Logics of Synonymy

Abstract

We investigate synonymy in the strong sense of content identity (and not just meaning similarity). This notion is central in the philosophy of language and in applications of logic. We motivate, uniformly axiomatize, and characterize several “benchmark” notions of synonymy in the messy class of all possible notions of synonymy. This class is divided by two intuitive principles that are governed by a no-go result. We use the notion of a scenario to get a logic of synonymy (SF) which is the canonical representative of one division. In the other division, the so-called conceptivist logics, we find, e.g., the well-known system of analytic containment (AC). We axiomatize four logics of synonymy extending AC, relate them semantically and proof-theoretically to SF, and characterize them in terms of weak/strong subject matter preservation and weak/strong logical equivalence. This yields ways out of the no-go result and novel arguments—independent of a particular semantic framework—for each notion of synonymy discussed (using, e.g., Hurford disjunctions or homotopy theory). This points to pluralism about meaning and a certain non-compositionality of truth in logic programs and neural networks. And it unveils an impossibility for synonymy: if it is to preserve subject matter, then either conjunction and disjunction lose an essential property or a very weak absorption law is violated.

Appendix: Proofs

In this appendix, we prove the theorems stated in the main text which essentially amounts to proving the characterization theorem (Theorem 2).

All our proofs are elementary. As mentioned, a methodological novelty is that we extend a technique used by [22, 23]: proving completeness results by developing an appropriate notion of a disjunctive normal form. That is, the idea of the proof is as follows: For each notion of synonymy, we find a corresponding notion of normal form that is (i) provably equivalent according to the synonymy and (ii) if two such forms satisfy the two characterizing properties, they are identical (modulo the order of literals and disjuncts). The theorems then follow: soundness is easy, and for completeness we move to the normal form of the two given sentences with the two properties; these normal forms hence have to be the identical, whence the original sentences are provably equivalent. The unifying, constructive, and theory-independent character of this proof has been discussed in Section 6.

Recall Definition 3 collecting the logics that we’ll be working with. If ${\mathcal{L}}$ is one of these logic obtained from AC by adding an axiom φ ≡ ψ, then we refer to φ ≡ ψ as the ${\mathcal{L}}$-axiom. Also note, by construction (see Definition 2), if φ is in disjunctive form, then L(φ) = {l : l literal in φ}. So, by Lemma 1, for any φ, $L(\varphi ) = L(\varphi _{{\max \limits }})$. Thus, we can—as we find it more convenient in this appendix—work with the conception of L(φ) as the set of literals of φ when φ is a disjunctive form, or as the set of literals of $\varphi _{{\max \limits }}$ when φ is arbitrary.

1.1 Step 1: Disjunctive Normal Forms

In this section, we do the first step: providing provably equivalent notions of normal form. In Section 6, we defined standard disjunctive normal forms. Now we define such a normal form for each logic. (Item (i) is due to [23].)

Definition 4 (Disjunctive forms)

Let φ = φ₁ ∨… ∨ φ_n be in standard disjunctive normal form. We say

• (i) φ is maximal if for all i ∈{1,…,n}, and for all literals l occurring in φ, there is a j ∈{1,…,n} such that L(φ_i ∧ l) = L(φ_j). In other words, φ_i ∧ l and φ_j have the same literals, that is, they are identical modulo order and repeats of the literals. (Roughly, this means that the set of disjuncts of φ is closed under adding literals of φ.) This will be the normal form of AC.

• (ii) φ is minimal if for all i,j ∈{1,…,n} with i≠j,

  $$ L(\varphi_{i}) \not\subseteq L(\varphi_{j}) \text{ and } L(\varphi_{j}) \not\subseteq L(\varphi_{i}). $$

  This will be the normal form of SF.

• (iii) φ is maximal positive if

  • (a) For every disjunct φ_i of φ, there is an $A \subseteq At(\varphi )$ and a minimal disjunct φ₀ of φ (i.e., there is no disjunct $\varphi _{0}^{\prime }$ of φ such that $L(\varphi _{0}^{\prime }) \subsetneq L(\varphi )$) such that L(φ_i) = L(φ₀) ∪ A, and

  • (b) if φ_i is a disjunct of φ and p ∈ At(φ), then φ_i ∧ p is a disjunct of φ (modulo the order of the literals).

  This will be the normal form of SFA.

• (iv) φ is maximal literal-contradiction closed if φ is maximal and if p,¬p ∈ L(φ), then p ∧¬p is a disjunct of φ. This will be the normal form of SCL.

• (v) φ is maximal atom-contradiction closed if φ is maximal and if p ∈ At(φ), then p ∧¬p is a disjunct of φ. This will be the normal form of SCA.

As mentioned, [23] shows that every formula φ is AC-provably equivalent to a standard maximal disjunctive normal form $\varphi _{{\max \limits }}$. We show the analogue for the new normal forms and extensions of AC. For this, we’ll need the following replacement rule.

