventilation according to which ‘withdrawing A ∩ B’ means exactly the latter. But this is a very strong postulate that has no counterpart in the core of default reasoning as understood by Friedman and Halpern. As Alchourrón, Gärdenfors and Makinson show, Ventilation implies Rational monotony (see Sect. 3 below).

  10. The Harper identity says that \(X\dot{-}A = X\cup(X*-A)\); cf. Gärdenfors (1988, p. 70). For the equivalence of (From \(\dot{-}\) to <BE) and (From * to <BE) we also need the implication from \(X*-(A\cap B)\subseteq B\) to \(X\subseteq B\) which is valid in basic belief change theory.

  11. Or if one just assumes ϕ A  ∨ ϕ B as the sole premise. Essentially the same idea is used as a method of deriving a plausibility relation from counterfactuals (with the ordering reversed) already by Lewis (1973, p. 54), and as a method of deriving a plausibility relation from a default inference relation by Lehmann and Magidor (1992, pp. 31, 45) and by Freund (1993, 1998). The point of the present paper is that this method can be put to work in much weaker settings than the ones envisaged by Lewis and Lehmann and Magidor, and even than the one envisaged by Freund.

  12. Because I needed additional principles linking contractions to revisions and default inference relations, my case for the reconstructive view would be stronger if we were interested just in contractions rather than default inference.—I will return to this topic in Sect. 5.

  13. In the absence of AND, the plausibility definition (From \({\vert\!\!\sim}\) to <BP) cannot be used. Hawthorne (1996, p. 200) gives a definition that appears to be suited for the logic of high conditional probability which characteristically lacks AND (as well as OR and CM). After appropriate notational adaptations, it reads thus:

    $$ A<B \quad \hbox{iff}\quad\,{\phi_{B}}{\vert\!\!\!\not\!\!\sim}{\bot}\, \hbox{and for all}\,\gamma\,\hbox{and}\,\delta,\,\hbox{if}\,\phi_{B}{\vee\gamma} \; {\vert\!\!\sim}{\delta}\,\hbox{then}\, {\phi_{B}\vee\gamma} \; {\vert\!\!\sim}{\neg\phi_{A}\wedge\delta} $$

    Hawthorne shows that on adding AND, this definition reduces to

    $$ A<B\quad\hbox{iff}\quad {\phi_{B}}{\vert\!\!\!\not\!\!\sim}{\bot}\,\hbox{and}\,\phi_{A}\vee\phi_{B}{\vert\!\!\sim}{\neg\phi_{A}} $$

    i.e., essentially (From \({\vert\!\!\sim}\) to <BP).

  14. The supplementary postulates are labelled postulates number 7 and 8 in the traditional numbering proposed by Alchourrón, Gärdenfors and Makinson. Until well into the 1990s, the research in nonmonotonic reasoning recognized finer distinctions than belief revision research.

  15. One could replace CP by the conditions

    $$ (\bot\hbox{Cond}) \quad \hbox{If}\,{\alpha\wedge\beta} \; {\vert\!\!\sim} {\bot},\, \hbox{then}\,{\alpha} \; {\vert\!\!\sim}{\neg\beta}\quad {\rm and} $$
    $$ (\bot\hbox{CM}) \quad \hbox{If}\,{\alpha} \; {\vert\!\!\sim}{\bot}, \,\hbox{then}\,{\alpha\wedge\beta} \; {\vert\!\!\sim}{\bot} $$

    Given REF and RW, both \(\bot\)Cond and \(\bot\)CM follow from CP. But even taken together, \(\bot\)Cond and \(\bot\)CM are still a lot weaker than CP. \(\bot\)Cond is a special case of Conditionalization, If \({\alpha\wedge\beta} \; {\vert\!\!\sim}{\gamma}\), then \({\alpha} \; {\vert\!\!\sim} {\beta\!\to\!\gamma},\) and \(\bot\)CM is a special case of CM. Note that \(\bot\)OR, If \({\alpha} \; {\vert\!\!\sim}{\bot}\) and \({\beta} \; {\vert\!\!\sim}{{\bot}}\), then \({\alpha\vee\beta} \; {\vert\!\!\sim}{\bot}\), would not be strong enough as a replacement of \(\bot\)Cond. It is not (relatively, given LLE, REF, RW and AND) equivalent to \(\bot\)Cond, even though OR is (relatively, given LLE, REF, RW and AND) equivalent to Conditionalization. Friedman and Halpern’s condition FH3 mentioned below, and also the conditions E\( \emptyset \)1 and E\( \emptyset \)2 discussed in Rott (2001, p. 235), are examples of similar technical conditions that are, I believe, of little intrinsic interest.

  16. They actually use plausibility measures, i.e., mappings that assign a member of a partially ordered set to each element of the domain. I will not attend to this difference in the present paper.

  17. Why? Because there is no “local” way of distinguishing between plausibility ties and plausibility incomparabilities. This can be seen from the basic idea of (From \({\vert\!\!\sim}\) to <BP): If the agent’s premise is ϕ A ∨ϕ B and she then disbelieves neither A nor B, this may be so either because A and B are tied or because they are incomparable in terms of plausibility. Also cf. Rott (1992, p. 50).

  18. Also see Dubois, Fargier and Prade (2004, pp. 34–35).

  19. See footnote 5.

  20. This is the OR rule for the non-vacuous case: If \({\alpha} \; {\vert\!\!\sim}{\gamma}\) and \({\beta} \; {\vert\!\!\sim}{\gamma}\) and \({\alpha\vee\beta} \; {\vert\!\!\!\not\!\!\sim}{\bot},\) then \({\alpha\vee\beta} \; {\vert\!\!\sim}{\gamma}.\) Of course, this contains the essence of the OR rule.

  21. Basic entrenchment relations were originally presented as relations between sentences rather than propositions. If we wanted to do the same here for the dual notion of plausibility, we would need a condition of Extensionality, like this one:

    $$ \hbox{If}\,\alpha \vdash \beta\,\hbox{and}\,\beta \vdash \alpha,\,\hbox{then}:\, \alpha<_{\rm BP} \; \gamma\,\hbox{iff}\,\beta<_{\rm BP} \; \gamma,\,\hbox{and}\,\gamma<_{\rm BP} \; \alpha\,\hbox{iff}\,\gamma<_{\rm BP} \; \beta. $$

  22. This condition could be strengthened to

    $$ \hbox{If}\,A<_{\rm BP} \; B\,\hbox{and}\,-A\cap B \subseteq C \subseteq A\cup B,\,\hbox{then}\, A<_{\rm BP} \; C $$

    with essentially the same proof, but then it would not be a weakened form of Continuing up (see below) any more.

