Proof. Let $\mathbb {M}=\langle W,v^+,v^-,\mu, e_1,e_2\rangle$ be a $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -model s.t. $e_2(\alpha )\neq 0$ . We construct a model $\mathbb {M}^*=\langle W,(v^*)^+,(v^*)^-,\mu, e^*_1,e^*_2\rangle$ where $e^*_1(\alpha )\neq 1$ . To do this, we define new $\mathsf {BD}$ valuations $(v^*)^+$ and $(v^*)^-$ on $W$ as follows.

\begin{align*} (v^*)^+&=W\setminus v^-&(v^*)^-&=W\setminus v^+ \end{align*}

It can be easily checked by induction on $\phi \in \mathscr {L}_{\mathsf {BD}}$ that

(1) \begin{align} |\phi |^+_{\mathbb {M}}&=W\setminus |\phi |^-_{\mathbb {M}^*}&|\phi |^-_{\mathbb {M}}&=W\setminus |\phi |^+_{\mathbb {M}^*} \end{align}

Now, since we can w.l.o.g. assume that $\mu$ is a (classical) probability measure on $W$ (recall Example 9), we have that

(2) \begin{align} e^*(\mathsf {Pr}\phi )=(1-\mu (|\phi |^-),1-\mu (|\phi |^+))=(1-e_2(\mathsf {Pr}\phi ),1-e_1(\mathsf {Pr}\phi )) \end{align}

It remains to show that $e^*(\alpha )=(1-e_2(\alpha ),1-e_1(\alpha ))$ for every $\alpha \in \mathscr {L}_{\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle }$ . We proceed by induction on formulas. The basis case of $\alpha =\mathsf {Pr}\phi$ holds by (2).

Consider $\alpha =\beta \rightarrow \beta '$ .

\begin{align*} e^*_1(\beta \rightarrow \beta ')&=\min (1,1-e^*_1(\beta )+e^*_1(\beta '))\\ &=\min (1,1-(1-e_2(\beta ))+(1-e_2(\beta ')))\,\,{(\text{ by } IH)}\\ &=\min (1,1-e_2(\beta ')+e_2(\beta ))\\ &=1-\max (0,e_2(\beta ')-e_2(\beta ))\\ &=1-e_2(\beta \rightarrow \beta ') \end{align*}

\begin{align*} e^*_2(\beta \rightarrow \beta ')&=\max (0,e^*_2(\beta ')-e^*_2(\beta ))\\ &=\max (0,1-e_1(\beta ')-(1-e_1(\beta )))\qquad \qquad {(\text{ by } IH)}\\ &=\max (0,e_1(\beta )-e_1(\beta '))\\ &=1-\min (1,1-e_1(\beta )+e_1(\beta ')) \end{align*}

The remaining cases of $\alpha =\neg \beta$ , $\alpha ={\sim }\beta$ , and $\alpha =\triangle \beta$ can be tackled similarly.

It is now clear that if $e(\alpha )=(1,y)$ for some $y\gt 0$ , we have $e^*(\alpha )=(1-y,0)$ , that is, $e^*_1(\alpha )\lt 1$ , as required. The result follows.

Proof. Let w.l.o.g. $\mathbb {M}=\langle W,v^+,v^-,\mu, e_1,e_2\rangle$ be a $\mathsf {BD}$ -model with $\pm$ -probability where $\mu$ is a classical probability measure and let $e(\alpha )=(x,y)$ . We show that in the $\mathsf {BD}$ -model $\mathbb {M}_{\mathbf {4}}=\langle W,v^+,v^-,\mu, e_1\rangle$ with four-probability $\mu$ and $e^{\mathbf {4}}$ induced by $\mu$ , $e^{\mathbf {4}}((\alpha ^\neg )^{\mathbf {4}})=x$ . This is sufficient to prove the result because if $\alpha ^\neg$ is not $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valid, $\mathsf {Pr}^{\mathsf {\unicode {x0141}}^2}_\triangle$ -valid $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valid either.