Lemma 2 (Replacement)

For$\mathcal {C} \in \{ \mathsf {AC} , \mathsf {SF} , \mathsf {SFA} \}$, the following rule is$\mathcal {C}$-admissible, that is, if the premise is$\mathcal {C}$-derivable, then the conclusion is$\mathcal {C}$-derivable. (Whenχ[φ] is a formula containing occurrences ofφ, thenχ[ψ] is the result of replacing all occurrences ofφbyψ.)

$$ \frac{\varphi \equiv \psi}{\chi[\varphi] \equiv \chi[\psi]} (R) $$

Proof

Most of the work has been done in [23]. It suffices to show that the following two rules are admissible.

$$ \frac{\varphi \equiv \psi}{\chi[\varphi] \equiv \chi[\psi]}\text{(PR)} \qquad \frac{\varphi \equiv \psi}{\neg \varphi \equiv \neg \psi}\text{(NR)} $$

where in (PR) the occurrences of φ in χ[φ] are not in the scope of ¬. The admissibility of (PR) is shown for AC by [23] and the proof also works for SF and SFA. For (NR), the proof is by induction on the proof of φ ≡ ψ. All the cases corresponding to the axioms and rules of AC are dealt with in [23]. So we only need to consider the cases where φ ≡ ψ is φ₀ ∨ (φ₀ ∧ ψ₀) ≡ φ₀ or φ₀ ∨ (φ₀ ∧ ψ₀) ≡ φ₀ ∨ (φ₀ ∧¬ψ₀). In these two cases we have to show that ¬φ ≡¬ψ is derivable. Indeed, it is easy to check that we have

$$ \begin{array}{@{}rcl@{}} \mathsf{SF} &\vdash& \neg (\varphi_{0} \vee (\varphi_{0} \wedge \psi_{0})) \equiv \neg \varphi_{0}\\ \mathsf{SFA} &\vdash& \neg (\varphi_{0} \vee (\varphi_{0} \wedge \psi_{0}))\equiv \neg (\varphi_{0} \vee (\varphi_{0} \wedge \neg \psi_{0})) \end{array} $$

using the distributivity and de Morgan axioms. □

This also holds for SCL and SCA, but this will follow immediately from the characterization theorem for these logics and we won’t need replacement for these logics in the proof. This is why we don’t prove it directly here, although it’s not too hard either.

Proposition 1 (Normal form forSF)

Every formulaφisSF-provably equivalent to a standard minimal disjunctive normal form$\varphi _{{\min \limits }}$.

Proof

As mentioned, there is a formula $\varphi ^{\prime }$ in standard disjunctive normal form that is AC-provably equivalent to φ and hence in particular SF-provably equivalent.

Next, we can delete—while preserving SF-provability—any disjunct φ_j occurring in $\varphi ^{\prime }$ if there already is a disjunct φ_i in $\varphi ^{\prime }$ with $L(\varphi _{i}) \subseteq L(\varphi _{j})$. This is because if there are such φ_j and φ_i, then, without loss of generality, φ_j = φ_i ∧ χ and (using the underlining to increase readability)

$$ \begin{array}{@{}rcl@{}} \varphi^{\prime}&=& \varphi_{1} \vee {\ldots} \vee \underline{\varphi_{i}} \vee {\ldots} \vee \underline{\varphi_{j}} \vee {\ldots} \vee \varphi_{n}\\ &=&\varphi_{1} \vee {\ldots} \vee \varphi_{i-1} \vee \underline{\varphi_{i}} \vee \varphi_{i+1} \vee {\ldots} \vee \varphi_{j-1} \vee \underline{ (\varphi_{i} \wedge \chi) } \vee \varphi_{j+1} \vee {\ldots} \vee \varphi_{n}\\ &\equiv&\underline{ \varphi_{i} \vee (\varphi_{i} \wedge \chi) }\vee\varphi_{1} \vee {\ldots} \vee \varphi_{i-1} \vee \varphi_{i+1} {\ldots} \vee \varphi_{j-1} \vee \varphi_{j+1} \vee {\ldots} \vee \varphi_{n}\\ &\equiv&\underline{ \varphi_{i} } \vee\varphi_{1} \vee {\ldots} \vee \varphi_{i-1} \vee \varphi_{i+1} {\ldots} \vee \varphi_{j-1} \vee \varphi_{j+1} \vee {\ldots} \vee \varphi_{n}\\ &\equiv&\varphi_{1} \vee {\ldots} \vee \underline{ \varphi_{i} } \vee {\ldots} \vee \varphi_{j-1} \vee \varphi_{j+1} \vee {\ldots} \vee \varphi_{n}, \end{array} $$

where we essentially used commutativity and the axiom φ ∨ (φ ∧ ψ) ≡ φ. Thus, we can reduce $\varphi ^{\prime }$ to a provably equivalent formula φ^∗ in minimal disjunctive form.

Finally, by commutativity, associativity, and idempotence we can reorder φ^∗ to make it standard (without changing minimality). Thus, we get a formula $\varphi _{{\min \limits }}$ that is provably equivalent to φ and in standard minimal disjunctive form. □

Proposition 2 (Normal form forSFA)

Every formulaφisSFA-provably equivalent to a standard maximal positive disjunctive normal formφ_pos.