  23. A counterexample similar to Example 2 for the dual notion of basic entrenchment is given in Rott (2003, p. 268, footnote 20). It is derived from an intuitive cycle in a belief revision function.

  24. This in fact is essentially the axiomatization given by Freund (1993, pp. 237–238). It is in turn essentially dual the to axiomatization of “generalized epistemic entrenchment” given in Rott (1992, p. 55).

  25. CP is only needed for part (ii) of Lemma 3; the proof shows that parts (v) and (vi) need only the weaker properties \(\bot\)CM and \(\bot\)Cond, respectively. Compare footnote 15.

  26. Personal communication, July 2011. Dubois, Fargier and Prade (2004, pp. 34–35) remark that Choice is much stronger than Qualitativeness, but they do not present a counterexample.

  27. See Friedman and Halpern (2001, p. 655) and Halpern (2003, p. 302).

  28. Used, modulo some translation, in Rott (2003, p. 264). Rott (2001, p. 254, 264) uses an analogue of \(\emptyset \nless [\![{\alpha}]\!] \) as the second disjunct.

  29. This equivalence is not valid for Friedman–Halpern plausibility relations.

  30. Kraus et al. (1990, p. 194, Definition 5.9 and Lemma 5.10). As mentioned in footnote 11, Lehmann and Magidor (1992) changed to the method of (From \( {\vert\!\!\sim} \) to <BP) in the context of “rational” systems of nonmonotonic reasoning, unfortunately without any comment. Hawthorne (1996, p. 201) and Halpern (1997, p. 13) overlooked this change.

  31. For this reason, whenever \({{\mathcal{P}}_{\mathrm{B}}({\vert\!\!\sim})}\) fails to be acyclic, so does \({{\mathcal{P}}_{\mathrm{FH}}({\vert\!\!\sim})}\).

  32. Suppose we interpret < abstractly as a relation that makes distinctions, in the sense that A < B implies that A and B can be told apart (if A < B, then, since < is irreflexive, A and B cannot be identical). Then the basic plausibility relation <BP is less discriminating or coarser than the Friedman–Halpern plausibility relation <FH. (But we should not put too much strain on this interpretation.)

  33. Thanks to Hannes Leitgeb for making me phrase my diagnosis in these words. Compare William James (1975, p. 30): “There can be no difference any-where that doesn’t make a difference elsewhere—no difference in abstract truth that doesn’t express itself in a difference in concrete fact and in conduct consequent upon that fact, imposed on somebody, somehow, somewhere and somewhen.”

  34. Homophonic translation is translation of one speaker’s expression by another speaker’s expresssion that reads or sounds the same. Since the two speakers may speak different languages or advocate fundamentally different theories, we should not rashly say that the two expressions mapped to each other by the translation are the same.

  35. This work also underlies the presentation of basic entrenchment in Rott (2003)

  36. It is understood here that some worlds are actually chosen. See the dual condition for entrenchments in Rott (2001, p. 251). Applying the connection (From <BP to <FH), this gives the following semantics for Friedman–Halpern plausibilities: D <FH E if and only if all of the chosen (most plausible) worlds of D ∪ E lie in E and some of them do not lie in D. Compare this with condition (2) in Friedman and Halpern (2001, p. 656).

  37. It is easy to show that the following condition corresponds precisely to the transitivity of <BP

    $$ (+) \quad \hbox{If}\,{\alpha\vee\beta} \; {\vert\!\!\sim}{\neg\alpha} \quad \hbox{and}\,{\beta\vee\gamma} \; {\vert\!\!\sim}{\neg\beta},\,\hbox{then}\, {{\alpha\vee\gamma}} \; {\vert\!\!\sim}{{\neg\alpha}} $$

    This condition is valid in preferential reasoning. Proof: From \( {{\alpha\vee\beta}} \; {\vert\!\!\sim}{{\neg\alpha}} \) and \( {{\neg\alpha\wedge\gamma}} \; {\vert\!\!\sim}{{\neg\alpha}}\) it follows by OR that \({\alpha\vee\beta\vee\gamma} \; {\vert\!\!\sim}{\neg\alpha}\). Similarly, we get from \({\beta\vee\gamma} \; {\vert\!\!\sim}{\neg\beta}\) that \( {{\alpha\vee\beta\vee\gamma}} \; {\vert\!\!\sim}{{\neg\beta}}\). By REF, AND and RW, we get \({\alpha\vee\beta\vee\gamma} \; {\vert\!\!\sim}{\alpha\vee\gamma}\). So by CM and LLE, \({\alpha\vee\gamma} \; {\vert\!\!\sim}{\neg\alpha}\). Since apparently (+) cannot be nicely simplified and is not of any intrinsic interest, it will not be mentioned here any more.

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Acknowledgments

I would like to thank David Etlin, Joe Halpern, Hannes Leitgeb, an anonymous referee of Erkenntnis, as well as audiences of two workshops on Halpern’s work (July 2011) and on “Philosophical Issues in Belief Revision, Conditionals and Possible Worlds Semantics” (September 2012) at the University of Konstanz, and of the Ninth Annual Formal Epistemology Workshop in Munich (May/June 2012) for helpful comments on earlier versions of this paper. They have led to substantial improvements.

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  1. Department of Philosophy, University of Regensburg, 93040, Regensburg, Germany

    Hans Rott

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Appendix: Proofs

Appendix: Proofs

Proof of Lemma 1

 

  • (Nonnegativity) Suppose for reductio that A <BP \( \emptyset \). Then A ∪ A <BP \( \emptyset \), and so by Choice easy A <BP A ∪ \( \emptyset \). But this means A <BP A, contradicting Irreflexivity.

  • (Asymmetry) Suppose for reductio that A <BP B and B <BP A, i.e., A ∪ A <BP B and B ∪ B <BP A. By Choice easy, we get A <BP A ∪ B and B <BP A ∪ B. This also means that A <BP B ∪ (A ∪ B) and B <BP A ∪ (A ∪ B). By Choice hard, it follows that A ∪ B <BP A ∪ B, contradicting Irreflexivity.

  • (Join right) Let A <BP B. This means that A ∪ A <BP B. By Choice easy and Choice hard, this is equivalent to A <BP A ∪ B.

  • (Meet right) Let A <BP B. By Join right, this is true just in case A <BP A ∪ B. Since A ∪ (− A ∩ B) = A ∪ B, this means that A <BP A ∪ (− A ∩ B). By Choice easy and Choice hard, this is true just in case A <BP − A ∩ B.

  • (GM-Dominance) Suppose for reductio that A <BP B and \(B\subseteq A. \) Since A = A ∪ B, we get A ∪ B <BP B. So by Join right, A ∪ B <BP (A ∪ B) ∪ B, contradicting Irreflexivity.