We proceed by induction on $\alpha ^\neg$ . For the basis case, we have that

\begin{align*} e_1(\mathsf {Pr}\phi )&=\mu (|\phi |^+)\\&=\mu (|\phi |^{\mathsf {b}}\cup |\phi |^{\mathsf {c}})\\& \begin{array}{l@{\qquad\qquad\qquad\qquad}c} =\mu (|\phi |^{\mathsf {b}})\cup \mu (|\phi |^{\mathsf {c}})\qquad & {(|\phi |^{\mathsf {b}} \text{ and } |\phi |^{\mathsf {c}} \text{ are disjoint })}\\ =e^{\mathbf {4}}(\mathsf {Bl}\phi )+e_{\mathbf {4}}(\mathsf {Cf}\phi ) & {(e^{\mathbf {4}} \text{ is induced by } \mu )}\\ =e^{\mathbf {4}}(\mathsf {Bl}\phi \oplus \mathsf {Cf}\phi )& {(e^{\mathbf {4}}(\mathsf {Bl}\phi )+e_{\mathbf {4}}(\mathsf {Cf}\phi )\leq 1)}\end{array} \end{align*}

The cases of connectives can be obtained by an application of the induction hypothesis.

Conversely, since the support of truth conditions in ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ coincide with the semantics of ${\mathsf {\unicode {x0141}}}_\triangle$ (cf. Definitions 11 and 12) and since $\alpha ^\neg$ is outer- $\neg$ -free, if $e(\alpha ^\neg )\lt 1$ for some $\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }$ -model $\mathbb {M}_{\mathbf {4}}=\langle W,v^+,v^-,\mu, e\rangle$ , then $e_1(\alpha ^\neg )\lt 1$ for $\mathbb {M}=\langle W,v^+,v^-,\mu, e_1,e_2\rangle$ with $e=e_1$ .

We proceed by induction on $\alpha ^\neg$ (recall that $e(\alpha )=e(\alpha ^\neg )$ in all $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -models). If $\alpha =\mathsf {Pr}\phi$ , then $e_1(\mathsf {Pr}\phi )=\mu (|\phi |^+)=\mu (|\phi |^{\mathsf {b}}\cup |\phi |^{\mathsf {c}})$ . But $|\phi |^{\mathsf {b}}$ and $|\phi |^{\mathsf {c}}$ are disjoint, whence $\mu (|\phi |^{\mathsf {b}}\cup |\phi |^{\mathsf {c}})=\mu (|\phi |^{\mathsf {b}})+\mu (|\phi |^{\mathsf {c}})$ , and since $\mu (|\phi |^{\mathsf {b}})+\mu (|\phi |^{\mathsf {c}})\leq 1$ , we have that $e_1(\mathsf {Bl}\phi \oplus \mathsf {Cf}\phi )=\mu (|\phi |^{\mathsf {b}})+\mu (|\phi |^{\mathsf {c}})=e_1(\mathsf {Pr}\phi )$ , as required.

The induction steps are straightforward since the semantic conditions of support of truth in ${\mathsf {\unicode {x0141}}}^2_\triangle$ coincide with the semantics of ${\mathsf {\unicode {x0141}}}_\triangle$ (cf. Definitions 12 and 11).

Proof. Assume w.l.o.g. that $\mathbb {M}=\langle W,v^+,v^-,\mu _{\mathbf {4}},e\rangle$ is a $\mathsf {BD}$ -model with a $\mathbf {4}$ -probability where $\mu _{\mathbf {4}}$ is a classical probability measure and $e(\beta )=x$ . It suffices to define a $\mathsf {BD}$ -model with $\pm$ -probability $\mathbb {M}^\pm =\langle W,v^+,v^-,\mu _{\mathbf {4}},e_1,e_2\rangle$ and show that $e_1(\beta ^\pm )=x$ . If $e(\beta )\lt 1$ , then $e(\beta ^\pm )\lt 1$ (and thus, is not valid). If $\beta ^\pm$ is not valid, we have that $e_1(\beta ^\pm )\lt 1$ by Lemma 17, whence $e(\beta )\lt 1$ , as well.