Proof

As mentioned, φ is AC-provably (and hence SFA-provably) equivalent to a formula $\varphi _{{\max \limits }}$ in maximal disjunctive normal form. Let φ₁,…,φ_r be the minimal disjuncts of $\varphi _{{\max \limits }}$. Then every disjunct $\varphi ^{\prime }$ of $\varphi _{{\max \limits }}$ is of the form $\varphi ^{\prime } = \varphi _{i} \wedge L$ (modulo ordering) for an i ≤ r and a (possibly empty) set L of literals occurring in φ. By using replacement (Lemma 2), the SFA-axiom, and idempotence several times, we can SFA-provably replace each $\varphi _{i} \vee \varphi ^{\prime }$ by φ_i ∨ (φ_i ∧ At(L)) and thus end up with a formula φ^∗ that still is SFA-provably equivalent to φ.¹ Clearly, φ^∗ satisfies (a), and it also satisfies (b): Let $\varphi ^{\prime }$ be a disjunct of φ^∗ and p ∈ At(φ^∗). Then $\varphi ^{\prime } = \varphi _{i} \wedge At(L)$ for an i ≤ r and a set L of literals occurring in $\varphi _{{\max \limits }}$, and p occurs in a literal l_p of $\varphi _{{\max \limits }}$ (since $At(\varphi ^{*}) = At(\varphi _{{\max \limits }}$). Then, by the maximality of $\varphi _{{\max \limits }}$, φ_i ∧ (L ∪{l_p}) is (modulo order) a disjunct of $\varphi _{{\max \limits }}$. By our replacement process, $\varphi _{i} \wedge At(L \cup \{ l_{p} \}) = \varphi _{i} \wedge (At(L) \cup \{ p \}) = \varphi ^{\prime } \wedge p$ is a disjunct of φ^∗. Consequently, φ^∗ is a maximal positive normal form of φ. □

Proposition 3 (Normal form forSCL)

Every formulaφisSCL-provably equivalent to a maximal literal-contradiction closed disjunctive normal formφ_lcl.

Proof

Given φ, form $\varphi _{{\max \limits }}$ (which can be done in AC which is contained in SCL). Then, for all $p , \neg p \in L(\varphi _{{\max \limits }})$, add the disjunct p ∧¬p to $\varphi _{{\max \limits }}$. Call the result $\varphi ^{\prime }$ which is still SCL-provably equivalent to φ by iterated application of the SCL-axiom (and applying the transitivity rule of AC). Then, again, form $\varphi ^{\prime }_{{\max \limits }}$ which is the required φ_lcl. □

Proposition 4 (Normal form forSCA)

Every formulaφisSCA-provably equivalent to a maximal atom-contradiction closed disjunctive normal formφ_acl.

Proof

As in Proposition 3 except for adding the disjunct p ∧¬p already if p ∈ At(φ). □

In Corollary 2 below, we prove that all of these normal forms are unique. (In the case of AC this was, as mentioned, already proven by [23].)

1.2 Step 2: Characterizing Normal Forms

In this section, we do the second step of the proof: showing that if two normal forms of a synonymy satisfy the two characterizing properties of that synonymy, they are identical (modulo the order and repeats of literals and disjuncts).

We first prove a lemma that we’ll need several times. Recall that φ ⇔_FDEψ was defined as: for all four-valued valuations v, v(φ) = v(ψ).

Lemma 3

Letφandψbe two sentences in disjunctive normal form such thatφ ⇔_FDEψ. Then:

• (i) For all disjunctsφ₀ofφ there is a disjunct ψ_∗ of ψ such that $L(\psi _{*}) \subseteq L(\varphi _{0})$.

• (ii) For all disjunctsφ_i of φ that are not disjuncts of ψ, there are disjunctsφ_jandψ_ksuch that$L(\varphi _{j}) = L(\psi _{k}) \subseteq L(\varphi _{i})$andφ_jis minimal inφ (i.e. no disjunct ofφis properly contained inφ_j).

Proof

Ad (i). Assume there is a disjunct φ₀ of φ such that for all disjuncts ψ_∗ of ψ we have $L(\psi _{*}) \not \subseteq L(\varphi _{0})$. We show φ⇎_FDEψ.

Indeed, consider the four-valued valuation v defined by

$$ v(p) := \left\{\begin{array}{ll} 1 &\text{, if } p \in L(\varphi_{0}) \text{ and } \neg p \not\in L(\varphi_{0}) \\ 0 &\text{, if } p \not\in L(\varphi_{0}) \text{ and }\neg p \in L(\varphi_{0}) \\ b &\text{, if } p \in L(\varphi_{0}) \text{ and } \neg p \in L(\varphi_{0}) \\ u &\text{, if } p \not\in L(\varphi_{0}) \text{ and } \neg p \not\in L(\varphi_{0}). \end{array}\right. $$

(Cf. the correspondence between valuations and sets of literals introduced in Section 5.) Then v(φ₀) ∈{1,b} since each literal in φ₀ is either 1 or b under v. Hence v(φ) ∈{1,b} since φ is a disjunction with disjunct φ₀.

Moreover, we claim that v(ψ) ∈{0,u}. It suffices to show that for each disjunct ψ_∗ of ψ we have v(ψ_∗) ∈{0,u}. Indeed, pick such a ψ_∗. By our assumption, there is a literal l in ψ_∗ that is not in φ₀. If l = q for an atomic sentence q, then q∉L(φ₀). So if ¬q ∈ L(φ₀), then v(q) = 0, and if ¬q∉L(φ₀), then v(q) = u. Hence v(q) = v(l) ∈{0,u}. Dually, if l = ¬q for an atomic q, then v(¬q) = v(l) ∈{0,u}. Since v(l) ∈{0,u}, also v(ψ_∗) ∈{0,u} since ψ_∗ is a conjunction with conjunct l.

