Abstract
To remedy the lack of precision attached to the concept of coherence, a plethora of probabilistic measures has been developed. To broaden the perspective, we do not focus on the differences between these quantitative but the differences between qualitative approaches to coherence by comparing three probabilistic definitions for the relation denoted by ‘coheres with’. To reveal the different logics underlying these relations, we introduce a considerable number of formal properties and examine whether the given coherence relations possess them. Among these properties are not only the classics reflexivity, symmetry and transitivity, but also a variety of features that concern propositions containing negation, conjunction and disjunction, as well as features that concern inconsistency, entailment and equivalence.
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Notes
Such a relation must not be confused with the binary coherence relation that is in the focus of Bovens and Hartmann’s (2003, chs. 1.4 and 2.2) account. Bovens and Hartmann are interested in the relation ‘is no less coherent than’, while we are interested in ‘coheres with’. Even if the latter relation’s scope is extended to sets of propositions, these relations remain independent of each other. For a set that is no less coherent than another set need not cohere with it, and a set that coheres with another set need not be no less coherent than it.
Cf. Achinstein’s (2001, chs. 3f.) discussion of these three approaches to confirmation.
In the last resort, it is possible to add the constraint that coherence relations obtain between x and y only if x ≠ y. This would merely entail that reflexivity is no longer among the possible properties of these relations. Note also that, to model cases of identical contents in a settheoretical manner, one may draw on multisets because they can contain the same proposition more than one time.
As shown by Schippers (2014), there are nonetheless screeningoff conditions that guarantee transitivity for each of the above relations.
The former constraint is sometimes labelled “cautious monotonicity’ (cf. Huber 2007).
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This study was supported by Deutsche Forschungsgemeinschaft (Grant Number SI 1731/11).
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Appendix
Appendix
 (REF):

\( P(xx) \) = 1, and \( P(x{\bar{x}}) \) = \( P({\bar{x}}x) \) = 0. Hence, the definientia of (IC), (AC) and therefore also (SC) are satisfied.
 (SYM):

(AC), (IC) and thus also (SC) require bidirectional confirmation in random order.
 (TRA):

This is a straightforward consequence of the fact that the underlying accounts of confirmation are not transitive (cf. Douven 2011; Roche and Shogenji 2014; Shogenji 2003).
 (INC):

\( P(xx) \) = \( P({\bar{x}}{\bar{x}}) \) = 1, and \( P(x{\bar{x}}) \) = \( P({\bar{x}}x) \) = 0. Hence, since x and \( {\bar{x}} \) are contingent and the probability function P is regular, \( P(x{\bar{x}}) \) < P(x) < \( P(xx) \) and \( P({\bar{x}}x) \) < P(\( {\bar{x}} \)) < \( P({\bar{x}}{\bar{x}}) \). The definientia of (IC) and (AC), and thus also (SC), are therefore violated.
 (INC′):

If x ⊥ y, then \( P(xy) \) = \( P(yx) \) = 0. Hence, \( P(xy) \) < P(x) and \( P(xy) \) < \( P(x\bar{y}) \), as well as \( P(yx) \) < P(y) and \( P(yx) \) < \( P(y{\bar{x}}) \). That is, the definientia of (IC) and (AC), and thus also (SC), are again violated.
 (NSC):

First, if x ~_{i} y, then \( P(xy) \) > P(x), which is for all contingent propositions equivalent to \( P(\bar{y}{\bar{x}}) \) > \( P(\bar{y}) \). Analogously, \( P(yx) \) > P(y) is equivalent to \( P({\bar{x}}\bar{y}) \) > P(\( {\bar{x}} \)). Thus, if x ~_{i} y, then also \( {\bar{x}} \) ~_{i} \( \bar{y} \), in order that ~_{i} satisfies (NSC). Secondly, probability distribution P_{1} in Table 6 shows that neither ~_{s} nor ~_{a} satisfies (NSC). According to this distribution, \( P_{1} (xy) \) ≈ .731 > .637 ≈ P_{1}(x) > .5, and \( P_{1} (yx) \) ≈ .812 > .707 ≈ P_{1}(y) > .5. Hence, x ~_{s} y and therefore x ~_{a} y. However, \( P_{1} (\bar{y}{\bar{x}}) \) ≈ .477 < .5 and hence neither \( {\bar{x}} \) ~_{a} \( \bar{y} \) nor \( {\bar{x}} \) ~_{s} \( \bar{y} \).
 (CON):