  • (Weak join left) Let A <BP C,  B <BP C and \(A\cup B\subseteq C. \) Since A and B are subsets of C, we have A <BP B ∪ C and B <BP A ∪ C, so by Choice hard A ∪ B <BP C.

  • (Weak continuing down) Let A <BP B and \(A\cap-B \subseteq C \subseteq A. \) Since A ∪ C = A, A ∪ C <BP B. So by Choice easy, C <BP A ∪ B. But A ∪ B = C ∪ B, so this means that C <BP C ∪ B. Thus by Choice hard C <BP B.

  • (Weak continuing up) Let A <BP B and \(B \subseteq C \subseteq A\cup B. \) By Choice easy, we get A <BP A ∪ B. Since A ∪ C = A ∪ B, we get A <BP A ∪ C. Thus by Choice hard A <BP C. □

Proof of Lemma 2

  1. (i)

    (BP-ix), Transitivity. Let A <BP B and B <BP C. Then, by Continuing up, A <BP B ∪ C and B <BP A ∪ C. By Choice hard, it follows that A ∪ B <BP C. So by Continuing down, A <BP C.Footnote 37

    (BP-x) Let A <BP C and B <BP C. By Continuing up, A <BP B ∪ C and B <BP A ∪ C. So by Choice hard, A ∪ B <BP C.

    (BP-xi) Let A <BP C and B <BP D. By Continuing up, A <BP B ∪ C ∪ D and B <BP A ∪ C ∪ D. So by Choice hard, A ∪ B <BP C ∪ D.

  2. (ii)

    As already noted, Choice easy follows from Continuing up and Continuing down. It remains to show that Choice hard is valid. Suppose that A <BP B ∪ C and B <BP A ∪ C. Then, by Continuing up, A <BP A ∪ B ∪ C and B <BP A ∪ B ∪ C. So, by BP-x, A ∪ B <BP A ∪ B ∪ C, and by BP-iii, A ∪ B <BP C.□

Proof of Lemma 3

Suppose that <BP is generated from \({\vert\!\!\sim}\) by (From \({\vert\!\!\sim}\) to <BP). Because of LLE and RW, <BP is well-defined. We will sometimes use LLE without mentioning it explicitly. In parts (v) and (vi), we use \(\bot\)CM and \(\bot\)Cond, respectively, rather than the stronger CP.

  1. (i)

    (Irreflexivity) This is trivial.

  2. (ii)

    (Minimality) Let \(A \not= \emptyset. \) This means that \(\phi_{A}\not\vdash \bot\). We want to show that \( \emptyset \) <BP A, that is, using (From \({\vert\!\!\sim}\) to <BP), \( \phi_{\emptyset}\vee\phi_{A}{\vert\!\!\sim}{\neg\phi_{\emptyset}}\) and \( {\phi_{\emptyset}}\vee\phi_{A} \; \vert\!\!\!\not\!\!\sim{\neg\phi_{A}}\). But \( \phi_{\emptyset} \) is \(\bot\), so by LLE and RW, this means that we need to show that \({\phi_{A}}{\vert\!\!\sim}{\top}\) and \({\phi_{A}} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{A}}\). The former follows from REF and RW. Now by CP, we know that \({\phi_{A}} \; {\vert\!\!\!\not\!\!\sim}{\bot}\). But this implies, by REF, AND and RW, that \({\phi_{A}} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{A}}\), as desired.

  3. (iii)

    (Choice easy) Let A ∪ B <BP C. We want to show that A <BP B ∪ C and B <BP A ∪ C. Using (From \({\vert\!\!\sim}\) to <BP), the supposition gives us

    $$ (\phi_{A}\vee\phi_{B})\vee\phi_{C}{\vert\!\!\sim}{\neg(\phi_{A}\vee\phi_{B})} \quad \hbox{and} \quad (\phi_{A}\vee\phi_{B})\vee\phi_{C} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{C}}. $$

    We need to show that

    $$ {\phi_{A}}\vee(\phi_{B}\vee\phi_{C}){\vert\!\!\sim}{\neg\phi_{A}} \quad \hbox{and} \quad {\phi_{A}}\vee(\phi_{B}\vee\phi_{C}) \; {\vert\!\!\!\not\!\!\sim} {\neg(\phi_{B}\vee \phi_{C})}, $$

    as well as

    $$ {\phi_{B}}\vee(\phi_{A}\vee\phi_{C}){\vert\!\!\sim}{\neg\phi_{B}} \quad \hbox{and} \quad {\phi_{B}\vee(\phi_{A}\vee\phi_{C})} \; {\vert\!\!\!\not\!\!\sim}{\neg(\phi_{A}\vee\phi_{C})}. $$

    This, too, follows straightforwardly from the supposition by repeated applications of LLE and RW.

  1. (iv)

    (Choice hard) Let A <BP B ∪ C and B <BP A ∪ C. We want to show that A ∪ B <BP C. Using (From \({\vert\!\!\sim}\) to <BP), the supposition gives us

    $$ {\phi_{A}}\vee(\phi_{B}\vee\phi_{C}){\vert\!\!\sim} {\neg\phi_{A}} \quad \hbox{and} \quad {\phi_{A}}\vee(\phi_{B}\vee\phi_{C}) \; {\vert\!\!\!\not\!\!\sim}{\neg(\phi_{B}\vee\phi_{C})}, $$

    as well as

    $$ {\phi_{B}}\vee(\phi_{A}\vee\phi_{C}){\vert\!\!\sim}{\neg\phi_{B}} \quad \hbox{and} \quad {\phi_{B}}\vee(\phi_{A}\vee\phi_{C}) \; {\vert\!\!\!\not\!\!\sim}{\neg(\phi_{A}\vee\phi_{C})}. $$

    We need to show that

    $$ (\phi_{A}\vee\phi_{B})\vee\phi_{C}{\vert\!\!\sim}{\neg(\phi_{A}\vee\phi_{B})} \quad \hbox{and} \quad (\phi_{A}\vee\phi_{B}) \vee\phi_{C}{\vert\!\!\!\not\!\!\sim}{\neg\phi_{C}}. $$

    This follows straightforwardly from the supposition by repeated applications of LLE, AND and RW.