Ad (ii). Fix a φ_i that is not disjunct of ψ. Consider the disjuncts φ_∗ of φ such that $L(\varphi _{*}) \subseteq L(\varphi _{i})$ and φ_∗ is not a disjunct of ψ. Since there are finitely many, we can pick a $\subseteq$-minimal one, say $\varphi _{i}^{*}$, that is, $L(\varphi _{i}^{*}) \subseteq L(\varphi _{i})$ and $\varphi _{i}^{*}$ is not a disjunct of ψ and

$$ \forall j \leq n: L (\varphi_{j}) \subsetneq L(\varphi_{i}^{*}) \Rightarrow L(\varphi_{j} ) \in C(\psi), $$

(1)

where C(ψ) is the set of L(ψ₀)’s of disjuncts ψ₀ of ψ. Since φ ⇔_FDEψ, we have by (i) that there is a ψ_k (k ≤ m) such that $L(\psi _{k}) \subseteq L(\varphi _{i}^{*})$. Among the disjuncts of ψ with this property, we can choose a minimal one $L(\psi _{k}^{*}) \subseteq L(\psi _{k})$, that is,

$$ \forall j \leq m: L (\psi_{j}) \subseteq L(\psi_{k}^{*}) \text{ and } L (\psi_{j}) \subseteq L(\varphi_{i}^{*}) \Rightarrow L(\psi_{j} ) = L(\psi_{k}^{*}). $$

(2)

Since ψ ⇔_FDEφ, we have by (i) that there is a φ_r such that $L(\varphi _{r}) \subseteq L(\psi _{k}^{*})$. Again, there is a minimal disjunct $L(\varphi _{r}^{*}) \subseteq L(\varphi _{r})$ (so no disjunct of φ is properly contained in $\varphi _{r}^{*}$). Since $L (\varphi _{i}^{*}) \neq L(\psi _{k})$ (otherwise $\varphi _{i}^{*}$ would be in ψ), we have

$$ L(\varphi_{r}^{*}) \subseteq L(\varphi_{r}) \subseteq L(\psi_{k}^{*}) \subseteq L(\psi_{k}) \subsetneq L(\varphi_{i}^{*}), $$

so by (1) we have $L (\varphi _{r}^{*}) \in C(\psi )$. So by (2), $L(\varphi _{r}^{*}) = L(\psi _{k}^{*}) \subseteq L(\varphi _{i})$. □

Proposition 5 (Identity ofSF-normal form)

Let φ and ψ be two sentences in standard minimal disjunctive normal form. Then

$$ \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \text{ ~ iff ~ } \varphi = \psi. $$

Proof

For the non-trivial direction, write

$$ \begin{array}{@{}rcl@{}} \varphi &=& \varphi_{1} \vee {\ldots} \vee \varphi_{r} \vee \varphi^{\prime}_{1} \vee {\ldots} \vee \varphi^{\prime}_{s}\\ \psi &=& \psi_{1} \vee {\ldots} \vee \psi_{u} \vee \psi^{\prime}_{1} \vee {\ldots} \vee \psi^{\prime}_{v}, \end{array} $$

where φ₁,…,φ_r are exactly those disjuncts of φ that are—modulo ordering—also disjuncts of ψ (so the remaining $\varphi ^{\prime }_{1} , {\ldots } , \varphi ^{\prime }_{s}$ aren’t disjuncts of ψ). Analogously, the unprimed disjuncts of ψ occur in φ, and the primed ones don’t.

We claim that primed disjuncts are extensions of unprimed ones, that is, for all j ≤ s, $\varphi _{j}^{\prime } = \varphi _{i} \wedge L$ (modulo ordering) for an i ≤ r and a set of literals L (hence L is a subset of the literals occurring in φ). Analogously for ψ.

Indeed, fix a $\varphi _{j}^{\prime }$. Since φ ⇔_FDEψ and $\varphi _{j}^{\prime }$ is not in ψ, we have by Lemma 3(ii) that there are disjuncts φ₀ and ψ_∗ (primed or unprimed) such that $L(\varphi _{0}) = L(\psi _{*}) \subseteq L(\varphi _{j}^{\prime })$. Hence φ₀ is in ψ and $\varphi _{j}^{\prime } = \varphi _{0} \wedge L$ for $L := L(\varphi _{j}^{\prime }) \setminus L(\varphi _{0})$, which shows the claim.

Now, since φ is minimal, no disjunct can be the extension of another one, hence the set of primed disjuncts is empty. The same goes for ψ. Thus, φ and ψ really look like this:

$$ \varphi = \varphi_{1} \vee {\ldots} \vee \varphi_{r} \text{~ ~ and ~ ~} \psi = \psi_{1} \vee {\ldots} \vee \psi_{u}, $$

and recall that the φ_i’s also occur as disjuncts in ψ and vice versa. Hence {L(φ₁),…,L(φ_r)} = {L(ψ₁),…,L(ψ_u)}. Since φ and ψ are standard, their order is fixed, so φ = ψ, as wanted. □

Proposition 6 (Identity ofAC-normal form)

Let φ and ψ be two sentences in standard maximal disjunctive normal form. Then

$$ L(\varphi) = L(\psi) \text{ and } \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \text{ ~ iff ~ } \varphi = \psi. $$

Proof

For the non-trivial direction, write φ = φ₁ ∨… ∨ φ_n and ψ = ψ₁ ∨… ∨ ψ_m. By standardness, it suffices to show that

$$ C(\varphi) := \{ L (\varphi_{1}) , {\ldots} , L (\varphi_{n}) \} = \{ L (\psi_{1}) , {\ldots} , L (\psi_{m}) \} =: C(\psi). $$

