Lewis’ Triviality for Quasi Probabilities

Abstract

According to Stalnaker’s Thesis (S), the probability of a conditional is the conditional probability. Under some mild conditions, the thesis trivialises probabilities and conditionals, as initially shown by David Lewis. This article asks the following question: does (S) still lead to triviality, if the probability function in (S) is replaced by a probability-like function? The article considers plausibility functions, in the sense of Friedman and Halpern, which additionally mimic probabilistic additivity and conditionalisation. These quasi probabilities comprise Friedman–Halpern’s conditional plausibility spaces, as well as other known representations of conditional doxastic states. The paper proves Lewis’ triviality for quasi probabilities and discusses how this has implications for three other prominent strategies to avoid Lewis’ triviality: (1) Adams’ thesis, where the probability function on the left in (S) is replaced by a probability-like function, (2) abandoning conditionalisation, where probability conditionalisation on the right in (S) is replaced by another propositional update procedure and (3) the approximation thesis, where equality in (S) is replaced by approximation. The paper also shows that Lewis’ triviality result is really about ‘additiveness’ and ‘conditionality’.

Appendices

Proofs

1.1 Restricted Standard Conditional Plausibility Spaces

Lemma 7

Let $F$ weakly $\displaystyle \oslash$-conditionalisable and A be $F$-consistent. Then $X \sqsubseteq X'$ and $F$ and $F_A$ are plausibilities over $X'$.

Proof

We may consider $X' = X \cup \{F_A(B) : B \in {\mathcal {A}}, F(A)> \mathbf {0}\}$. $X'$ is partially ordered by assumption.

• Pl1: Let $A \subseteq B$ and C$F$-consistent. Thus $A \cap C \subseteq B \cap C$. Hence $F(A \cap C) \le _X F(B \cap C)$ (Pl1). And since $\displaystyle \oslash$ is monotone in the first argument, we have $F(A|C) = F(A \cap C) \displaystyle \oslash F(C) \le _X F(B \cap C) \displaystyle \oslash F(C) = F(B|C)$.

• Pl2: first w.r.t. $\mathbf {0},\mathbf {1}$. $F(\varOmega |A)= F(\varOmega \cap A) \displaystyle \oslash F(A) = F(A \cap A) \displaystyle \oslash F(A)= F(A|A)= F(\varOmega ) = \mathbf {1}$ by success. $F(\emptyset |A) = F(\emptyset ) = \mathbf {0}$ (by same $\mathbf {0}$). By Pl1 (and for $C\in {\mathcal {A}}$, s.t. $F(C)> \mathbf {0}$) we have for all $B \in {\mathcal {A}}$, $\mathbf {0}\le F_C(B) \le \mathbf {1}$, so that $\mathbf {0}, \mathbf {1}$ are the minmum and maximum of $X'$, which thereby is pointed and $\mathbf {0}_C = \mathbf {0}, \mathbf {1}_C = \mathbf {1}$. This proves Pl2.

• $X \sqsubseteq X'$. $X \subseteq X'$ by definition, $X'$ is pointed by the above remark and $X'$ has the same $\mathbf {0}$ and $\mathbf {1}$ as X. Finally, $\le '$ restricted to X is $\le$ by the above definition of $X'$.

• $F=F(.|\varOmega )$ and $F(\varOmega )=F(\varOmega |\varOmega )> \mathbf {0}$ by spreading. Thus by the above, $F$ can also be seen as a plausibility over $X'$. $\square$

Theorem 6

$S=\{\langle \varOmega , X_A, F_A \rangle :A \in {\mathcal {A}}, F(A)>_{\varOmega } \mathbf {0}_{\varOmega }\}$ with $F= F(.|\varOmega )$ and $X=X_{\varOmega }$ is a proper conditional plausibility space satisfying C3.1 iff $F$ is weakly $\displaystyle \oslash$-conditionalisable. S satisfies additionally C3.2 iff $F$ additionally satisfies coincidence. (For C3.1, C3.2, see Theorem 3.)

Proof

$(\Rightarrow )$ :

Since we restrict to $F_A$’s such that $F(A)>0$, by standardness all $X_A$ have the same $\mathbf {1}$ and $\mathbf {0}$. Denote ${\mathrm{Im}}(F)= \text{ Im }[F(.|\varOmega )]$ and consider $X' = \bigcup _{F_A \in S} \text{ Im }(F_A)$.

0.:

$F=F(.|\varOmega )$ by convention and $F(.|\varOmega ) \in S$ is spreading by assumption and thus $F$ is a spreading plausibility function over $X=X_{\varOmega }$.