  2. (v)

    (Continuing up) Let A <BP B. We want to show that A <BP B ∪ C. Using (From \( {\vert\!\!\sim} \) to <BP), the supposition gives us

    $$ {\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{\neg\phi_{A}} \quad \hbox{and} \quad \phi_{A}\vee\phi_{B} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{B}}. $$

    We need to show that

    $$ {\phi_{A}\vee\phi_{B}\vee\phi_{C}}{\vert\!\!\sim}{\neg\phi_{A}} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}\vee\phi_{C}} \; {\vert\!\!\!\not\!\!\sim} {\neg(\phi_{B}\vee\phi_{C})}. $$

    By REF and RW, we have that \({\neg\phi_{A}\wedge\phi_{C}}{\vert\!\!\sim}{\neg\phi_{A}}\). So by OR, we get

    $$ (\phi_{A}\vee\phi_{B})\vee(\neg\phi_{A}\wedge\phi_{C}){\vert\!\!\sim} {\neg\phi_{A}}, $$

    which simplifies to

    $$ {\phi_{A}\vee\phi_{B}}\vee\phi_{C}{\vert\!\!\sim}{\neg\phi_{A}}. $$

    Now suppose for reductio that also

    $$ {\phi_{A}\vee\phi_{B}}\vee\phi_{C}{\vert\!\!\sim}{\neg(\phi_{B}\vee\phi_{C})}. $$

    Then by AND and RW, \(\phi_{A}\vee\phi_{B}\vee\phi_{C}{\vert\!\!\sim}{{\bot}}. \) But then by \(\bot\)CM, which is weaker than CP, and LLE, \({\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{\bot}\) and by RW \( {\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{\neg\phi_{B}},\) and we have a contradiction.

  3. (vi)

    (Continuing down) Let A ∪ C <BP B. We want to show that A <BP B. Using (From \({\vert\!\!\sim} \) to <BP), the supposition gives us

    $$ {\phi_{A}\vee\phi_{B}\vee\phi_{C}} {\vert\!\!\sim}{\neg(\phi_{A}\vee\phi_{C})} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}\vee\phi_{C}} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{B}}. $$

    We need to show that

    $$ {\phi_{A}}\vee\phi_{B}{\vert\!\!\sim}\neg\phi_{A} \quad \hbox{and} \quad {\phi_{A}}\vee\phi_{B} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{B}}. $$

    From the supposition, we get by two different applications of RW that both

    $$ {\phi_{A}\vee\phi_{B}\vee\phi_{C}}{\vert\!\!\sim}{\neg\phi_{A}} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}\vee\phi_{C}}{\vert\!\!\sim} {\phi_{A}}\vee\phi_{B}\vee \neg\phi_{C}. $$

    So by CM and LLE, \( {\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{\neg\phi_{A}}. \) Now suppose for reductio that also \({\phi_{A}}\vee\phi_{B}{\vert\!\!\sim} {\neg\phi_{B}}\). Then by REF, AND and RW, \({\phi_{A}\vee\phi_{B}}{\vert\!\!\sim}{\bot}.\) By LLE and \(\bot\)Cond, which is weaker than CP, we get

    $$ (\phi_{A}\vee\phi_{B}\vee\phi_{C})\wedge (\phi_{A}\vee\phi_{B}){\vert\!\!\sim}{\bot} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}\vee\phi_{C}}{\vert\!\!\sim} {\neg(\phi_{A}\vee\phi_{B})}. $$

    So finally, by RW, \({\phi_{A}\vee\phi_{B}\vee\phi_{C}} {\vert\!\!\sim}{\neg\phi_{B}},\) and we have a contradiction. □

Proof of Lemma 4

  1. (i)

    (LLE) Let \(\alpha\vdash\beta,\,\beta\vdash\alpha\) and \( {{\alpha}} \; {\vert\!\!\sim}{{\gamma}}, \) that is, \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!] \) or \( [\![{\alpha}]\!] = \emptyset\). For \( {{\beta}}{\vert\!\!\sim}{{\gamma}}\) we have to show that \( [\![{\beta\wedge\neg\gamma}]\!] < [\![{\beta\wedge\gamma}]\!] \) or \( [\![{\beta}]\!] = \emptyset. \) But this is immediate since \(\alpha\vdash\beta\) and \(\beta\vdash\alpha\) imply that \( [\![{\beta}]\!] = [\![{\alpha}]\!] ,\,[\![{\beta\wedge\gamma}]\!] = [\![{\alpha\wedge\gamma}]\!] \) and \( [\![{\beta\wedge\neg\gamma}]\!] = [\![{\alpha\wedge\neg\gamma}]\!]\).

  2. (ii)

    (REF) For \( {{\alpha}} \; {\vert\!\!\sim}{{\alpha}}\) we need to show that \( [\![{\alpha\wedge\neg\alpha}]\!] < [\![{\alpha\wedge\alpha}]\!] \) or \( [\![{\alpha}]\!] = \emptyset. \) But the former reduces to \(\emptyset < [\![{\alpha}]\!]\). This is fulfilled if \( [\![{\alpha}]\!] \not= \emptyset, \) by Minimality.

  3. (iii)

    (CP) Let \( {{\alpha}} \; {\vert\!\!\sim}{{\bot}}\). Then \( [\![{\alpha\wedge\neg\bot}]\!] < [\![{\alpha\wedge\bot}]\!] \) or \( [\![{\alpha}]\!] = \emptyset. \) But the former means that \( [\![{\alpha}]\!] < \emptyset, \) contrary to BP-i which we derived from Irreflexivity and Choice easy. So \( [\![{\alpha}]\!] = \emptyset\) which means that \(\alpha \vdash \bot, \) as desired.

  4. (iv)

    (RW) Let \( {{\alpha}} \; {\vert\!\!\sim}{{\beta}}\) and \(\beta\vdash\gamma\). Then \( [\![{\alpha\wedge\neg\beta}]\!] < [\![{\alpha\wedge\beta}]\!] \) or \( [\![{\alpha}]\!]=\emptyset. \) If the latter, we have \( {{\alpha}} \; {\vert\!\!\sim}{{\gamma}}\) by definition. So assume the former. We need to show that \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!]\). Since \( [\![{\alpha\wedge\neg\beta}]\!] = [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] \cup [\![{\alpha\wedge\neg\gamma}]\!] , \) we have \( [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] \cup [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\beta}]\!]\). So by Choice easy, \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] \cup [\![{\alpha\wedge\beta}]\!]\). Equivalently, \([\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!]\), and this is what we wanted to show.