Assume for contradiction that there is a φ_i (for i ≤ n) such that L(φ_i)∉C(ψ) (the other case is analogous). Since φ ⇔_FDEψ, we have by Lemma 3(ii) that there are $L(\varphi _{j}) = L(\psi _{k}) \subseteq L(\varphi _{i})$. Write L := L(φ_i) ∖ L(φ_j), so L(φ_i) = L(φ_j) ∪ L. Since φ and ψ have the same literals, the literals in L also occur in ψ. So, since φ_j is a disjunct of ψ and ψ is in maximal normal form, L(φ_j) ∪ L is a disjunct of ψ, in contradiction to L(φ_i)∉C(ψ). □

Proposition 7 (Identity ofSFA-normal form)

Let φ and ψ be two sentences in standard maximal positive disjunctive normal form. Then

$$ At(\varphi) = At(\psi) \text{ and } \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \text{ ~ iff ~ } \varphi = \psi. $$

Proof

For the non-trivial direction, write φ = φ₁ ∨… ∨ φ_n and ψ = ψ₁ ∨… ∨ ψ_m. As in Proposition 6 before, it suffices to show that C(φ) = C(ψ). Assume for contradiction that L(φ_i) violates this claim. Since φ is maximally positive we have, by clause (iii)(a) of Definition 4, that φ_i = φ₀ ∧ A for a minimal disjunct φ₀ of φ and an $A \subseteq At(\varphi )$.

We claim that L(φ₀) ∈ C(ψ). If not, then by Lemma 3(ii) there are $L(\varphi _{j}) = L(\psi _{k}) \subseteq L(\varphi _{0})$ with φ_j being minimal in φ. By minimality of φ₀, L(φ₀) = L(φ_j) = L(ψ_k) ∈ C(ψ).

Since φ_i = φ₀ ∧ A with φ₀ being a disjunct of ψ and $A \subseteq At(\varphi ) = At(\psi )$, we have, by clause (iii)(b) of Definition 4, that φ_i ∈ C(ψ), contradiction. □

Proposition 8 (Identity ofSCL-normal form)

Let φ and ψ be two sentences in standard maximal literal-contradiction closed disjunctive normal form. Then

$$ L(\varphi) = L(\psi) \text{ and } \varphi \Leftrightarrow_{\mathsf{C}} \psi \text{ ~ iff ~ } \varphi = \psi. $$

Proof

For the non-trivial direction, assume L(φ) = L(ψ) and φ ⇔_Cψ. As before, given a disjunct φ₀ of φ = φ₁ ∨… ∨ φ_n, we need to show that φ₀ also is a disjunct of ψ = ψ₁ ∨… ∨ ψ_m. For contradiction, assume otherwise. Note that,

$$ \text{for all $i \leq m$, } ~ L (\psi_{i}) \not\subseteq L(\varphi_{0}), $$

(3)

because otherwise, by maximality and L(φ) = L(ψ), φ₀ already is a disjunct of ψ. We consider two cases.

Case 1: φ₀ is not classically satisfiable. Then, since φ₀ is a conjunction of literals, there is p ∈ At(φ₀) such that p,¬p ∈ φ₀. Hence p,¬p ∈ L(φ) = L(ψ). Hence, since ψ is closed under literal-contradictions, p ∧¬p is a disjunct of ψ that is contained in φ₀, in contradiction to (3).

Case 2: φ₀ is classically satisfiable. We construct a classical valuation v : {p₀,p₁,…}→{0, 1} making φ₀ true (in signs, v⊧φ₀) and every ψ_i false (in signs, v⊮ψ_i), in contradiction to φ ⇔_Cψ.

Since φ₀ is assumed to be satisfiable, fix a valuation v₀ making it true. We construct v inductively: We perform m steps (corresponding to ψ₁,…,ψ_m) such that at the end of step i we have a valuation v_i making φ₀ true and ψ₁ ∨… ∨ ψ_i false. We can then choose v := v_m.

Step 1. Find a literal l in ψ₁ which is not in φ₀—which must exists since, by (3), $L (\psi _{1}) \not \subseteq L(\varphi _{0})$. Define v₁ to be like v₀ except that it makes l false (then v₁⊧φ₀ and v₁⊮ψ₁, as required).

Step i + 1. Assume v_i has been constructed such that v_i⊧φ₀ and v_i⊮ψ₁ ∨… ∨ ψ_i. We build v_i+ 1. We consider two cases.²

Case (a). There is a literal l in L(ψ_i+ 1) ∖(L(φ₀) ∪ L(ψ₁) ∪… ∪ L(ψ_i)). Then, as in step 1, define v_i+ 1 to be like v_i except that it makes l false (then v_i+ 1⊧φ₀ and v_i+ 1⊮ψ₁ ∨… ∨ ψ_i and v_i+ 1⊮ψ_i+ 1, as required).

Case (b). $L(\psi _{i+1}) \subseteq L(\varphi _{0}) \cup L(\psi _{1}) \cup {\ldots } \cup L(\psi _{i})$. For the remainder of the proof, we abuse notation and write the conjunctive normal form χ for the set L(χ). Thus, we have (where ${\varphi _{0}^{c}}$ denotes all literals not in φ₀):

$$ \psi_{i+1} = (\psi_{i+1} \cap \varphi_{0} ) \cup (\psi_{i+1} \cap \psi_{1} \cap {\varphi_{0}^{c}} ) \cup \ldots \cup (\psi_{i+1} \cap \psi_{i} \cap {\varphi_{0}^{c}} ). $$

Since $\psi _{i+1} \not \subseteq \varphi _{0}$, at least one of the sets $\psi _{i+1} \cap \psi _{j} \cap {\varphi _{0}^{c}}$ (for j ≤ i) is non-empty. Let j₁,…,j_r ≤ i be those j for which there is a literal $l_{j} \in \psi _{i+1} \cap \psi _{j} \cap {\varphi _{0}^{c}} \neq \emptyset$. Now, either there is a valuation w making all literals of the set $L := \{l_{j_{1}} , {\ldots } , l_{j_{r}} \}$ false, or there isn’t.