We proceed by induction on $\beta$ . If $\beta =\mathsf {Bl}\phi$ , then $e(\mathsf {Bl}\phi )=\mu _{\mathbf {4}}(|\phi |^{\mathsf {b}})$ . Now observe that $\mu _{\mathbf {4}}(|\phi |^+)=\mu (|\phi |^{\mathsf {b}}\cup |\phi |^{\mathsf {c}})=\mu _{\mathbf {4}}(|\phi |^{\mathsf {b}})+\mu _{\mathbf {4}}(|\phi |^{\mathsf {c}})$ since $|\phi |^{\mathsf {b}}$ and $|\phi |^{\mathsf {c}}$ are disjoint. But $\mu _{\mathbf {4}}(|\phi |^+)\!=\!e_1(\mathsf {Pr}\phi )$ and $\mu _{\mathbf {4}}(|\phi |^{\mathsf {c}})\!=\!\mu _{\mathbf {4}}(|\phi \!\wedge \!\neg \phi |^+)$ since $|\phi \!\wedge \!\neg \phi |^+\!=\!|\phi |^{\mathsf {c}}$ . Thus, $\mu _{\mathbf {4}}(|\phi |^{\mathsf {b}})=e_1(\mathsf {Pr}\phi \ominus \mathsf {Pr}(\phi \wedge \neg \phi ))$ as required.

Other basis cases of $\mathsf {Cf}\phi$ , $\mathsf {Uc}\phi$ , and $\mathsf {Db}\phi$ can be tackled similarly. The induction steps are straightforward since the support of truth in ${\mathsf {\unicode {x0141}}}^2_\triangle$ coincides with semantical conditions in ${\mathsf {\unicode {x0141}}}_\triangle$ .

Proof. To simplify the presentation of the proof, we will further assume that $\unicode{x1D4C1}(\mathsf {Y}\phi )=\unicode{x1D4C1}(\phi )+2$ for every $\mathsf {Y}\in \{\mathsf {Pr},\mathsf {Bl},\mathsf {Cf},\mathsf {Uc},\mathsf {Db}\}$ . We begin with $\alpha$ . From Definition 22, it is clear that $\unicode{x1D4C1}(\alpha ^\neg )={\mathscr {O}}(\unicode{x1D4C1}(\alpha ))$ . It remains to show by induction on $\alpha ^\neg$ that $\unicode{x1D4C1}((\alpha ^\neg )^{\mathbf {4}})={\mathscr {O}}(\unicode{x1D4C1}(\alpha ^\neg ))$ .

The basis case of $\alpha ^\neg =\mathsf {Pr}\phi$ is simple: $\unicode{x1D4C1}(\mathsf {Pr}\phi )=\unicode{x1D4C1}(\phi )+2$ , whence

\begin{align*} \unicode{x1D4C1}((\mathsf {Pr}\phi )^{\mathbf {4}})=\unicode{x1D4C1}(\mathsf {Bl}\phi \oplus \mathsf {Cf}\phi )=2\cdot \unicode{x1D4C1}(\phi )+7={\mathscr {O}}(\unicode{x1D4C1}(\alpha ^\neg )) \end{align*}

Here, $7$ is the sum of lengths of modalities, $\oplus$ , and outer brackets in $(\mathsf {Pr}\phi )^{\mathbf {4}}$ .

Now let $\alpha ^\neg =(\alpha _1\rightarrow \alpha _2)$ . Then $\unicode{x1D4C1}(\alpha ^\neg )=\unicode{x1D4C1}(\alpha _1)+\unicode{x1D4C1}(\alpha _2)+3$ (we count outer brackets here). Since $(\alpha _1\rightarrow \alpha _2)^{\mathbf {4}}=\alpha ^{\mathbf {4}}_1\rightarrow \alpha ^{\mathbf {4}}_2$ , we have $\unicode{x1D4C1}(\alpha ^{\mathbf {4}}_1\rightarrow \alpha ^{\mathbf {4}}_2)=\unicode{x1D4C1}(\alpha ^{\mathbf {4}}_1)+\unicode{x1D4C1}(\alpha ^{\mathbf {4}}_2)+3$ . We apply the induction hypothesis and obtain that $\unicode{x1D4C1}(\alpha ^{\mathbf {4}}_1\rightarrow \alpha ^{\mathbf {4}}_2)={\mathscr {O}}(\unicode{x1D4C1}(\alpha _1))+{\mathscr {O}}(\unicode{x1D4C1}(\alpha _2))+3={\mathscr {O}}(\unicode{x1D4C1}(\alpha _1\rightarrow \alpha _2))$ .

The cases of $\alpha =\triangle \alpha '$ and $\alpha ={\sim }\alpha '$ can be dealt with in the same manner.