1.:

For $\langle x,y \rangle \in {\mathrm{Im}}(F)^2$$F$-consistent, i.e., such that there are A, C with $x=F(A \cap C)$ and $y = F(C) > \mathbf {0}$, define $x \displaystyle \oslash y = F(A|C)$. Then $\displaystyle \oslash$ is a partial function $X^2 \longrightarrow X'$ (using C3.1 twice), defined (and total) over consistent couples in ${\mathrm{Im}}(F)$ (by the above definition) and $X,X_A \sqsubseteq X'$ (by definition of $X'$). $\displaystyle \oslash$ is monotone in first argument by C3.1.

2.:

As a consequence $F(A|C)$ is defined for C$F$-consistent and $F(A|C) = F(A \cap C) \displaystyle \oslash F(C)$.

3.:

Weak vacuity by convention.

4.:

Success: Since $F(\varOmega A) =F(AA)$ and $F(A)= F(A)>0$. Thus $F(A|A) = F(\varOmega |A) =F(\varOmega |\varOmega ) = F(\varOmega )$ (C3.1 twice).

5.:

$F(\emptyset |\varOmega ) = \mathbf {0}_{\varOmega } = \mathbf {0}=\mathbf {0}_A = F(\emptyset |A)$ by standardness.

$(\Leftarrow )$ :

1.:

Partial conditional plausibility space: Consider $S=\{\langle \varOmega , X_A, F_A \rangle : F(A)> \mathbf {0}, A \in {\mathcal {A}}\}$. By weak vacuity $F= F(.|\varOmega )$ which is assumed spreading and thus $F\in S$. Additionally, $F_A$ and $F$ are plausibilities over $X'$ for A$F$-consistent (see Lemma 7) and we can take $X_A=X' = X_{\varOmega }$.

2.:

Standard: Follows from $X_A=X'$.

3.:

C3.1: follows from monotonicity in the first argument of $\displaystyle \oslash$ and weak vacuity.

• C3.2 implies that we can extend $\displaystyle \oslash$ such that $F(A|BC) = F(AB|C) \displaystyle \oslash F(B|C)$, for $F(B|C)> \mathbf {0}$ and for C$F$-consistent. Therefore BC is $F$-consistent: $F(B|C)>\mathbf {0}$ is defined and thus $F(BC) \displaystyle \oslash F(C)>\mathbf {0}$. But then $F(BC)> \mathbf {0}$, else, i.e., if $F(BC) = \mathbf {0}$, then we would have $F(B|C)= \mathbf {0}$ by the fact that $\mathbf {0}$ is the left-identity of $\displaystyle \oslash$. Thus $F(.|BC)$ is in effect defined. Consider now an $F$-consistent C and an $F_C$-consistent B. Then by the above equation we obtain coincidence:

  $$\begin{aligned} (F_C)_B(A)= F(AB|C) \displaystyle \oslash F(B|C) = F(A|BC)= F_{BC}(A) \end{aligned}$$

• Conversely. Suppose coincidence holds. Then for the above C and B, we obtain

  $$\begin{aligned} F(AB|C) \displaystyle \oslash F(B|C) = (F_C)_B(A) = F_{BC}(A)= F(A|BC) \end{aligned}$$

  Thus for such C, B, we obtain the implication in C3.2, since $\displaystyle \oslash$ is a function. $\square$

1.2 The Algebraic Structure of $\oplus$ and $\displaystyle \oslash$

Lemma 8

Let $F$ be a separable plausibility over the algebra ${\mathcal {A}}$. Then $\oplus$ forms an ortho-complemented category over ${\mathrm{Im}}(F)$, with $\mathbf {1}$ being $\oplus$-absorbing (as in Lemma 1).

Proof

• 1. $\mathbf {0}$ is the (unique) $\oplus$-Identity over${\mathrm{Im}}(F)$: Let $b \in {\mathrm{Im}}(F)\subseteq X$. Thus there is $B \in {\mathcal {A}}$ with $F(B)=b$.

  $$\begin{aligned} \begin{array}{rlr} \mathbf {0}\oplus b &{} = F(\emptyset ) \oplus F(B) &{} \text{(Pl2, } \text{ assumption) } \\ &{} = F(\emptyset \cup B) &{} \text{(separable) }\\ &{} = F(B) &{} \text{(algebra: } \emptyset \text{ identity } \text{ for } \cup \text{) }\\ &{} = b &{} \text{(assumption) } \end{array} \end{aligned}$$

  The same reasoning can be applied to obtain the left identity $b \oplus \mathbf {0}= b$. Uniqueness: Suppose (for reductio) there is $\mathbf {0}' \in {\mathrm{Im}}(F)$ such that $\mathbf {0}' \ne \mathbf {0}$ and which also satisfies $\mathbf {0}' \oplus b = b$ and $b \oplus \mathbf {0}' = b$ for all $b \in {\mathrm{Im}}(F)$. Then in particular $\mathbf {0}\oplus \mathbf {0}' = \mathbf {0}$ as well as $\mathbf {0}\oplus \mathbf {0}' =\mathbf {0}'$. Thus $\mathbf {0}' = \mathbf {0}$. Contradiction.