If x ~_{s} y, then \( P(yx) \) > P(y) and \( P(yx) \) > \( P(\bar{y}x) \). Hence, \( P(\bar{y}x) \) < \( P(\bar{y}) \) and \( P(\bar{y}x) \) < \( P(y{\bar{x}}) \)}, and therefore x ≁_{i} \( \bar{y} \) and x ≁_{a} \( \bar{y} \) and thus also x ≁_{s} \( \bar{y} \).
 (CON′):

Consider probability distribution P_{6} in Table 7. Here \( P_{6}(xy) \) = 1 > .5 = P_{6}(x) and \( P_{6}(yx) \) = .5 > .25 = P_{6}(y); hence, x ~_{i} y. Furthermore, \( P_{6}(zx) \) = .5 > .25 = P_{6}(z) and \( P_{6}(xz) \) = 1 > .5 = P_{6}(x), so that also x ~_{i} z. Nonetheless, y ⊥ z because P_{6}(y ∧ z) = 0, with the result that ~_{i} violates (CON′). On the other hand, if x ~_{a} y, then \( P(yx) \) > .5 and hence \( P(\bar{y}x) \) ≤ .5. Hence, since \( P(\bar{y}x) \) = P(\( \bar{y} \) ∧ \( zx \)) + P(\( \bar{y} \) ∧ \( \bar{z}x \)), P(\( \bar{y} \) ∧ \( zx \)) ≤ .5. Finally, since y ⊥ z, z implies \( \bar{y} \), in order that P(\( \bar{y} \) ∧ \( zx \)) = \( P(zx) \) ≤ .5. That is, x ≁_{a} z and x ≁_{s} z, entailing that both relations satisfy (CON′).
 (WAC):

As to incremental coherence, since P is regular and x is contingent, P(\( xx \) ∧ y) = 1 > P(x). Due to the symmetry of incremental confirmation, it also holds that P(x ∧ \( yx \)) > P(x ∧ y); hence, x ~_{i} x ∧ y. As to absolute coherence, since P is regular and x is contingent, P(\( xx \) ∧ y) = 1 > P(\( {\bar{x}}x \) ∧ y). Moreover, P(x ∧ \( yx \)) = \( P(yx) \) and P(\( \overline{x \wedge y} x \)) = P(\( {\bar{x}} \) ∨ \( \bar{y}x \)) = \( P(\bar{y}x) \). Since x ~_{a} y, we also get \( P(yx) \) > \( P(\bar{y}x) \), so that P(x ∧ \( yx \)) = \( P(yx) \) > \( P(\bar{y}x) \) = P(\( \overline{x \wedge y} x \)). Hence, x ~_{a} x ∧ y. Therefore, all three accounts satisfy (WAC).
 (WOC):

As to incremental coherence, since P is regular and x is contingent, P(x ∨ \( yx \)) = 1 > P(x). Due to the symmetry of incremental confirmation, we thus get x ~_{i} x ∨ y. As to absolute coherence, P(x ∨ \( yx \)) = 1 > 0 = P(\( \overline{x \vee y} x \)). Furthermore, since x ~_{a} y, \( P(xy) \) > \( P({\bar{x}}y) \), and therefore P(x ∧ y) > P(\( {\bar{x}} \) ∧ y). Hence, P(x) = P(x ∧ y) + P(x ∧ \( \bar{y} \)) > P(\( {\bar{x}} \) ∧ y) + P(x ∧ \( \bar{y} \)) > P(\( {\bar{x}} \) ∧ y). Consequently, P(\( xx \) ∨ y) = P(x)/P(x ∨ y) > P(\( {\bar{x}} \) ∧ y)/P(x ∨ y) = P(\( {\bar{x}}x \) ∨ y). This means that x ~_{a} x ∨ y and x ~_{s} x ∨ y, so that all three accounts satisfy (WOC).
 (WCC):

According to probability distribution P_{2} in Table 6, P_{2}(x ∧ \( yx \)) ≈ .516 > .5 > .188 ≈ P_{2}(x ∧ y), and P_{2}(\( xx \) ∧ y) = 1 > .5 > P_{2}(x). Hence, x ~_{i} x ∧ y and x ~_{a} x ∧ y, so that also x ~_{s} x ∧ y. However, P_{2}(\( xy \)) ≈ .353 < 0.363 ≈ P_{2}(x) < .5. Therefore, neither x ~_{i} y nor x ~_{a} y, entailing that x ~_{s} y does not hold, too.
 (WCD):

Consider again distribution P_{2}. P_{2}(\( xx \) ∨ y) ≈ .514 > .5 > .363 ≈ P_{2}(x), and P_{2}(x ∨ \( yx \)) = 1 > .707 ≈ P_{2}(x ∨ y) > .5. Thus, x ~_{i} x ∨ y and x ~_{a} x ∧ y, implying that also x ~_{s} x ∨ y But since P_{2}(\( xy \)) < P_{2}(x) < .5, x and y are assessed incoherent on all three accounts.
 (EOC):