Consider now $\beta \in \mathscr {L}_{\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }}$ . We show by induction on $\beta$ that $\unicode{x1D4C1}(\beta ^\pm )={\mathscr {O}}(\unicode{x1D4C1}(\beta ))$ . Assume that $\beta =\mathsf {Bl}\phi$ . Then $(\mathsf {Bl}\phi )^\pm =\mathsf {Pr}\phi \ominus \mathsf {Pr}(\phi \wedge \neg \phi )$ . Observe that $\mathsf {Pr}\phi \ominus \mathsf {Pr}(\phi \wedge \neg \phi )$ is a shorthand for ${\sim }(\mathsf {Pr}\phi \rightarrow \mathsf {Pr}(\phi \wedge \neg \phi ))$ . Thus, we have

\begin{align*} \unicode{x1D4C1}({\sim }(\mathsf {Pr}\phi \rightarrow \mathsf {Pr}(\phi \wedge \neg \phi )))=3\cdot \unicode{x1D4C1}(\phi )+12={\mathscr {O}}(\unicode{x1D4C1}(\mathsf {Bl}\phi )) \end{align*}

Note again that $12$ is the sum of the lengths of modalities, $\wedge$ , $\neg$ , $\rightarrow$ , $\sim$ , and brackets in ${\sim }(\mathsf {Pr}\phi \rightarrow \mathsf {Pr}(\phi \wedge \neg \phi ))$ . The cases of other modal atoms are similar. The cases of $\beta =\beta _1\rightarrow \beta _2$ , $\beta =\triangle \beta '$ , and $\beta ={\sim }\beta '$ can be considered in the same way as we did $(\alpha ^\neg )^{\mathbf {4}}$ . The result now follows since modalities do not nest.

Proof. Soundness can be established by the routine check of the validity of the axioms. For example for ${\mathbf {4}}\mathsf {equiv}$ , observe that if $\phi$ and $\chi$ are equivalent in $\mathsf {BD}$ , then $|\phi |^+=|\chi |^+$ and $|\phi |^-=|\chi |^-$ , whence $|\phi |^{\mathsf {x}}=|\chi |^{\mathsf {x}}$ for every $\mathsf {x}\in \{\mathsf {b},\mathsf {d},\mathsf {c},\mathsf {u}\}$ . Thus, $e(\mathsf {X}\phi )=e(\mathsf {X}\chi )$ for each $\mathsf {X}\in \{\mathsf {Bl},\mathsf {Db},\mathsf {Cf},\mathsf {Uc}\}$ , from where $\mathsf {X}\phi \leftrightarrow \mathsf {X}\chi$ is valid. ${\mathbf {4}}\mathsf {contr}$ , ${\mathbf {4}}\mathsf {neg}$ , and ${\mathbf {4}}\mathsf {mon}$ are straightforward translations of $\mathbf {contr}$ , $\mathbf {neg}$ , and $\mathbf {BCmon}$ . ${\mathbf {4}}\mathsf {ex}$ is the translation of $\pm \mathsf {ex}$ , whence are valid by Theorem24.

Last, consider ${\mathbf {4}}\mathsf {part1}$ and ${\mathbf {4}}\mathsf {part2}$ . We have $\sum \limits _{\mathsf {x}\in \{\mathsf {b},\mathsf {d},\mathsf {c},\mathsf {u}\}}\!\!\!\!|\phi |^{\mathsf {x}}=1$ . Hence, $\sum \limits _{\mathsf {X}\in \{\mathsf {Bl},\mathsf {Db},\mathsf {Cf},\mathsf {Uc}\}}\!\!\!\!e(\mathsf {X}\phi )=1$ . Thus, we have that ${\mathbf {4}}\mathsf {part1}$ and ${\mathbf {4}}\mathsf {part2}$ are valid.