  5. (v)

    (AND) Let \( {{\alpha}} \; {\vert\!\!\sim}{{\beta}}\) and \( {{\alpha}}{\vert\!\!\sim} \; {{\gamma}}\). Then \( [\![{\alpha\wedge\neg\beta}]\!] < [\![{\alpha\wedge\beta}]\!] \) and \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!] , \) or \( [\![{\alpha}]\!] = \emptyset. \) If the latter, we have \( {{\alpha}} \; {\vert\!\!\sim}{{\beta\wedge\gamma}}\) by definition. So suppose the former. We have to show that \( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] < [\![{\alpha\wedge\beta\wedge\gamma}]\!] . \) By Choice easy, \( [\![{\alpha\wedge\neg\beta}]\!] < [\![{\alpha}]\!] \) and \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha}]\!] . \) So by BP-vi, \( [\![{\alpha\wedge\neg\beta}]\!] \cup [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha}]\!] , \) or equivalently, \( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] < [\![{\alpha}]\!] . \) By BP-iv, \( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] < -( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] ) \cap [\![{\alpha}]\!] , \) or equivalently, \( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] < [\![{\alpha\wedge\beta\wedge\gamma}]\!] , \) which is what we wanted to show. (Note that BP-iv and BP-vi follow already from Choice easy and Choice hard.)

  6. (vi)

    (OR) Let \( {{\alpha}} \; {\vert\!\!\sim}{{\gamma}}\) and \( {{\beta}} \; {\vert\!\!\sim}{{\gamma}}\). Then \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!] \) or \( [\![{\alpha}]\!] = \emptyset, \) and \( [\![{\beta\wedge\neg\gamma}]\!] < [\![{\beta\wedge\gamma}]\!] \) or \( [\![{\beta}]\!] = \emptyset. \) If \( [\![{\alpha}]\!] = \emptyset\) or \( [\![{\beta}]\!] = \emptyset, \) then β or α is logically equivalent to α ∨ β, and we get \( {{\alpha\vee\beta}} \; {\vert\!\!\sim}{{\gamma}}\) by LLE which we have already proved. So suppose that \( [\![{\alpha}]\!] \not= \emptyset\) and \( [\![{\beta}]\!] \not= \emptyset. \) We have to show that \( [\![{(\alpha\vee\beta)\wedge\neg\gamma)}]\!] < [\![{(\alpha\vee\beta)\wedge\gamma}]\!] . \) By Continuing up, we get \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\beta\wedge\neg\gamma}]\!] \cup [\![{\alpha\wedge\gamma}]\!] \cup [\![{\beta\wedge\gamma}]\!] \) and \( [\![{\beta\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\neg\gamma}]\!] \cup [\![{\alpha\wedge\gamma}]\!] \cup [\![{\beta\wedge\gamma}]\!] . \) So by Choice hard, \( [\![{\alpha\wedge\neg\gamma}]\!] \cup [\![{\beta\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!] \cup [\![{\beta\wedge\gamma}]\!] , \) and this is equivalent to \( [\![{(\alpha\vee\beta)\wedge\neg\gamma)}]\!] < [\![{(\alpha\vee\beta)\wedge\gamma}]\!] , \) which is what we wanted to show.

  7. (vii)

    (CM) Let \( {{\alpha}} \; {\vert\!\!\sim}{{\beta}}\) and \( {{\alpha}} \; {\vert\!\!\sim}{{\gamma}}\). Then \( [\![{\alpha\wedge\neg\beta}]\!] < [\![{\alpha\wedge\beta}]\!] \) and \( [\![{\alpha\wedge\neg\gamma}]\!] < [\![{\alpha\wedge\gamma}]\!] , \) or \( [\![{\alpha}]\!] = \emptyset. \) If the latter, we have \( [\![{\alpha\wedge\gamma}]\!] = \emptyset, \) and we get \( {{\alpha\wedge\gamma}}{\vert\!\!\sim}{{\beta}}\) by definition. So suppose the former. We have to show that \( [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] < [\![{\alpha\wedge\beta\wedge\gamma}]\!] . \) We first copy the proof of AND and derive \( [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] < [\![{\alpha\wedge\beta\wedge\gamma}]\!] . \) Now we note that \( [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] \subseteq [\![{\alpha\wedge\neg(\beta\wedge\gamma)}]\!] , \) apply Continuing down and get \( [\![{\alpha\wedge\neg\beta\wedge\gamma}]\!] < [\![{\alpha\wedge\beta\wedge\gamma}]\!] , \) as desired. □

Proof of Observation 1

  1. (a)

    Let \( {{\alpha}} \; {\vert\!\!\sim}'{{\beta}}, \) where \({ {\vert\!\!\sim} ' = {\mathcal{I}}_{\mathrm{FH}}({\mathcal{P}}_{\mathrm{FH}}( {\vert\!\!\sim} ))}\). Then by (From <FH to \( {\vert\!\!\sim} \)),

    $$ [\![{\alpha\wedge\neg\beta}]\!] <_{\rm FH} \; [\![{\alpha\wedge\beta}]\!] \quad \hbox{or}\,\emptyset \nless_{\rm FH} [\![{\alpha}]\!]. $$

    That is, by (From \( {\vert\!\!\sim} \) to <FH)

    $$ \begin{aligned} &{(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)} {\vert\!\!\sim}{\alpha\wedge\beta} \quad \hbox{and} \quad {(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)} \; {\vert\!\!\!\not\!\!\sim}{\alpha\wedge\neg\beta},\\ &\quad \hbox{or} \, {\alpha} \; {\vert\!\!\!\not\!\!\sim}{\alpha},\,\hbox{or}\, {{\alpha}} \; {\vert\!\!\sim}{{\bot}} \end{aligned} $$

    That is, by LLE and REF,

    $$ {\alpha} \; {\vert\!\!\sim}{\alpha\wedge\beta}\quad \hbox{and} \quad {\alpha} \; {\vert\!\!\!\not\!\!\sim}{\alpha\wedge\neg\beta},\, \hbox{or}\,{\alpha} \; {\vert\!\!\sim}{\bot} $$

    By the reflexivity condition REF, AND and RW, this means

    $$ {{\alpha}} \; {\vert\!\!\sim}{{\beta}} \quad \hbox{and} \quad {\alpha} \; {\vert\!\!\!\not\!\!\sim}{\neg\beta},\, \hbox{or}\,{\alpha}{\vert\!\!\sim}{\bot} $$

    Finally, by AND and RW, this reduces to

    $$ {\alpha} \; {\vert\!\!\sim}{\beta} \,\hbox{or}\,{\alpha} \; {\vert\!\!\sim}{\bot} $$

    and finally, by RW again, to \({\alpha} \; {\vert\!\!\sim}{\beta}\). But this means that \({\vert\!\!\sim}^{\prime}\) is identical to \( {\vert\!\!\sim}\).