• 2. Existence of$\oplus$complements over${\mathrm{Im}}(F)$: Let $a \in {\mathrm{Im}}(F)\subseteq X$, then there is $A \in {\mathcal {A}}$, such that $F(A)=a$. Thus there is $\overline{a} \in X$ such that $F(\overline{A}) = \overline{a}$ and $a, \overline{a}$ are $\oplus$ complements to each other:

  $$\begin{aligned} \begin{array}{rlr} a \oplus \overline{a} &{} = F(A) \oplus F(\overline{A}) &{} \text{(assumption) }\\ &{} = F( A\cup \overline{A}) &{} \text{(separable) }\\ &{} = F(\varOmega ) &{} \text{(algebra) }\\ &{} = \mathbf {1}&{} \text{(Pl2) } \end{array} \end{aligned}$$

• 3. Ortho-complements over${\mathrm{Im}}(F)$: Let $a \in {\mathrm{Im}}(F)\subseteq X$. Then there is $A \in {\mathcal {A}}$ such that $F(A)=a$. And $F(\overline{A}) = \overline{a}$ as well as $F(\overline{\overline{A}})=\overline{\overline{a}}$, and from algebra:

  $$\begin{aligned} \overline{\overline{a}}=F(\overline{\overline{A}}) = F(A)=a \end{aligned}$$

• 4. $\oplus$ is associative over disjoint triples in ${\mathrm{Im}}(F)$: Let $a,b,c \in {\mathrm{Im}}(F)\subseteq X$ be a disjoint triple. Thus there are $A,B,C \in {\mathcal {A}}$, such that $F(A) =a, F(B)=b, F(C) =c$ and A, B, C are mutually disjoint. Therefore $A \cup B$ and C are disjoint. Similarly A and $B \cup C$ are disjoint. Thus:

  $$\begin{aligned} \begin{array}{rlr} (a \oplus b) \oplus c &{} =(F(A) \oplus F(B)) \oplus F(C) &{} \text{(assumption) }\\ &{} = F(A \cup B) \oplus F(C) &{} \text{(separable) }\\ &{} = F((A \cup B) \cup C) &{} \text{(separable) }\\ &{} = F(A \cup (B \cup C)) &{} \text{(algebra: } \cup \text{ associative) }\\ &{} = F(A) \oplus F( B \cup C) &{} \text{(separable) }\\ &{} = F(A) \oplus (F( B) \oplus F(C)) &{} \text{(separable) }\\ &{} = a \oplus (b \oplus c) &{} \text{(assumption) } \end{array} \end{aligned}$$

• 5. $\oplus$is commutative for disjoint couples over${\mathrm{Im}}(F)$: Let $a,b \in {\mathrm{Im}}(F)\subseteq X$ be disjoint couples. Thus there are disjoint $A,B \in {\mathcal {A}}$, such that $F(A)=a$ and $F(B)=b$. Thus:

  $$\begin{aligned} \begin{array}{rlr} a \oplus b &{} =F(A) \oplus F(B) &{} \text{(assumption) }\\ &{} = F(A \cup B) &{} \text{(separable) }\\ &{} = F(B \cup A) &{} \text{(algebra: } \cup \text{ commutes) }\\ &{} = F(B) \oplus F(A) &{} \text{(seperable) }\\ &{} = b \oplus a &{} \text{(assumption) } \end{array} \end{aligned}$$

• 6. $\mathbf {1}$ is $\oplus$-absorbing over disjoint couples of${\mathrm{Im}}(F)$: assume $\langle a, \mathbf {1}\rangle \in X^2$ is a disjoint couple. Then there are disjoint $A,B \in {\mathcal {A}}$ such that $F(A)=a$, $F(B)=\mathbf {1}$ (note B need not be $\varOmega$ and therefore A need not be $\emptyset$). Thus:

  $$\begin{aligned} \begin{array}{rlr} \mathbf {1}&{}= F(B) &{} \text{(assumption) }\\ {} &{}\le _X F(A \cup B) &{} \text{(Pl1) }\\ &{}\le _X F(\varOmega ) &{} \text{(Pl1) }\\ {} &{}= \mathbf {1}&{} \text{(Pl2) } \end{array} \end{aligned}$$