Let us now prove the completeness. We reason by contraposition. Assume that $\Xi \nvdash _{\mathscr {H}\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }}\alpha$ . Now, observe that $\mathscr {H}\mathbf {4}\mathsf {Pr}^{\mathsf {\unicode {x0141}}_\triangle }$ proofs are, actually, ${\mathsf {\unicode {x0141}}}_\triangle$ proofs with additional probabilistic axioms. Let $\Xi ^*$ stand for $\Xi$ extended with probabilistic axioms built over all pairwise non-equivalent $\mathscr {L}_{\mathsf {BD}}$ formulas constructed from $\texttt {Prop}[\Xi \cup \{\alpha \}]$ . Clearly, $\Xi ^*\nvdash _{\mathscr {H}\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }}\alpha$ either. Moreover, $\Xi ^*$ is finite as well since $\mathsf {BD}$ is tabular (and whence, there exist only finitely many pairwise non-equivalent $\mathscr {L}_{\mathsf {BD}}$ formulas over a finite set of variables). Now, by the weak completeness of ${\mathsf {\unicode {x0141}}}_\triangle$ (Proposition 28), there exists an ${\mathsf {\unicode {x0141}}}_\triangle$ valuation $e$ s.t. $e[\Xi ^*]=1$ and $e(\alpha )\neq 1$ .

It remains to construct a $\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }$ -model $\Xi ^*\models _{\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}_\triangle }}}\alpha$ using $e$ . We proceed as follows. First, we set $W=2^{\mathsf {Lit}[\Xi ^*\cup \{\alpha \}]}$ , and for every $w\in W$ define $w\in v^+(p)$ iff $p\in w$ and $w\in v^-(p)$ iff $\neg p\in w$ . We extend the valuations to $\phi \in \mathscr {L}_{\mathsf {BD}}$ in the usual manner. Then, for $\mathsf {X}\phi \in \texttt {Sf}[\Xi ^*\cup \{\alpha \}]$ , we set $\mu _{\mathbf {4}}(|\phi |^{\mathsf {x}})=e(\mathsf {X}\phi )$ according to modality $\mathsf {X}$ .

It remains to extend $\mu _{\mathbf {4}}$ to the whole $2^W$ . Observe, however, that any map from $2^W$ to $[0,1]$ that extends $\mu _{\mathbf {4}}$ is, in fact, a $\mathbf {4}$ -probability. Indeed, all requirements from Definition 6 concern only the extensions of formulas. But the model is finite, $\mathsf {BD}$ is tabular, and $\Xi ^*$ contains all the necessary instances of probabilistic axioms and $e[\Xi ^*]=1$ , whence all constraints on the formulas are satisfied.

Proof. The proof follows Bílková et al. (Reference Bílková, Frittella, Kozhemiachenko, Das and Negri2021, Theorem 1). For soundness, we show that if $\langle v_1,v_2\rangle$ satisfies the premise of a rule, then it also satisfies one of its conclusions. We consider the case of $\triangle \!\geqslant _1$ (the rules for other connectives are tackled by Bílková et al. (Reference Bílková, Frittella, Kozhemiachenko, Das and Negri2021); the remaining rules for $\triangle$ can be dealt with analogously).

Let $\triangle \phi \geqslant _1i$ be satisfied by $\langle v_1,v_2\rangle$ . Then, $v_1(\triangle \phi )\geqslant i$ . We have two cases: (i) $i=0$ and (ii) $i\gt 0$ . In the first case, the left conclusion is satisfied. In the second case, $v_1(\triangle \phi )=1$ . Hence, $v_1(\phi )=1$ , that is, the right conclusion is satisfied.

For completeness, note from Bílková et al. (Reference Bílková, Frittella, Kozhemiachenko, Das and Negri2021, Proposition 1) that $\phi$ is ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ -valid iff $v_1(\phi )=1$ for every $v$ . Let us now show that every complete open branch can be satisfied. Assume that $\mathscr {B}$ is a complete open branch. We construct the satisfying valuation as follows. Let $\triangledown \!\in \!\{\leqslant _1,\geqslant _1,\leqslant _2,\geqslant _2\}$ and $p_1,\ldots, p_m$ be the propositional variables appearing in the atomic labelled formulas in $\mathscr {B}$ .