If there isn’t, $l_{j_{1}} \vee {\ldots } \vee l_{j_{r}}$ is a tautology, whence there is a p such that $p , \neg p \in L \subseteq \psi _{i+1}$. Hence, any valuation makes ψ_i+ 1 false. Thus, we choose v_i+ 1 := v_i and we have v_i+ 1⊧φ₀ and v_i+ 1⊮ψ₁ ∨… ∨ ψ_i and v_i+ 1⊮ψ_i+ 1, as required.

So assume there is a valuation $w \models \neg l_{j_{1}} \wedge {\ldots } \wedge \neg l_{j_{r}}$. Define v_i+ 1 to be like v_i except that on L it is like w. We claim v_i+ 1 has the required properties. Indeed, v_i+ 1⊧φ₀ because v_i⊧φ₀ and v_i+ 1 only differs from v_i on the set L which is disjoint from At(φ₀) (since each l ∈ L is in ${\varphi _{0}^{c}}$). Moreover, v_i+ 1⊮ψ_i+ 1 since for $l_{j_{1}} \in L$ we have $l_{j_{1}} \in \psi _{i+1}$ (by construction of L) and $v_{i+1} (l_{j_{1}}) = w(l_{j_{1}}) = 0$. Finally, we fix an j ≤ i and show that v_i+ 1⊮ψ_j. If $\psi _{i+1} \cap \psi _{j} \cap {\varphi _{0}^{c}} \neq \emptyset$. Then, by construction, l_j ∈ L, so v_i+ 1(l_j) = w(l_j) = 0, whence v_i+ 1⊮ψ_j. So assume $\psi _{i+1} \cap \psi _{j} \cap {\varphi _{0}^{c}} = \emptyset$. Since v_i⊮ψ_j, there is a literal l ∈ ψ_j such that v_i(l) = 0. Moreover, l∉φ₀ since v_i⊧φ₀. Since $\psi _{i+1} \cap \psi _{j} \cap {\varphi _{0}^{c}} = \emptyset$ we hence have $l \not \in \psi _{i+1} \supseteq L$. Hence v_i+ 1(l) = v_i(l) = 0, so v_i+ 1⊮ψ_j. □

Proposition 9 (Identity ofSCA-normal form)

Let φ and ψ be two sentences in standard maximal atom-contradiction closed disjunctive normal form. Then

$$ At(\varphi) = At(\psi) \text{ and } \varphi \Leftrightarrow_{\mathsf{C}} \psi \text{ ~ iff ~ } \varphi = \psi. $$

Proof

For the non-trivial direction, assume At(φ) = At(ψ) and φ ⇔_Cψ.

We claim that L(φ) = At(φ) ∪¬At(φ) (where ¬At(φ) := {¬p : p ∈ At(φ)}). Indeed, “$\subseteq$” is trivial, so assume l ∈ At(φ) ∪¬At(φ). Write p for the atom of the literal l. Then p ∈ At(φ). Since φ is closed under atom-contradictions, p ∧¬p is a disjunct of φ. Hence p,¬p ∈ L(φ), whence l ∈ L(φ).

Hence L(φ) = At(φ) ∪¬At(φ) = At(ψ) ∪¬At(ψ) = L(ψ). So by Proposition 8 (which applies since φ and ψ are atom- and, hence in particular, literal-contradiction closed), φ = ψ. □

1.3 Proofs of the Theorems

In this section, we put everything together and prove the theorems.

Theorem 2 (i). For all sentences φ and ψ, SF ⊩ φ ≡ ψ iff ⟦φ⟧ = ⟦ψ⟧ (iff φ ⇔_FDEψ).

Proof

Soundness (left to right). It is readily checked that if φ ≡ ψ is an axiom of SF, then φ ⇔_FDEψ. Moreover, it is also readily checked that if $\varphi ^{\prime } \equiv \psi ^{\prime }$ is the result of applying one of the rules to φ ≡ ψ, and if φ ⇔_FDEψ, that then also $\varphi ^{\prime } \Leftrightarrow _{\mathsf {FDE}} \psi ^{\prime }$.

Completeness (right to left). Assume φ ⇔_FDEψ. By Proposition 1, there are normal forms $\varphi _{{\min \limits }}$ and $\psi _{{\min \limits }}$ of φ and ψ, respectively. By soundness,

$$ \varphi_{{\min}} \Leftrightarrow_{\mathsf{FDE}} \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \Leftrightarrow_{\mathsf{FDE}} \psi_{{\min}}, $$

so, by Proposition 5, $\varphi _{{\min \limits }} = \psi _{{\min \limits }}$, so we can prove in SF that

$$ \varphi \equiv \varphi_{{\min}} \equiv \psi_{{\min}} \equiv \psi, $$

so ⊩_SFφ ≡ ψ. □

Curiously, while maximal disjunctive forms were needed for the completeness of truthmaker semantics for AC, we need minimal disjunctive forms for the completeness of scenario semantics for SF.