  $a \oplus \mathbf {1}= F(A) \oplus F(B) = F(A \cup B) = \mathbf {1}$ follows by antisymmetry of $\le _X$. $\square$

Lemma 9

If $F$ is a weakly $\displaystyle \oslash$-conditionalisable plausibility, then for $a \in {\mathrm{Im}}(F)$, $\displaystyle \oslash$ satisfies the properties (1–4) of Lemma 2 ($\mathbf {1}$$\displaystyle \oslash$-right Identity, $\displaystyle \oslash$-Inverses, $\mathbf {1}$ unique, $\mathbf {0}$ left absorbing). If $F$ is $\displaystyle \oslash$-conditionalisable, then $\displaystyle \oslash$ also satisfies (5) $\displaystyle \oslash$-cancelling.

Proof

• 1. $\mathbf {1}$$\displaystyle \oslash$-right identity: Let $a \in {\mathrm{Im}}(F)$. Thus there is $A \in {\mathcal {A}}$ such that $a=F(A) =F(A | \varOmega ) = F(A \cap \varOmega ) \displaystyle \oslash F(\varOmega )= F(A) \displaystyle \oslash \mathbf {1}=a \displaystyle \oslash \mathbf {1}$ (by weak vacuity and $A \cap \varOmega =A$).

• 2. $\displaystyle \oslash$-Inverses for$F$-consistent$a \in {\mathrm{Im}}(F)$: Let $a \in {\mathrm{Im}}(F)$$F$-consistent. Thus there is $A \in {\mathcal {A}}$ such that $a = F(A)>_X \mathbf {0}$. But $F_A$ has the same $\mathbf {1}$ as $F$ (from success). Thus $\mathbf {1}= F(\varOmega )= F(\varOmega |A)=F(A \cap \varOmega ) \displaystyle \oslash F(A) = F(A) \displaystyle \oslash F(A)$.

• 3. $\mathbf {1}$Uniqueness: Assume (for reductio) there is another $\mathbf {1}' \in {\mathrm{Im}}(F)$ such that $a \displaystyle \oslash \mathbf {1}' = a$ for all $a \in {\mathrm{Im}}(F)$. But then $\mathbf {1}' \displaystyle \oslash \mathbf {1}' = \mathbf {1}'$. Yet $\mathbf {1}' \displaystyle \oslash \mathbf {1}' = \mathbf {1}$ by Inverses. Thus $\mathbf {1}' = \mathbf {1}$.

• 4. $\mathbf {0}$ is left $\displaystyle \oslash$-absorbing for$a\in {\mathrm{Im}}(F)F$-consistent: Let $a \in {\mathrm{Im}}(F)$$F$-consistent. Thus there is $A \in {\mathcal {A}}$ such that $F(A)=a$ and $F(.|A)$ is defined ($F$-consistency). $\mathbf {0}\displaystyle \oslash a = F(\emptyset ) \displaystyle \oslash F(A) = F(\emptyset \cap A) \displaystyle \oslash F(A)= F(\emptyset | A) = F(\emptyset ) = \mathbf {0}$ (by Pl2, conditionalisation, closure and the fact that $F$ and $F(.|A)$ have the same $\mathbf {0}$).

• 5. Cancelling: Can be proven from coincidence, see the if-and Lemma 11. But since it is not directly required, we simply state it. $\square$

1.3 Pseudo-Inverse

Given $\displaystyle \oslash$, let us define $x \otimes b= a$, whenever $a \displaystyle \oslash b$ is defined and $a\displaystyle \oslash b =x$. $\otimes$ is then a partial function ${\mathrm{Im}}(F)(\displaystyle \oslash ) \times {\mathrm{Im}}(F)\longrightarrow {\mathrm{Im}}(F)$, defined for all $x \in {\mathrm{Im}}(F)(\displaystyle \oslash )$ and $b \in {\mathrm{Im}}(F)$ such that $b=F(B)$, B$F$-consistent and $x= F(A|B)$ for some A, and then $x \otimes b = F(A \cap B)$. Thus, intuitively $\otimes$ is defined over conditional plausibility values $x=F(A|B)$ and corresponding plausibility values $b=F(B)$. Note then, that $\otimes$ satisfies the fundamental equation for $\langle a,b\rangle$ an $F$-consistent couple:

$$\begin{aligned} (a \displaystyle \oslash b) \otimes b =a \end{aligned}$$

(A.1)

In this sense, $\otimes$ partially inverses $\displaystyle \oslash$. From this, we obtain

Lemma 10

(pseudo-inverses) Let $F$ be a $\displaystyle \oslash$-conditionalisable plausibility. For $x\in {\mathrm{Im}}(F)$, $F$-consistent we have

• 1. $\mathbf {1}\otimes x = x = x \otimes \mathbf {1}$.