Let $\{p_1\triangledown j_1,\ldots, p_m\triangledown j_n\}$ and $\{k_1\leq l_1,\ldots, k_q\leq l_q\}$ be the sets of all atomic labelled formulas and all numerical constraints in $\mathscr {B}$ . Notice that one variable might appear in many atomic labelled formulas, hence we might have $m \neq n$ . Since $\mathscr {B}$ is complete and open, the following system of linear inequalities over the set of variables $\{x_{p_1}^L,x_{p_1}^R,\ldots, x_{p_m}^L,x_{p_m}^R\}$ must have at least one solution under the constraints listed:

\begin{equation*}\unicode{x1D4C9}(p_1\triangledown i_1),\ldots, \unicode{x1D4C9}(p_m\triangledown i_n),k_1\leq l_1,\ldots, k_q\leq l_q\end{equation*}

Let $c=(c^L_1,c^R_1,\ldots, c^L_m,c^R_m)$ be a solution to the above system of inequalities such that $c^L_j$ ( $c^R_j$ ) is the value of $x_{p_j}^L$ ( $x_{p_j}^R$ ). Define valuations as follows: $v_1(p_j)=c^L_j$ and $v_2(p_j)=c^R_j$ .

It remains to show by induction on $\phi$ that all formulas present at $\mathscr {B}$ are satisfied by $v_1$ and $v_2$ . The basis case of variables holds by the construction of $v_1$ and $v_2$ . We consider the case of $\triangle \phi \geqslant _1i$ (the cases of other variables can be dealt with similarly).

Assume that $\triangle \phi \geqslant _1i\in \mathscr {B}$ . Since $\mathscr {B}$ is complete and open, we have that either (i) $i\leq 0\in \mathscr {B}$ or (ii) $\{\phi \geqslant _1j,j\geq 1\}\subseteq \mathscr {B}$ . In the first case, since there is a solution for $\mathscr {B}^{\unicode{x1D4C9}}$ , we have that $i=0$ , and thus $\triangle \phi \geqslant _1i$ is satisfied by any $v_1$ . In the second case, we have that $\{\phi \geqslant _1j,j\geq 1\}$ is satisfied by the induction hypothesis, whence $v_1(\phi )=1$ , and thus, $v_1(\triangle \phi )=1$ , as required.

Proof. We construct $\alpha ^*$ as follows. First, we take every modal atom $\mathsf {Pr}\phi$ (recall that $\phi \in \mathscr {L}_{\mathsf {BD}}$ ) and replace $\phi$ with its $\mathsf {NNF}$ . Second, we replace every literal $\neg p$ occurring in $\mathsf {Pr}(\mathsf {NNF}(\phi ))$ with a corresponding fresh variable $p'$ . That is, $\neg p$ is replaced with $p'$ , $\neg q$ with $q'$ , etc. Outer-layer connectives remain the same. Observe, that this increases the number of symbols in $\alpha$ at most linearly: indeed, the transformation of $\phi$ into $\mathsf {NNF}(\phi )$ is linear (recall Remark 3) and then we replace every occurrence of $\neg p$ with a fresh variable $p^*$ . It remains to check that validity is preserved.

Let $\alpha$ be not $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valid, $\mathsf {Pr}^{\mathsf {\unicode {x0141}}^2}_\triangle$ -model. $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -model. By Lemma 17, this is equivalent to $e_1(\alpha )\neq 1$ in some $\mathbb {M}=\langle W,v^+,v^-,\mu, e_1,e_2\rangle$ . Now, let $\mathbb {M}^*=\langle W,{v^*}^+,{v^*}^-,\mu, e^*_1,e^*_2\rangle$ s.t. ${v^*}^+(p)=v^+(p)$ and ${v^*}^+(p')=v^-(p)$ . It suffices to show that $e_1(\alpha )=e^*_1(\alpha ^*)$ .

We proceed by induction on $\alpha$ . Let $\alpha =\mathsf {Pr}\phi$ for some $\phi \in \mathscr {L}_{\mathsf {BD}}$ . We have that $e_1(\alpha )=\mu (|\phi |^+)=\mu (|\mathsf {NNF}(\phi )|^+)$ . We check that $|\mathsf {NNF}(\phi )|^+_{\mathbb {M}}=|\mathsf {NNF}(\phi )^*|^+_{\mathbb {M}^*}$ by induction on $\mathsf {NNF}(\phi )$ .