  2. (b)

    Let \( {{\alpha}} \; {\vert\!\!\sim}^{\prime}{\beta}\) with \({ {\vert\!\!\sim}^{\prime}= {\mathcal{I}}({\mathcal{P}}_{\mathrm{FH}}({\vert\!\!\sim})). }\) Then by (From < to \( {\vert\!\!\sim}),\,\)

    $$ [\![{\alpha\wedge\neg\beta}]\!]<_{\rm FH} [\![{\alpha\wedge\beta}]\!] \, \hbox{or} \, [\![{\alpha}]\!] = \emptyset. $$

    That is, by (From \( {\vert\!\!\sim} \) to <FH)

    $$ {{(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)}} \; {\vert\!\!\sim}{{\alpha\wedge\beta}} \,\hbox{and}\, {{(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)}} \; {{\vert\!\!\!\not\!\!\sim}}{{\alpha\wedge\neg\beta}}, \,\hbox{or}\,\alpha\vdash \bot $$

    That is, by LLE, REF, AND and RW,

    $$ {{\alpha}} \; {\vert\!\!\sim}{{\beta}} \quad \hbox{and} \quad {{\alpha}} \; {{\vert\!\!\!\not\!\!\sim}}{{\neg\beta}}, \, \hbox{or}\, \alpha\vdash \bot $$

    By AND, RW and CP, this reduces to

    $$ {{\alpha}} \; {\vert\!\!\sim}{{\beta}} \quad {\text{or}} \quad \alpha\vdash \bot $$

    and finally, by REF and RW again, to \( {{\alpha}} \; {\vert\!\!\sim}{{\beta}}. \) But this means that \( {\vert\!\!\sim}^{\prime}\) is identical to \( {\vert\!\!\sim}\).□

Proof of Observation 2

Let \( {{\alpha}} \; {\vert\!\!\sim}^{\prime}{{\beta}}\). Then by (From < to \( {\vert\!\!\sim} \)),

$$ [\![{\alpha\wedge\neg\beta}]\!] <_{\rm BP} [\![{\alpha\wedge\beta}]\!], \quad \hbox{or} \quad [\![{\alpha}]\!] = \emptyset $$

Thus, by (From \( {\vert\!\!\sim} \) to <BP),

$$ \begin{aligned} &{{(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)}}{\vert\!\!\sim} {{\neg(\alpha\wedge\neg\beta)}} \quad \hbox{and} \quad {{(\alpha\wedge\beta)\vee(\alpha\wedge\neg\beta)}} \; {{\vert\!\!\!\not\!\!\sim}}{{\neg(\alpha\wedge\beta)}},\\ &\quad\quad \hbox{or}\, \alpha\vdash \bot \end{aligned} $$

that is, by LLE,

$$ {{\alpha}} \; {\vert\!\!\sim}{{\neg\alpha\vee\beta}} \quad \hbox{and} \quad {{\alpha}} \; {{\vert\!\!\!\not\!\!\sim}}{{\neg\alpha\vee\neg\beta}}, \,\hbox{or}\,\alpha\vdash \bot $$

By REF, AND and RW, this is equivalent to

$$ {{\alpha}} \; {\vert\!\!\sim}{{\beta}} \quad \hbox{and} \quad {{\alpha}}\;{{\vert\!\!\!\not\!\!\sim}}{{\neg\beta}},\,\hbox{or}\,\alpha\vdash \bot. $$

By REF, AND, RW and CP, this is equivalent to \( {{\alpha}} \; {\vert\!\!\sim}{{\beta}}. \) But this means that \( {\vert\!\!\sim}^{\prime}\) is identical to \( {\vert\!\!\sim}\).

Proof of Observation 3

Let \(A <_{\rm BP}^{\prime} B. \) Then by (From \({\vert\!\!\sim} \) to <BP),

$$ {\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{{\neg\phi_{A}}} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}} \; {\vert\!\!\!\not\!\!\sim}{\neg\phi_{B}} $$

Thus, by (From <BP to \( {\vert\!\!\sim} \)),

$$ \begin{aligned} &(A\cup B)\cap-A <_{\rm BP}(A\cup B)\cap-A\,\hbox{or}\,A\cup B = \emptyset,\\ &\quad\quad \hbox{and}\,(A\cup B)\cap-B \nless_{\rm BP} (A\cup B)\cap-B \quad \hbox{and} \quad A\cup B \not= \emptyset \end{aligned} $$

This condition reduces to

$$ A <_{\rm BP} -A\cap B \quad \hbox{and} \quad B \nless_{\rm BP}\,\,A\cap-B \quad \hbox{and} \quad A\cup B \not= \emptyset $$

which by BP-iv and Asymmetry is equivalent to

$$ A <_{\rm BP} B, \quad \hbox{and} \quad A\not= \emptyset\,\hbox{or}\,B \not= \emptyset $$

and by Irreflexivity this reduces to A <BP B.□

Proof of Observation 4

Let \(A <_{\rm FH}^{\prime} B. \) Then by (From \( {\vert\!\!\sim} \) to <FH),

$$ {\phi_{A}\vee\phi_{B}{\vert\!\!\sim}}{\phi_{B}} \quad \hbox{and} \quad {\phi_{A}\vee\phi_{B}}{\vert\!\!\!\not\!\!\sim}{\phi_{A}} $$

Thus, by (From <FH to \( {\vert\!\!\sim} \)),

$$ \begin{aligned} &(A\cup B)\cap-B <_{\rm FH} (A\cup B)\cap B \quad {\text{or}} \quad \emptyset \nless_{\rm FH} A\cup B,\,\hbox{and}\\ &\quad\quad (A\cup B)\cap-A \nless_{\rm FH} (A\cup B)\cap A \quad \hbox{and} \quad \emptyset <_{\rm FH} A\cup B \end{aligned} $$

This condition reduces to

$$ (+)\quad A\cap-B >_{\text {FH}} B \quad \hbox{and} \quad -A\cap B \nless_{\rm FH} A \quad \hbox{and} \quad \emptyset >_{\rm FH} A\cup B $$

This is not reducible to

$$ A <_{\rm FH} \; B. $$

Both directions fail, as we will show with the help of two counterexamples. For the rest of this proof, let the set of possible worlds be W = {u, v, w}.

Example 3

of Sect. 4.4.2 shows that (+) does not imply A <FH B. Remember that in this example, < is a transitive and virtually connected relation over the subsets of W such that

$$ \emptyset < \{v\} \sim \{w\} \sim \{v,w\} < \{u\} \sim \{u,v\} \sim \{u,w\} < \{u,v,w\} $$

where A∼ B iff neither A < B nor B < A. We verified before that Qualitativeness is satisfied and that we indeed have an FH plausibility relation <FH =  < . But now consider A = {u, v} and B = {u, w}. We have A ∩ −B <FH B and \(-A\cap B \nless_{\rm FH} \; A\) and \( \emptyset \) <FH A ∪ B, so (+) is satisfied, but not A <FH B.