• 2. $\mathbf {0}\otimes x = \mathbf {0}$.

Proof

Since $x \in {\mathrm{Im}}(F)$ is $F$-consistent there is $A \in {\mathcal {A}}$ such that $F(A)=x$ and A is $F$-consistent, so that $F_A$ is defined.

• 1. $\mathbf {1}\otimes x$ is defined for $F(A)=x$ and A$F$-consistent: Chose $B = \varOmega$. Then $z=F(A \cap B) = F(A \cap \varOmega ) = F(A)=x$. Thus we can write $\mathbf {1}= x \displaystyle \oslash x = z \displaystyle \oslash x$ ($\displaystyle \oslash$-inverses). Hence $\mathbf {1}\otimes x = (x \displaystyle \oslash x) \otimes x = (z \displaystyle \oslash x) \otimes x = z = x$ by the fundamental equation. Similarly $x \otimes \mathbf {1}$ is defined for $x \in {\mathrm{Im}}(F)$: Let $F(A)=x \in {\mathrm{Im}}(F)$. Thus $x = x \displaystyle \oslash 1$ (by 1 right $\displaystyle \oslash$-identity) and $x = F(A|\varOmega )$. Hence $x \otimes \mathbf {1}= x$ by the fundamental equation.

• 2. $\mathbf {0}\otimes x$ is defined for A$F$-consistent and $F(A)=x$: Chose $B = \overline{A}$. Then $A \cap B = \emptyset$ and thus $\mathbf {0}= F(A \cap B) = z$ (Pl2). Hence we can rewrite $\mathbf {0}= z \displaystyle \oslash x$. Thus $\mathbf {0}\otimes x = (z \displaystyle \oslash x) \otimes x = z = \mathbf {0}$ by the fundamental equation. $\square$

1.4 Triviality

In all what follows, we assume $F$ to be a quasi probability. The following lemma proves the plausibilistic analogue to what Douven called the Generalised Stalnaker (hypo)thesis:

Lemma 11

(if-and) Let $F$ be a $\displaystyle \oslash$-conditionalisable plausibility and $A,B \in {\mathcal {A}}$ such that $A \cap B$ is $F$-consistent and assume ${\mathrm{S}}(F_B)$ (as in Lemma 3). Then

$$\begin{aligned} F(A > C| B) = F(C| A \cap B). \end{aligned}$$

(A.2)

Proof

Let $F$ be a $\displaystyle \oslash$-conditionalisable plausibility, $A \cap B$$F$-consistent (i.e., $F(A\cap B) >_X \mathbf {0}$) and $F_B$ satisfies the Stalnaker Thesis. Then B as well as A are $F$-consistent (Pl1). Thus $F_B$ and $F_{A \cap B}$ are defined and thus the Stalnaker Thesis makes sense for $F_B$ and both sides of the equation are defined. Furthermore, we obtain that A remains $F_B$-consistent since $\displaystyle \oslash$ is monotone in the first argument. In effect, consider $F_B(A)= F(A \cap B) \displaystyle \oslash F(B)$. We have $F(A \cap B) >_X \mathbf {0}= F(\emptyset \cap B)$. By monotonicity in the first argument $F(A \cap B) \displaystyle \oslash F(B) >_X F(\emptyset \cap B) \displaystyle \oslash F(B) = \mathbf {0}$. Thus A remains $F_B$-consistent. Hence coincidence applies, i.e., $(F_B)_A = F_{A \cap B}$. Thus

$$\begin{aligned} \begin{array}{rlr} F(A> C| B) &{} = F_B(A > C) &{} (B \text{ is } F\text{-consistent })\\ &{} = F_B(C|A) &{} ({\mathrm{S}}(F_B), )\\ &{} = (F_B)_A(C) &{} (A \text{ is } F_B\text{-consistent })\\ &{} = F(C |A \cap B) &{} (\text{ coincidence }, F \text{ is } A \cap B \text{ conditionalisable }) \end{array} \end{aligned}$$