If $\mathsf {NNF}(\phi )=p$ , then $p^*=p$ and $v^+(p)={v^*}^+(p)$ by construction. If $\mathsf {NNF}(\phi )=\neg p$ , then $(\neg p)^*=p'$ and $|\neg p|^+=v^-(p)={v^*}^+(p')$ by construction. If $\mathsf {NNF}(\phi )=\chi \wedge \psi$ , then $\mathsf {NNF}(\phi )=(\chi \wedge \psi )^*=\chi ^*\wedge \psi ^*$ . By the induction hypothesis, $|\chi |^+=|\chi ^*|^+$ and $|\psi |^+=|\psi ^*|^+$ . Hence, $|\chi \wedge \psi |^+_{\mathbb {M}}=|(\chi \wedge \psi )^*|^+_{\mathbb {M}^*}$ . The case of $\mathsf {NNF}(\phi )=\chi \vee \psi$ can be considered in the same way.

It follows that $e_1(\mathsf {Pr}\phi )=e^*_1(\mathsf {Pr}(\mathsf {NNF}(\phi )^*))$ . The cases of ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ connectives can be proven by a straightforward application of the induction hypothesis.

For the converse, assume that $\alpha ^*$ is not $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valid. Hence, $e_1(\alpha ^*)\neq 1$ in some model $\mathbb {M}=\langle W,v^+,v^-,\mu, e_1,e_2\rangle$ by Lemma 17. We define $\mathbb {M}^\bullet =\langle W,{v^\bullet }^+,{v^\bullet }^-,\mu, e^\bullet _1,e^\bullet _2\rangle$ with ${v^\bullet }^-(p)=v^+(p')$ and ${v^\bullet }^+(p)=v^+(p)$ . We can now show that $e_1(\alpha ^*)=e^\bullet _1(\alpha )$ (i.e., $e^\bullet _1(\alpha )\neq 1$ ) as in the previous case. The result is as follows.

Proof. Recall that $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ and $\mathbf {4}\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}_\triangle }$ can be linearly embedded into one another (Theorems24 and 25 and Lemma 26). Thus, it remains to provide a non-deterministic polynomial algorithm that decides whether a formula is falsifiable (i.e., non-valid) for one of these logics. We choose $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ since it has only one modality.

Let $\alpha \in \mathscr {L}_{\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle }$ be over modal atoms $\mathsf {Pr}\phi _1$ , …, $\mathsf {Pr}\phi _n$ . By Lemma 41, we can w.l.o.g. assume that $\alpha$ does $\mathsf {Pr}^{\mathsf {\unicode {x0141}}^2}_\triangle$ -valuation $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valuation for $\alpha$ .

First, we replace every modal atom $\mathsf {Pr}\phi _i$ with a fresh variable $q_{\phi _i}$ . Denote the new formula $\alpha ^-$ . It is clear that $\unicode{x1D4C1}(\alpha ^-)={\mathscr {O}}(\unicode{x1D4C1}(\alpha ))$ . We construct a $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ tableau beginning with $\{\alpha ^-\leqslant _1c,c\lt 1\}$ . As $\alpha ^-$ does not contain $\neg$ , every branch gives us an a system of linear inequalities that has a solution iff $\alpha ^-$ is ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ -falsifiable: $\alpha ^-$ is ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ -falsifiable iff at least one system of inequalities corresponding to a tableau branch has a solution. Clearly, if $\alpha ^-$ is not ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ -falsifiable, it is not $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -falsifiable either.

Now let $\mathfrak {Br}=\{\mathscr {B}_1,\ldots, \mathscr {B}_w\}$ be the set of all open branches in the $\mathscr {T}\left ({\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}\right )$ tableau for $\alpha ^-$ . We guess a $\mathscr {B}\in \mathfrak {Br}$ and consider the following system of (in)equalities:

\begin{align*} \qquad {(\mathrm {LI}(1)^{\mathscr {B}})} z_1\triangledown t_1,\ldots, z_n\triangledown t_n,k_1\leq k'_1,\ldots, k_r\leq k'_r,m_1\geq 1,\ldots, m_s\geq 1,m'_1\leq 0,\ldots, m'_t\leq 0 \end{align*}

Here, $z_i$ ’s correspond to the values of $q_{\phi _i}$ ’s in $\alpha ^-$ and $t_i$ ’s are linear polynomials that label $q_{\phi _i}$ ’s. Numerical constraints give us $k$ ’s, $k'$ ’s, $m$ ’s, and $m'$ ’s. Denote the number of inequalities and the number of variables in ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) with $l_1$ and $l_2$ , respectively. It is clear that $l_1={\mathscr {O}}(\unicode{x1D4C1}(\alpha ^-))$ and $l_2={\mathscr {O}}(\unicode{x1D4C1}(\alpha ^-))$ .