Example 5

is described in the main text after the statement of Observation 4. The relevant relation is represented by the string

$$ \emptyset < \{v\} \sim \{w\} \sim \{v,w\} < \{u\} \sim \{u,v\} < \{u,w\} \sim \{u,v,w\} $$

This relation shows that A <FH B does not imply (+). It is easy to see that < is an FH plausibility relation. The only non-trivial comparison we have to check for Qualitativeness concerns the three disjoint sets {u},  {v} and {w}, and we can again verify that

$$ \{w\}<\{u\}\cup\{v\} \,\, \hbox{and}\,\{v\}<\{u\}\cup\{w\} \quad \hbox{taken together imply} \quad \{v\}\cup\{w\}<\{u\} $$

Thus Qualitativeness is satisfied here, we indeed have an FH plausibility relation <FH =  < . Consider again A = {u, v} and B = {u, w}. Clearly, we have A <FH B, but not \(-A\cap B \nless_{\rm FH} A, \) violating (+).□

Proof of Lemma 6

  1. (i)

    Let <FH be an FH plausibility relation, \({<_{\rm BP} ={\mathcal{P}}_{\mathrm{B}}(<_{\rm FH} )}\) and A <BP B. By definition, this means that A <FH − A ∩ B. So by (FH-Dominance), A <FH B. If A and B are disjoint propositions, then −A ∩ B is identical to B, so \({<_{\rm BP} ={\mathcal{P}}_{\mathrm{B}}(<_{\rm FH})}\) agrees by definition with <FH on disjoint propositions.

  2. (ii)

    Let <BP be a basic plausibility relation, \({<_{\rm FH} ={\mathcal{P}}_{\mathrm{FH}}(<_{\rm BP})}\) and A <BP B. We want to show that A <FH B, that is, by definition, A ∩ −B <BP B and \(-A\cap B \nless_{\rm BP} \; A\). But the former follows from A <BP B by Weak continuing down, BP-vii. From A <BP B, we also get A <BP − A ∩ B by BP-iv. So \(-A\cap B \nless_{\rm BP} \; A\), by Asymmetry, BP-ii, which is what we needed.

    If A and B are disjoint propositions, then A ∩ −B is identical to A, and −A ∩ B is identical to B. So the clause for the definition of A <FH B with \({<_{\rm FH} ={\mathcal{P}}_{_{\mathrm{FH}}}(<_{\rm BP} )}\) reduces to A <BP B and \(B \nless_{\rm BP} \; A. \) Since <BP satisfies Asymmetry, this reduces to A <BP B. Thus \({<_{\rm FH} ={\mathcal{P}}_{_{\mathrm{FH}}}(<_{\rm BP} )}\) agrees with <BP on disjoint propositions.□

Proof of Observation 5

(FH0a) First we need to show Irreflexivity. Suppose for reductio that both A <FH B and B <FH A, that is

  1. (i)

    A ∩ −B <BP B and \(-A \cap B \nless_{\rm BP} \; A\) and

  2. (ii)

    B ∩ −A <BP A and \(-B \cap A \nless_{\rm BP} \; B\)

This is clearly contradictory.

(FH0b) Next we show (Transitivity). Let A <FH B and B <FH C, that is

  1. (i)

    A ∩ −B <BP B and \(-A \cap B \nless_{\rm BP} \; A\) and

  2. (ii)

    B ∩ −C <BP C and \(-B \cap C \nless_{\rm BP} \; B\)

In order to show that A <FH C, we need to show that

  1. (iii)

    A ∩ −C <BP C and \(-A \cap C \nless_{\rm BP} \; A\).

Let us first prove the first conjunct of (iii). From (i) and (ii) we get, using BP-xi,

$$ (A \cap -B) \cup (B \cap -C) <_{\rm BP} B\cup C $$

By Meet right, BP-iv, we get

$$ (A \cap -B) \cup (B \cap -C) <_{\rm BP} -((A \cap -B) \cup (B \cap -C)) \cap (B\cup C) $$

or equivalently,

$$ (A \cap -B) \cup (B \cap -C) <_{\rm BP} (-A \cup B) \cap (-B \cup C) \cap (B\cup C) $$

The right-hand side is a subset of C, so we get, using Continuing up,

$$ (A \cap -B) \cup (B \cap -C) <_{\rm BP} C $$

The left-hand side is a superset of A ∩ −C, so by Continuing down,

$$ A\cap -C <_{\rm BP} C $$

which is the first conjunct of (iii). Let us now turn to the second conjunct of (iii). Suppose for reductio that also −A ∩ C <BP A. Taken together with \( A\cap -C <_{\rm BP} C, \) which we have just established, this gives us, with the help of BP-xi,

$$ (A\cap -C) \cup (-A \cap C) <_{\rm BP} \; A \cup C $$

Now we use Meet right, BP-iv, and get

$$ (A\cap -C) \cup (-A \cap C) <_{\rm BP} -((A\cap -C) \cup (-A \cap C)) \cap (A \cup C) $$

which is equivalent to

$$ (A\cap -C) \cup (-A \cap C) <_{\rm BP} (-A\cup C) \cap (A \cup -C) \cap (A \cup C) $$

or simply

$$ (A\cap -C) \cup (-A \cap C) <_{\rm BP} \; A \cap C $$

Now we join this with A ∩ −B <BP B from (i), using BP-xi and get

$$ (A \cap -B) \cup (A\cap -C) \cup (-A \cap C) <_{\rm BP} \; B \cup (A \cap C) $$

Applying Meet right, BP-iv, once more, we get

$$ (A \cap -B) \cup (A\cap -C) \cup (-A \cap C) <_{\rm BP} -((A \cap -B) \cup (A\cap -C) \cup (-A \cap C)) \cap \quad\quad (B \cup (A \cap C)) $$

which is equivalent to

$$ (A \cap -B) \cup (A\cap -C) \cup (-A \cap C) <_{\rm BP} ((-A \cup B) \cap (-A\cup C) \cap (A \cup -C)) \cap \quad\quad (B \cup (A \cap C)) $$

The right-hand side of this inequality is a subset of (− A ∪ B) ∩ (A ∪ B) and thus of B, so by Continuing up, we get

$$ (A \cap -B) \cup (A\cap -C) \cup (-A \cap C) <_{\rm BP} \; B $$

The left-hand side of the inequality includes −B ∩ C, so by Continuing down, we infer −B ∩ C <BP B But this violates (ii). So we have found a contradiction and proved the second conjunct of (iii). Therefore \( <_{\rm FH} = {{\mathcal{P}}}_{\mathrm{FH}}(<_{\rm BP} ) \) is transitive.