Note that thereby (line 2 to 4) we have indirectly proven the cancellation property: $(x \displaystyle \oslash b) \displaystyle \oslash (y \displaystyle \oslash b) = x \displaystyle \oslash y$. For this use coincidence and $x = F(A \cap B \cap C), y = F(A \cap B)$ and $b = F(B)$. $\square$

Lemma 12

(accept) Let $F$ be a $\displaystyle \oslash$-conditionalisable plausibility and $A,C \in {\mathcal {A}}$ such that $A \cap C$ is $F$-consistent and assume ${\mathrm{S}}(F_C)$ (as in Lemma 4). Then

$$\begin{aligned} F(A > C| C) \otimes F(C) = F(C). \end{aligned}$$

(A.3)

Proof

Let $F, A,C$ as described, we may thus apply the if-and Lemma. Note that $F(A> C|C)= F((A > C) \cap C) \displaystyle \oslash F(C)$ so that the left hand side of the above equation is defined. Note also, that C and $A \cap C$ are $F$-consistent, so that $F(.|A \cap C)$ and $F(.|C)$ are defined. Since $F$ is $\displaystyle \oslash$-conditionalisable, the law of $\displaystyle \oslash$-inverses holds and $\mathbf {1}$ is the left $\otimes$-identity. Thus

$$\begin{aligned} \begin{array}{lr} F(A > C| C) \otimes F(C) = F(C| A \cap C) \otimes F(C) &{} \text{(if-and } \text{ Lemma) }\\ \quad = [~F(C \cap (A \cap C)) \displaystyle \oslash F(A \cap C)~] \otimes F(C) &{} (\text{ conditionalisable })\\ \quad = [~F(A \cap C) \displaystyle \oslash F(A \cap C) ~] \otimes F(C) &{} (\text{ algebra })\\ \quad = \mathbf {1}\otimes F(C) &{} (\displaystyle \oslash \text{-inverses })\\ \quad = F(C) &{} (\mathbf {1}\otimes \text{-left } \text{ Identity) } \end{array} \end{aligned}$$

$\square$

Lemma 13

(reject) Let $F$ be a $\displaystyle \oslash$-conditionalisable and $A,\overline{C} \in {\mathcal {A}}$ such that $A \cap \overline{C}$ is $F$-consistent and assume ${\mathrm{S}}(F_{\overline{C}})$ (as in Lemma 5). Then

$$\begin{aligned} F(A > C| \overline{C}) \otimes F(\overline{C}) = \mathbf {0}. \end{aligned}$$

(A.4)

Proof

Let $F, A, \overline{C}$ as described, we may thus apply the if-and Lemma 11. Note that $F(A> C|\overline{C})= F((A > C) \cap \overline{C}) \displaystyle \oslash F(\overline{C})$, so that the left hand side of the above equation is defined. Note also, that $\overline{C}$ and $A \cap \overline{C}$ are $F$-consistent and hence $F(.|A \cap \overline{C})$ and $F(.|\overline{C})$ are defined and by conditionalisability have the same $\mathbf {0}$ as $F$, so that $\mathbf {0}$ is $\displaystyle \oslash$-left absorbing. Finally $\mathbf {0}$ is $\otimes$-left absorbing. Thus

$$\begin{aligned} \begin{array}{lr} F(A > C| \overline{C}) \otimes F(\overline{C}) =F(C| A \cap \overline{C}) \otimes F(\overline{C}) &{} \text{(if-and } \text{ Lemma) }\\ \quad =[~F(C \cap (A \cap \overline{C})) \displaystyle \oslash F(A \cap \overline{C}) ~] \otimes F(\overline{C}) &{} (\text{ conditionalisable })\\ \quad = [~F(\emptyset ) \displaystyle \oslash F(A \cap \overline{C}) ~] \otimes F(\overline{C}) &{} \text{(algebra) }\\ \quad = [~\mathbf {0}\displaystyle \oslash F(A \cap \overline{C})~] \otimes F(\overline{C}) &{} (\text{ Pl2, } \text{ minimum })\\ \quad = \mathbf {0}\otimes F(\overline{C}) &{} (\mathbf {0}~\text{ is } \displaystyle \oslash \text{-left } \text{ absorbing })\\ \quad = \mathbf {0}&{} (\mathbf {0} \text{ is } \otimes \text{-left } \text{ absorbing }) \end{array} \end{aligned}$$