We need to check whether $z_i$ ’s are coherent as probabilities of $\phi _i$ ’s. that is that there is a probability measure $\mu$ on $2^{\texttt {Prop}(\alpha )}$ s.t. $\mu (|\phi _i|^+)=z_i$ . Recall again that because of Klein et al. (Reference Klein, Majer and Rafiee Rad2021, Theorem 4), we can assume that the probabilities are classical. Moreover, we can assume that there are $n$ propositional variables in $\alpha$ (we can always add new superfluous variables or modal atoms to $\alpha$ to make their numbers equal).

Now, introduce $2^n$ variables $u_v$ indexed by $n$ -letter words over $\{0,1\}$ . These denote whether the variables of $\phi _i$ ’s are true under $v^+$ and thus correspond to subsets of $\texttt {Prop}(\alpha )$ . For example, if $\texttt {Prop}(\alpha )=\{p_1,p_2,p_3,p_4\}$ , then $u_{1001}$ encodes $\{p_1,p_4\}$ . Note that $\neg$ does not occur in $\alpha ^-$ and thus we care only about $e_1$ and $v^+$ . Furthermore, while $n$ is the number of $\phi _i$ ’s, we can add extra modal atoms or variables to make it also the number of variables. We let $a_{i,v}=1$ when $\phi _i$ is true under $v^+$ and $a_{i,v}=0$ otherwise. Now add new equations to $\mathrm{LI}(1)^{\mathscr {B}}$ , namely

(LI) \begin{align} \sum _{v}u_v&=1&\sum _{v}(a_{i,v}\cdot u_v)&=z_i \end{align}

The new problem – ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) $\cup$ ( $\mathrm {LI}(2\exp )^{\mathscr {B}}$ ) – has a solution over $[0,1]$ iff $\alpha$ is $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -satisfiable since ( $\mathrm {LI}(2\exp )^{\mathscr {B}}$ ) encodes a measure on $\texttt {Prop}(\alpha )$ while the existence of a solution for ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) ensures that $\alpha$ is ${\mathsf {\unicode {x0141}}}^2_{(\triangle, \rightarrow )}$ -satisfiable. Furthermore, although there are $l_2+2^n+n$ variables in ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) $\cup$ ( $\mathrm {LI}(2\exp )^{\mathscr {B}}$ ), it has no more than $l_1+n+1$ (in)equalities. Thus by Fagin et al. (Reference Fagin, Halpern and Megiddo1990, Lemma 2.5), it has a solution with at most $l_1+n+1$ non-zero entries. We guess a list $L$ of at most $l_1+n+1$ words $v$ (its size is $n\cdot (l_1+n+1)$ ). We can now compute the values of $a_{i,v}$ ’s for $i\leq n$ and $v\in L$ and obtain a new system of inequalities:

(LI(2poly) ) \begin{align} \sum _{v\in L}u_v&=1&\sum _{v\in L}(a_{i,v}\cdot u_v)&=z_i \end{align}

It is clear that ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) $\cup$ ( $\mathrm {LI}(2\mathrm {poly})^{\mathscr {B}}$ ) is of polynomial size w.r.t. $\unicode{x1D4C1}(\alpha ^-)$ and thus can be solved in polynomial time. Moreover, ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) $\cup$ ( $\mathrm {LI}(2\mathrm {poly})^{\mathscr {B}}$ ) has a solution iff the values of $\mathsf {Pr}\phi _i$ ’s occurring on $\mathscr {B}$ are coherent as probabilities. If there is no solution for ( $\mathrm {LI}(1)^{\mathscr {B}}$ ) $\cup$ ( $\mathrm {LI}(2\mathrm {poly})^{\mathscr {B}}$ ), we guess another open branch from the tableau and repeat the procedure. If there is no open branch $\mathscr {B}'\in \mathfrak {Br}$ s.t. its corresponding system of inequalities $\mathrm {LI}(1)^{\mathscr {B}'}\cup \mathrm {LI}(2\mathrm {poly})^{\mathscr {B}'}$ has a solution, then $\alpha$ is $\mathsf {Pr}^{{\mathsf {\unicode {x0141}}}^2}_\triangle$ -valid as well by Lemma 41.