  • (FH1) Now we verify that \({<_{\rm FH} = {\mathcal{P}}_{\mathrm{FH}}(<_{\rm BP} )}\) satisfies FH-Dominance. So suppose that \(A\subseteq B. \)

First, we show that if C <FH A, then C <FH B. That is, if C ∩ −A <BP A and \(-C \cap A \nless_{\rm BP} \; C, \) then C ∩ −B <BP B and \(-C \cap B \nless_{\rm BP} \; C. \) But C ∩ −A <BP A entails C ∩ −B <BP B due to (Continuing up) and Continuing down, since \(C\cap -B \subseteq C\cap -A. \) Similarly, \(-C \cap A \nless_{\rm BP} \; C\) and Continuing down entail that \(-C \cap B \nless_{\rm BP} \; C. \)

Second, we show that if B <FH C, then A <FH C. That is, if B ∩ −C <BP C and \(-B \cap C \nless_{\rm BP} \; B, \) then A ∩ −C <BP C and \(-A \cap C \nless_{\rm BP} \; A. \) But again, B ∩ −C <BP C implies A ∩ −C <BP C due to Continuing down, and \(-B \cap C \nless_{\rm BP} \; B\) implies \(-A \cap C \nless_{\rm BP} \; A\) due to Continuing up and Continuing down.

(FH2) Now we verify that \({<_{\rm FH} = {\mathcal{P}}_{_{\mathrm{FH}}}(<_{\rm BP} )}\) satisfies Qualitativeness. Since <BP satisfies the unrestricted condition Choice hard and <FH agrees with <BP over disjoint sets, by Lemma 6(d), it follows immediately that <FH satisfy Choice over disjoint sets, and that just means that it satisfies Qualitativeness.

(FH3) Lastly, we need to show that \( \emptyset \) <FH A ∪ B implies that either \( \emptyset \) <FH A or \( \emptyset \) <FH B. Since <FH agrees with <BP over disjoint sets, by Lemma 6(d), it is sufficient to show that \( \emptyset \) <BP A ∪ B implies that either \( \emptyset \) <BP A or \( \emptyset \) <BP B. But the former implies that \(A\cup B\not=\emptyset, \) by Irreflexivity. Hence either \(A\not=\emptyset\) or \(B\not=\emptyset. \) So by Minimality, either \( \emptyset \) <BP A or \( \emptyset \) <BP B. □

Proof of Observation 6

Let <FH be a qualitative FH relation and \({<_{\rm BP} = {\mathcal{P}}_{\mathrm{B}}(<_{\rm FH} )}\).

  • (Irreflexivity) Suppose A <BP A, that is, by definition, A <FH − A ∩ A. This means that A <FH \( \emptyset \), contradicting FH1 and Irreflexivity for <FH.

  • (Minimality) Suppose that \(A\not=\emptyset. \) We need to show that \( \emptyset \) <BP A, that is, by definition \( \emptyset \) <FH − \( \emptyset \) ∩ A, which is \( \emptyset \) <FH A. This is guaranteed if, and only if, <FH satisfies Minimality.

  • (Continuing up) Let A <BP B. By definition, this means that A <FH − A ∩ B. By FH1, this implies that A <FH − A ∩ (B ∪ C), which means, by definition again, that A <BP B ∪ C.

  • (Continuing down) Let A ∪ C <BP B. By definition, this means that A ∪ C <FH − (A ∪ C) ∩ B. By FH1, this implies that A <FH − A ∩ B, which means, by definition again, that A <BP B.

Choice easy follows from Continuing up and Continuing down.

  • (Choice hard). Suppose that A <BP B ∪ C and B <BP A ∪ C. We need to show that A ∪ B <BP C. By the definition of \({{\mathcal{P}}_{_{\mathrm{B}}}(<_{\rm FH} ), }\) our supposition means that

  1. (i)

    A <FH − A ∩ (B ∪ C) and

  2. (ii)

    B <FH − B ∩ (A ∪ C).

Condition (i) can be equivalently expressed as

  • (\(\hbox{i}^{\prime}\)) A <FH (−A ∩ B) ∪ (− A ∩ −B ∩ C)

Consider also the condition

  • (\(\hbox{ii}^{\prime}\)) −A ∩ B <FH A ∪ (−A ∩ −B ∩ C)

This condition is entailed by (ii), since −A ∩ B is a subset of B and −B ∩ (A ∪ C) is a subset of A ∪ (−A ∩ −B ∩ C), and <FH satisfies FH-Dominance. Clearly, the sets A, −A ∩ B and −A ∩ −B ∩ C are pairwise disjoint. Now we can apply the Qualitativeness of <FH and conclude from (\(\hbox{i}^{\prime}\)) and (\(\hbox{ii}^{\prime}\)) that

  • (\(\hbox{iii}^{\prime}\)) A ∪ (−A ∩ B) <FH − A ∩ −B ∩ C

But A ∪ (−A ∩ B) is identical to A ∪ B, and −A ∩ −B ∩ C is identical to −(A ∪ B) ∩ C, so (\(\hbox{iii}^{\prime}\)) is equivalent to

  1. (iii)

    A ∪ B <FH − (A ∪ B) ∩ C

By the definition of \({{\mathcal{P}}_{_{\mathrm{B}}}(<_{\rm FH} ), }\) this just means that A ∪ B <BP C, which is what we needed to show.□

Proof of Observation 7

(i) Assume that \(A<_{\rm BP}^{\prime} \; B\). This means, by (From <FH to <BP),

$$ A <_{\rm FH} -A\cap \; B $$

By Lemma 6(d), this is equivalent to

$$ A <_{\rm BP} -A\cap \; B $$

which by BP-iv just means A <BP B as desired. (ii) Assume that \({<_{\rm FH}^{\prime} = {\mathcal{P}}_{\mathrm{FH}}({\mathcal{P}}_{\mathrm{B}}(<_{\rm FH} )), }\) and assume that

$$ A<_{\rm FH}^{\prime} \; B. $$

This means, by (From <BP to <FH),

$$ A\cap-B <_{\rm BP} B \quad \hbox{and}\,-A\cap B \nless_{\rm BP} \; A, $$

By Lemma 6(c), this is equivalent to

$$ A\cap-B <_{\rm FH} B \quad \hbox{and}\,-A\cap B \nless_{\rm FH} \; A. $$

By FH1, the first conjunct of this is implied by A <FH B, and the second conjunct implies \(B\nless_{\rm FH} \; A. \) But this is all we can say. As counterexamples against both directions we can use Example 3 and Example 5, exactly in the same way as they are used in the proof of Observation 4. Thus there is no reduction to the target sentence A <FH B.□

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Rott, H. Two Concepts of Plausibility in Default Reasoning. Erkenn 79 (Suppl 6), 1219–1252 (2014). https://doi.org/10.1007/s10670-013-9548-y

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