$\square$

Lemma 14

(totality) Let $F$ be a quasi probability ($\oplus$-separable and $\displaystyle \oslash$-conditionalisable plausibility) and let $C,\overline{C}$ be $F$-consistent (as in Lemma 6). Then

$$\begin{aligned} F(A) = \oplus [~F(A|C) \otimes F(C) ~,~F(A|\overline{C}) \otimes F(\overline{C})~]. \end{aligned}$$

(A.5)

Proof

Let $F$ and $A,C \in {\mathcal {A}}$ as claimed and $\otimes$ as previously defined. Setting $a=F(A \cap C)$, $a' = F(A \cap \overline{C})$, $c = F(C), c' = F(\overline{C})$, we obtain $(\sharp )$: $F(A \cap C)= (F(A \cap C) \displaystyle \oslash F(C)) \otimes F(C) = F(A|C) \otimes F(C)$ and $F(A\cap \overline{C})= (F(A \cap \overline{C}) \displaystyle \oslash F(\overline{C})) \otimes F(\overline{C}) = F(A|\overline{C}) \otimes F(\overline{C})$. Then

$$\begin{aligned} \begin{array}{rlr} F(A) &{} =F((A \cap C) \cup (A \cap \overline{C}) )&{} (\text{ algebra })\\ &{} =\oplus [~F(A \cap C)~,~F(A \cap \overline{C}) ~]&{} (\text{ separable })\\ &{}= \oplus [~F(A|C) \otimes F(C) ~,~F(A|\overline{C}) \otimes F(\overline{C})~]&{} (\sharp ) \end{array} \end{aligned}$$

$\square$

Theorem 7

(triviality) Assume ${\mathrm{S}}(\mathbb {C})$ and ${\mathrm{C}}(\mathbb {C})$ for the class of plausibilities $\mathbb {C}$. Let $F$ be a quasi probability ($\displaystyle \oslash$-conditionalisable and $\oplus$-separable plausibility) in $\mathbb {C}$. Additionally, let $A,C \in {\mathcal {A}}$ such that $F(A\cap C), F(A \cap \overline{C})>_X \mathbf {0}$ (as in Theorem 4). Then

$$\begin{aligned} F(A > C) = F(C). \end{aligned}$$

(A.6)

Proof

Since $A \cap C$ and $A \cap \overline{C}$ are $F$-consistent, $A,C,\overline{C}$ are $F$-consistent. Thus by seperability and conditionalisability, the Lemma 14 of totality applies for the partition $\{C, \overline{C}\}$. Additionally, since $F$ is a $\displaystyle \oslash$-conditionalisable plausibility, we have A is $F_C$-consistent and $F_{\overline{C}}$-consistent (see previous argument). Thus $F_C$ as $F_{\overline{C}}$ are defined and by ${\mathrm{C}}(\mathbb {C})$, they are in $\mathbb {C}$. By ${\mathrm{S}}(\mathbb {C})$ the Stalnaker Thesis holds for them. Thus we also have the conditions of the accept and reject Lemma 12, 13. Hence:

$$\begin{aligned} \begin{array}{lr} F(A> C) = \oplus ( F(A> C| C) \otimes F(C)~, ~F(A > C| \overline{C}) \otimes F(\overline{C})) &{} (\text{ totality } \text{ Lemma })\\ \quad = \oplus (F(C)~,~\mathbf {0}) &{} (\text{ accept, } \text{ reject } \text{ Lemma })\\ \quad = F(C) &{} (\mathbf {0}\oplus \text{-Identity }) \end{array} \end{aligned}$$

$\square$

Corollary 5

For any quasi probability $F\in \mathbb {C}$, $A,C \in {\mathcal {A}}$ such that $F(AC), F(A \overline{C})>_X \mathbf {0}$: ${\mathrm{S}}(\mathbb {C})$ and ${\mathrm{C}}(\mathbb {C})$ imply $F(C|A) =F(A > C)=F(C)$.

Call a plausibility function $F$trivial, if there are no 3 disjoint propositions $C,D,E \in {\mathcal {A}}$ such that they are $F$-possible (this is, $F(C), F(D), F(E)>_X \mathbf {0}$). We now prove Triviality (Corollary 4 in the text).

Corollary 6

(Lewis-like triviality) There is no class $\mathbb {C}$ of quasi probability functions, such that ${\mathrm{S}}(\mathbb {C})$ and ${\mathrm{C}}(\mathbb {C})$, unless all $F\in \mathbb {C}$ are trivial.