Since x is contingent and P is regular, P(\( xx \) ∧ y) = 1 > P(x). Therefore, by the symmetry of incremental confirmation, P(x ∧ \( yx \)) > P(x ∧ y), in order that ~_{i} satisfies (EOC). On the other hand, as is shown by distribution P_{3} in Table 6, there are situations in which P_{3}(x ∧ \( yx \)) = P_{3}(x ∧ \( yy \)) ≈ .332 < P_{3}(x ∧ y) ≈ 0.147 < .5. Hence, neither x ~ x ∧ y nor y ~ x ∧ y for either ~_{a} or ~_{s.}
 (EOD):

Since x ∨ y is contingent and P is regular, P(x ∨ \( yx \)) = 1 > P(x ∨ y). Hence, by symmetry also P(\( xx \) ∨ y) > P(x), entailing that x ~_{i} x ∨ y. But distribution P_{4} in Table 6 proves possible that P(\( xx \) ∨ y) = P(\( yx \) ∨ y) = .5, so that neither x ~ x ∨ y nor y ~ x ∨ y for either ~_{a} or ~_{s.}
 (CSC):

Consider probability distribution P_{7} in Table 7. P_{7}(\( xy \)) ≈ .165 > .163 ≈ P_{7}(x), in order that by symmetry x ~_{i} y. However, P_{7}(x ∧ \( zy \) ∧ z) ≈ .003 < .006 ≈ P_{7}(x ∧ z), implying that x ∧ z ~_{i} y ∧ z does not hold. Furthermore, according to distribution P_{8} in the same table, P_{8}(\( xy \)) ≈ .613 > .5 > .419 ≈ P_{8}(x) and P_{8}(\( yx \)) = 1 > .684 ≈ P_{8}(y) > .5; therefore x ~_{s} y. But P_{8}(x ∧ \( zy \) ∧ z) ≈ .333 < .5, with the result that both ~_{a} and ~_{s} violate (CSC).
 (DWC):

To show that ~_{a} satisfies (DWC), assume that x ∨ z ≁_{a} y ∨ z because P(x ∨ \( zy \) ∨ z) = P(x ∧ y ∧ \( zy \) ∨ z) + P(x ∧ y ∧ \( \bar{z}y \) ∨ z) + P(x ∧ \( \bar{y} \) ∧ \( zy \) ∨ z) + P(\( {\bar{x}} \) ∧ y ∧ \( zy \) ∨ z) + P(\( {\bar{x}} \) ∧ \( \bar{y} \) ∧ \( zy \) ∨ z) ≤ .5. Since P(x ∧ \( \bar{y} \) ∧ \( \bar{z}y \) ∨ z) = P(\( {\bar{x}} \) ∧ \( \bar{y} \) ∧ \( \bar{z}y \) ∨ z) = 0, this implies that the remaining probability P(\( {\bar{x}} \) ∧ y ∧ \( \bar{z}y \) ∨ z) > .5. Hence, P(\( {\bar{x}} \) ∧ y ∧ \( \bar{z}y \)) ≥ P(\( {\bar{x}} \) ∧ y ∧ \( \bar{z}y \) ∨ z) > .5. But this entails that P(x ∧ y ∧ \( zy \)) + P(x ∧ y ∧ \( \bar{z}y \)) = \( P(xy) \) ≤ .5, so that x ≁_{a} y. Since an analogous argument applies if x ∨ z ≁_{a} y ∨ z because P(y ∨ \( zx \) ∨ z) ≤ .5, ~_{a} satisfies (DWC). Furthermore, distribution P_{7} in Table 7 proves that ~_{i} does not meet this constraint. Here, x ~_{i} y but P_{7}(x ∨ \( zy \) ∨ z) ≈ .730 < .732 ≈ P_{7}(x ∨ z), so that x ∨ z ≁_{i} y ∨ z. One can easily find similar distributions showing that ~_{s} also violates (DWC).
 (CMC):

We have seen that (CSC) is violated by all coherence relations, that is, it is possible that x ~ y but x ∧ z ≁ y ∧ z. Since reflexivity implies that z ~ z, it is also possible that x ~ y and z ~ z but x ∧ z ≁ y ∧ z. Hence, if z′ = z, then it is possible that x ~ y and z ~ z′ while x ∧ z ~ y ∧ z′. Hence, (CMC) is also violated by all coherence relations.
 (DMC):