Theorem 2 (ii). For all sentences φ and ψ, we have

$$ AC \vdash \varphi \equiv \psi \text{ iff } \left\{\begin{array}{ll} L(\varphi_{{\max}}) = L(\psi_{{\max}}) &\text{, and} \\ \varphi \Leftrightarrow_{\mathsf{FDE}} \psi &\text{.} \end{array}\right. $$

Proof

Let’s start with the left-to-right direction. We show AC ⊩ φ ≡ ψ implies ⇔_FDEψ by contraposition (though it could also be shown directly by induction): If φ⇎_FDEψ, then, by Theorem 1, SF⊯φ ≡ ψ, so in particular AC⊯φ ≡ ψ (since SF is an extension of AC). And AC ⊩ φ ≡ ψ implies $L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})$ because of the following. If AC ⊩ φ ≡ ψ, then $\mathsf {AC} \vdash \varphi _{{\max \limits }} \equiv \varphi \equiv \psi \equiv \psi _{{\max \limits }}$. So $\varphi _{{\max \limits }}$ and $\psi _{{\max \limits }}$ are two sentences in standard maximal normal form that are AC-equivalent to φ. We know that a sentence’s standard maximal normal form is unique in AC. (A purely syntactic proof of this fact was given in [2], and [23] gave a semantic proof using his truthmaker semantics.) Hence $\varphi _{{\max \limits }} = \psi _{{\max \limits }}$, and, in particular, $L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})$. (Of course, this could also be shown directly by induction on AC-proofs.)

For the other direction, assume $L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})$ and φ ⇔_FDEψ. We have $\mathsf {AC} \vdash \varphi _{{\max \limits }} \equiv \varphi$ and $\mathsf {AC} \vdash \psi _{{\max \limits }} \equiv \psi$. Moreover, we’ve seen in the left-to-right direction that AC-equivalence entails FDE-equivalence, so we have

$$ \varphi_{{\max}} \Leftrightarrow_{\mathsf{FDE}} \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \Leftrightarrow_{\mathsf{FDE}} \psi_{{\max}}. $$

Hence, applying Proposition 6 to $\varphi _{{\max \limits }}$ and $\psi _{{\max \limits }}$ we get that $\varphi _{{\max \limits }} = \psi _{{\max \limits }}$, whence AC indeed proves $\varphi \equiv \varphi _{{\max \limits }} \equiv \psi _{{\max \limits }} \equiv \psi$. □

Further equivalences are given by the following.

Corollary 1. The following are equivalent:

• (i) AC ⊩ φ ≡ ψ.

• (ii) For every state model M, we have [φ]_M = [ψ]_M.

• (iii) [φ]_C = [ψ]_C where C is the canonical scenario model based on sets of scenarios.

Proof Proof (Sketch)

(i)⇒(ii) is the soundness theorem of [23]. (ii)⇒(iii) is trivial. So it remains to show (iii)⇒(i).

Indeed, assume [φ]_C = [ψ]_C. Since φ is AC-provably equivalent to its standard maximal disjunctive form $\varphi _{{\max \limits }}$, and ψ to $\psi _{{\max \limits }}$, we have, by the just mentioned soundness, that $[\varphi _{{\max \limits }}]_{C} = [\psi _{{\max \limits }}]_{C}$.

Analogous to [23], it is straightforward to show the following. (a) For a conjunctive form φ = l₁ ∧… ∧ l_n we have, by induction on n, that |φ|_C = {L_φ} where $L_{\varphi } = \{ s_{l_{1}} , {\ldots } , s_{l_{n}} \}$.³ (b) For a disjunctive form φ = φ₁ ∨… ∨ φ_n, $|\varphi |_{C} = \{ L_{\varphi _{1}} , {\ldots } , L_{\varphi _{n}} \}$. (c) If φ is a maximal disjunctive normal form, then |φ|_C = [φ]_C. (d) If φ and ψ are maximal standard disjunctive forms, then [φ]_C = [ψ]_C iff φ = ψ.

Hence $\varphi _{{\max \limits }} = \psi _{{\max \limits }}$. Whence AC proves $\varphi \equiv \varphi _{{\max \limits }} \equiv \psi _{{\max \limits }} \equiv \psi$. □

Theorem 2 (iii). For all sentences φ and ψ, we have

$$ \mathsf{SFA} \vdash \varphi \equiv \psi \text{ iff } \left\{\begin{array}{ll} At(\varphi) = At(\psi) &\text{, and} \\ \varphi \Leftrightarrow_{\mathsf{FDE}} \psi &\text{.} \end{array}\right. $$

Proof

The left-to-right direction is immediate by induction on SFA-proofs: For SFA-axioms φ ≡ ψ we have that At(φ) = At(ψ) and φ ⇔_FDEψ, and these two properties are preserved by the SFA-rules.