Proof

Assume there is a class $\mathbb {C}$ of quasi probability functions such that ${\mathrm{S}}(\mathbb {C})$ and ${\mathrm{C}}(\mathbb {C})$. Assume (for reductio) that $\mathbb {C}$ contains a non-trivial quasi probability $F$. Therefore, there are disjoint $C,D,E \in {\mathcal {A}}$ such that $F(C), F(D), F(E) >_X \mathbf {0}$. Consider $A = C \cup D$. Thus $AC = C$ and $A \overline{C} =D$. Hence $F(AC),F(A\overline{C})>0$, as well as $F(A), F(C), F(\overline{C})>_X \mathbf {0}$ (Pl1).

Since $F\in \mathbb {C}$ and $A,C \in {\mathcal {A}}$, we have $A > C \in {\mathcal {A}}$ (from ${\mathrm{S}}(\mathbb {C})$) and because $F$ is only defined over elements of ${\mathcal {A}}$). Thus by the Corollary 5 to the triviality Theorem 7:

$$\begin{aligned} F(C|A) = F(C). \end{aligned}$$

However, note that $F(C|A) = F(CA) \displaystyle \oslash F(A) = F(C) \displaystyle \oslash (F(C) \oplus F(D)) \ne F(C)$. Otherwise $F(C) \oplus F(D) =\mathbf {1}$ (uniqueness of $\mathbf {1}$). But then $(F(C) \oplus F(D)) \oplus F(E) = \mathbf {1}$ and thus $\mathbf {1}\oplus F(E) = \mathbf {1}$ and therefore $F(E) = \mathbf {0}$ (uniqueness of $\mathbf {0}$). But then it is not possible that $F(E)>_X \mathbf {0}$. This contradicts the assumption. $\square$

Comparative Remarks

The main differences between the approach of Morgan and Mares (1995) and Morgan (1999) and the one presented here are the following ones. In Morgan and Mares (1995) $P(C| \varGamma )$ is a previously understood function, defined for a sentenceC and a set$\varGamma$ of sentences. The generalised Stalnaker Thesis is directly assumed and instead of exploiting the equality $P(C) = P(C|A)$ to prove triviality, they exploit the inequality $P(C) \le P(C|A)$ (as I did in Theorem 5). In their framework this is the inequality $P(C|\varGamma ) \le P(C| \varGamma \cup \{A\})$ [theorem 2.4 in Morgan and Mares (1995) and theorem V.6 in Morgan (1999)]. They also prove a slightly different triviality result, namely $P(A|\varGamma )= P(B|\varGamma )$ for all $A,B, \varGamma$ such that $P(A|\varGamma ), P(B|\varGamma )$ are not maximal [theorem 3.1 in Morgan and Mares (1995) and theorem V.7 in Morgan (1999)]. The framework targets Adams’ Thesis and is built to allow formulating a non-standard probabilistic semantics. The logical relation between their assumptions and the assumptions made here is not clean cut, because of the difference in the result and the difference in the form of the conditional plausibility.

Roughly, the differences are as follows: Their assumptions (C1–C5)⁵⁶ used to prove a triviality in Morgan and Mares (1995) are on the one hand ‘stronger’ and on the other hand ‘weaker’ than the assumptions made here. In particular, their C1 assumes a total order, C2 is a success postulate, from C3 a version of $\displaystyle \oslash$-conditionalisability can be arrived at, C4 is a version of the generalised Stalnaker thesis and C5 is a weak assumption on negation. The assumptions are ‘stronger’, in the sense that neither totality, nor a form of C3, nor the generalised Stalnaker thesis are assumed here. Note also that if we consider their functions up to logical equivalence then C1–C5 imply a version of $\displaystyle \oslash$-conditionalisability. On the other hand, especially C3 cannot be derived from the assumptions made here—it would correspond to the cancelling property, but for arguments for which this property is not necessarily defined. The assumptions are ‘weaker’, because their P may treat logically equivalent sentences (or sets of sentences) in a different manner and C5 is a weaker assumption than separability, i.e., it can be derived from the latter.

Similar things can still be said about the weaker assumptions in Morgan (1999), i.e., only a transitive relation is assumed (CCP1), the success postulate is weakened with (CCP2) and the negation postulate as well with (CCP5). However, the generalised Stalnaker thesis is still directly assumed with (CCP4) and a weak version of C3 is present in (CCP3) which is still not derivable from the assumptions made here. Note also that the triviality obtained from replacing (CCP4) by a simple Stalnaker Thesis (CCP4’) still assumes a special form of conditionalisation (CCP0) which is built on the general iteration property $P_{\varSigma }(A| \varGamma )= P(A | \varGamma \cup \varSigma )$ and which I only assume in special cases (in the property of coincidence).