By analogy with (CMC), the fact that ~_{s} and ~_{i} violate (DWC) entails that they also violate (DMC). To show that, although ~_{a} satisfied (DWC), it does not satisfy (DMC), let y be \( \bar{y} \), z be y and z′ be \( {\bar{x}} \), and assume that x ~_{a} \( \bar{y} \) and y ~_{a} \( {\bar{x}} \). Probability distribution P_{9} in Table 7 provides an example because all relevant conditional probabilities for these pairs equal 1. However, it is clearly not the case that x ∨ y ~_{a} \( \bar{y} \) ∨ \( {\bar{x}} \).
 (LAC):

Probability distribution P_{10} in Table 7 shows that none of the coherence relations satisfies (LAC). P_{10}(\( xz \)) ≈ .586 > .555 ≈ P_{10}(x) > .5, P_{10}(\( zx \)) ≈ .581 > .551 ≈ P_{10}(z) > .5, P_{10}(\( yz \)) ≈ .619 > .587 ≈ P_{10}(y) > .5 and P_{10}(\( zy \)) ≈ .580 > .551 ≈ P_{10}(z) > .5. Hence, x ~_{s} z and y ~_{s} z. However, P_{10}(x ∧ \( yz \)) ≈ .450 < .480 ≈ P_{10}(x ∧ \( yz \)) < .5. Therefore, neither x ∧ y ~_{i} z nor x ∧ y ~_{a} z, and a fortiori not x ∧ y ~_{s} z.
 (LOC):

An example showing that none of the accounts satisfies (LOC) is given by probability distribution P_{11}. P_{11}(\( xz \)) ≈ .535 > .532 ≈ P_{11}(x) > .5, P_{11}(\( zx \)) ≈ .501 > .5 > .498 ≈ P_{11}(z), P_{11}(\( yz \)) ≈ .531 > .527 ≈ P_{11}(y) > .5 and P_{11}(\( zy \)) ≈ .502 > .5 > .498 ≈ P_{11}(z). Accordingly, x ~_{s} z and y ~_{s} z. But P_{11}(\( zx \) ∨ y) ≈ .493 < .498 ≈ P_{11}(z) < .5. Hence, it is not the case that x ∨ y ~ z for each of the three accounts.
 (RAC):

Due to the symmetry of the coherence relations, (RAC) is equivalent to (LAC).
 (ROC):

Due to the symmetry of the coherence relations, (ROC) is equivalent to (LOC).
 (LLE):

This is a straightforward consequence of the fact that the laws governing probability are not affected when propositions are replaced by equivalent propositions.
 (LRE):

Ditto.
 (CEQ):

If x and y are equivalent, then \( P(xy) \) and \( P(yx) \) equal 1 and are thus smaller than .5. Moreover, since x and y are contingent and P is regular, it is also entailed that \( P(xy) \) > \( P({\bar{x}}y) \) and \( P(yx) \) > \( P(\bar{y}x) \). Hence, x ~ y for all three relations.
 (CEN):

If x entails y, then \( P(yx) \) = 1. Hence, since y is contingent and P is regular, it is also implied that \( P(yx) \) > P(y). By symmetry we obtain \( P(xy) \) > P(x), with the result that x ~_{i} y. On the other hand, consider probability distribution P_{5} in Table 6. Here x entails y because P_{5}(\( yx \)) = 1 with x and y being contingent. However, P_{5}(\( xy \)) = 1/9, so that neither x ~_{a} y nor x ~_{s} y.
 (LMO):

According to probability distribution P_{12} from Table 7, P_{12}(\( xy \)) ≈ .798 > .5 > .474 ≈ P_{12}(x) and P_{12}(\( yx \)) ≈ .604 > .5 > .359 ≈ P_{12}(y). Therefore, x ~ y for all three coherence relations. Furthermore, y ⊢ z because P_{12}(\( zy \)) = 1 with z and y being contingent. However, x ~ z holds for no relation because P_{12}(\( xz \)) ≈ .352 < .474 ≈ P_{12}(x) < .5.
 (RMO):

Consider probability distribution P_{13}. P_{13}(\( xy \)) = 1 > .751 > .5 ≈ P_{13}(x) and P_{13}(\( yx \)) ≈ .670 > .503 ≈ P_{13}(y) > .5. Hence, x ~ y on all three accounts. Moreover, z ⊢ x because P_{12}(\( xz \)) = 1 with x and z being contingent. Nonetheless, we have P_{13}(\( yz \)) = 0 < .5 < .503 ≈ P_{13}(y), implying that y ~ z obtains for none of the accounts.
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Schippers, M., Siebel, M. The logic of coherence. Synthese 198, 7697–7714 (2021). https://doi.org/10.1007/s11229020025421
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DOI: https://doi.org/10.1007/s11229020025421
Keywords
 Probability
 Formal epistemology
 Coherence