For the other direction, assume At(φ) = At(ψ) and φ ⇔_FDEψ. By Proposition 2, there are φ_pos and ψ_pos in standard maximal positive disjunctive form such that SFA ⊩ φ_pos ≡ φ and SFA ⊩ ψ_pos ≡ ψ. Moreover, we’ve seen in the left-to-right direction that SFA-equivalence entails having the same atoms and FDE-equivalence, so we have

$$ \begin{array}{@{}rcl@{}} &At(\varphi_{\text{pos}}) = At(\varphi) = At(\psi) = At(\psi_{\text{pos}}) \text{, and } \\ &\varphi_{\text{pos}} \Leftrightarrow_{\mathsf{FDE}} \varphi \Leftrightarrow_{\mathsf{FDE}} \psi \Leftrightarrow_{\mathsf{FDE}} \psi_{\text{pos}}. \end{array} $$

Hence we have by Proposition 7 that φ_pos = ψ_pos. Hence SFA indeed proves φ ≡ φ_pos ≡ ψ_pos ≡ ψ. □

Theorem 2 (iv). For all sentences φ and ψ, we have

$$ \mathsf{SCL} \vdash \varphi \equiv \psi \text{ iff } \left\{\begin{array}{ll} L(\varphi_{{\max}}) = L(\psi_{{\max}}) &\text{, and} \\ \varphi \Leftrightarrow_{\mathsf{C}} \psi &\text{.} \end{array}\right. $$

Proof

The left-to-right direction is shown by induction on SCL-proofs: That SCL ⊩ φ ≡ ψ implies φ ⇔_C is immediate, so let’s consider the subject matter condition. Concerning the axioms, we only need to check the new SCL-axiom. So given φ and $p , \neg p \in L(\varphi ) = L(\varphi _{{\max \limits }})$ we need to show L(φ) = L(φ ∨ (p ∧¬p)), that is, $L(\varphi _{{\max \limits }}) = L ([\varphi \vee (p \wedge \neg p)]_{{\max \limits }} )$. Indeed, we have $\mathsf {AC} \vdash \varphi \equiv \varphi _{{\max \limits }}$, so by AC’s rule for disjunction $\mathsf {AC} \vdash \varphi \vee (p \wedge \neg p) \equiv \varphi _{{\max \limits }} \vee (p \wedge \neg p)$. Hence AC proves $[\varphi \vee (p \wedge \neg p)]_{{\max \limits }} \equiv \varphi \vee (p \wedge \neg p) \equiv \varphi _{{\max \limits }} \vee (p \wedge \neg p)$. Since AC-synonymy implies L-identity, $L([\varphi \vee (p \wedge \neg p)]_{{\max \limits }}) = L(\varphi _{{\max \limits }} \vee (p \wedge \neg p) ) = L(\varphi _{{\max \limits }}) \cup \{p , \neg p \} = L(\varphi _{{\max \limits }})$. Concerning the rules, we only have the rules of AC of which we know that they preserve L-identity.

For the other direction, assume $L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})$ and φ ⇔_Cψ. By Proposition 3, there are φ_lcl and ψ_lcl in standard maximal literal-contradiction closed disjunctive form such that SCL ⊩ φ_lcl ≡ φ and SCL ⊩ ψ_lcl ≡ ψ. Moreover, from the left-to-right direction we know

$$ \begin{array}{@{}rcl@{}} &L(\varphi_{\text{lcl}} ) = L(\varphi ) = L(\psi ) = L(\psi_{\text{lcl}} ) \text{, and } \\ &\varphi_{\text{lcl}} \Leftrightarrow_{\mathsf{C}} \varphi \Leftrightarrow_{\mathsf{C}} \psi \Leftrightarrow_{\mathsf{C}} \psi_{\text{lcl}}. \end{array} $$

Hence we have by Proposition 8 that φ_lcl = ψ_lcl. So SCL indeed proves φ ≡ φ_lcl ≡ ψ_lcl ≡ ψ. □

Theorem 2 (v). For all sentences φ and ψ, we have

$$ \mathsf{SCA} \vdash \varphi \equiv \psi \text{ iff } \left\{\begin{array}{ll} At(\varphi) = At(\psi) &\text{, and} \\ \varphi \Leftrightarrow_{\mathsf{C}} \psi &\text{.} \end{array}\right. $$

Proof

The left-to-right direction is immediate by induction on SCA-proofs.

For the other direction, assume At(φ) = At(ψ) and φ ⇔_Cψ. By Proposition 4, there are φ_acl and ψ_acl in standard maximal atom-contradiction closed disjunctive form such that SCA ⊩ φ_acl ≡ φ and SCA ⊩ ψ_acl ≡ ψ. Moreover, from the left-to-right direction we know

$$ \begin{array}{@{}rcl@{}} &At (\varphi_{\text{acl}} ) = At (\varphi ) = At (\psi ) = At (\psi_{\text{acl}} ) \text{, and } \\ &\varphi_{\text{acl}} \Leftrightarrow_{\mathsf{C}} \varphi \Leftrightarrow_{\mathsf{C}} \psi \Leftrightarrow_{\mathsf{C}} \psi_{\text{acl}}. \end{array} $$

Hence we have by Proposition 9 that φ_acl = ψ_acl. So SCA indeed proves φ ≡ φ_acl ≡ ψ_acl ≡ ψ. □

Corollary 2 (Uniqueness of normal form)

Let$\mathcal {C}$be one of the systemsAC, SFA, SCL, SCA, or SF. Then every sentenceφhas auniquestandard disjunctive normal form$\varphi _{\mathcal {C}}$with the properties corresponding to the system$\mathcal {C}$(e.g. maximal, maximal positive, etc.).

Proof

Assume $\varphi _{\mathcal {C}}$ and $\varphi _{\mathcal {C}}^{\prime }$ are standard normal forms of φ in the system $\mathcal {C}$. Then $\mathcal {C} \vdash \varphi _{\mathcal {C}} \equiv \varphi _{\mathcal {C}}^{\prime }$. Apply the $\mathcal {C}$-soundness theorem and then the $\mathcal {C}$-characterization lemma to get $\varphi _{\mathcal {C}} = \varphi _{\mathcal {C}}^{\prime }$. □