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 nogo 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 socalled conceptivist logics, we find, e.g., the wellknown system of analytic containment (AC). We axiomatize four logics of synonymy extending AC, relate them semantically and prooftheoretically to SF, and characterize them in terms of weak/strong subject matter preservation and weak/strong logical equivalence. This yields ways out of the nogo 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 noncompositionality 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.
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Notes
To be a bit more precise: Say \(\varphi ^{\prime } = \varphi _{i} \wedge \bar {p} \wedge \neg q_{1} \wedge {\ldots } \wedge \neg q_{m}\). Then, by maximality, \(\varphi _{i} \wedge \bar {p} \wedge \neg q_{1} \wedge {\ldots } \wedge \neg q_{m1}\) is a disjunct of \(\varphi _{{\max \limits }}\), too. By the SFAaxiom, SFA proves that
$$ \begin{array}{@{}rcl@{}} (\varphi_{i} \wedge \bar{p} \wedge \neg q_{1} \wedge {\ldots} \wedge \neg q_{m1} )\vee ((\varphi_{i} \wedge \bar{p} \wedge \neg q_{1} \wedge {\ldots} \wedge \neg q_{m1}) \wedge \neg q_{m})\\ \equiv(\varphi_{i} \wedge \bar{p} \wedge \neg q_{1} \wedge {\ldots} \wedge \neg q_{m1} ) \vee((\varphi_{i} \wedge \bar{p} \wedge \neg q_{1} \wedge {\ldots} \wedge \neg q_{m1} ) \wedge q_{m}), \end{array} $$so we can replace the formula to the left of ≡, which is modulo order a subformula of \(\varphi _{{\max \limits }}\), by the formula to the right and obtain an SFAequivalent formula φ_{1}.
We continue this process with \(\varphi _{i} \wedge (\bar {p} \wedge q_{m}) \wedge \neg q_{1} \wedge {\ldots } \wedge \neg q_{m1}\) by using the disjunct \(\varphi _{i} \wedge (\bar {p} \wedge q_{m}) \wedge \neg q_{1} \wedge {\ldots } \wedge \neg q_{m2}\) that was in the original \(\varphi _{{\max \limits }}\) and still is in φ_{1}. So we can SFAprovably replace ¬q_{m− 1} by q_{m− 1} and obtain φ_{2}. We continue until we replaced all the ¬q_{j}’s by q_{j}’s.
And if this replacement process applied to another \(\varphi ^{\prime \prime } = \varphi _{i} \wedge \bar {p} \wedge \bar {\neg r}\) also requires a disjunct \(\varphi _{i} \wedge \bar {p} \wedge \neg q_{1} \wedge {\ldots } \wedge \neg q_{k} \), then we first add a copy of this disjunct to the current φ_{j} (which we SFAprovably can do by idempotence) and then use one of them to replace ¬q_{k} with q_{k}.
To not be overly tedious, we omit a fully detailed proof of this fact.
We’ll see below that it is not necessary, but we could consider separately the trivial case where already v_{i}⊮ψ_{i+ 1}. There we set v_{i+ 1} := v_{i}. By assumption, we have v_{i+ 1}⊧φ_{0} and v_{i+ 1}⊮ψ_{1} ∨… ∨ ψ_{i} ∨ ψ_{i+ 1}, as required.
Recall from Section 5 that in the canonical scenario model we defined s_{l} := {l} for literals l.
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Acknowledgements
For inspiring discussions and comments, I’m grateful to Franz Berto, Kit Fine, Peter Hawke, Johannes Korbmacher, Karolina Krzyzanowska, Michiel van Lambalgen, Hannes Leitgeb, Martin Lipman, Dean McHugh, Aybüke Özgün, Giorgio Sbardolini, Tom Schoonen, an anonymous referee of this journal, and the audiences of the workshop “Hyperintensional Logics and Truthmaker Semantics” at Ghent University and the Logic of Conceivability Seminar at the University of Amsterdam. This work is part of the research programme “Foundations of Analogical Thinking” with project number 32220017, which is financed by the Netherlands Organisation for Scientific Research (NWO).
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Appendix: Proofs
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 theoryindependent 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 = φ_{1} ∨… ∨ 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 φ_{0} 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(φ_{0}) ∪ 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.

(a)

(iv)
φ is maximal literalcontradiction 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 atomcontradiction 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 ACprovably 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.)
Proof
Most of the work has been done in [23]. It suffices to show that the following two rules are admissible.
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 φ_{0} ∨ (φ_{0} ∧ ψ_{0}) ≡ φ_{0} or φ_{0} ∨ (φ_{0} ∧ ψ_{0}) ≡ φ_{0} ∨ (φ_{0} ∧¬ψ_{0}). In these two cases we have to show that ¬φ ≡¬ψ is derivable. Indeed, it is easy to check that we have
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 formulaφisSFprovably 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 ACprovably equivalent to φ and hence in particular SFprovably equivalent.
Next, we can delete—while preserving SFprovability—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)
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 formulaφisSFAprovably equivalent to a standard maximal positive disjunctive normal formφ_{pos}.
Proof
As mentioned, φ is ACprovably (and hence SFAprovably) equivalent to a formula \(\varphi _{{\max \limits }}\) in maximal disjunctive normal form. Let φ_{1},…,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 SFAaxiom, and idempotence several times, we can SFAprovably 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 SFAprovably equivalent to φ.^{Footnote 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 formulaφisSCLprovably equivalent to a maximal literalcontradiction 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 SCLprovably equivalent to φ by iterated application of the SCLaxiom (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 formulaφisSCAprovably equivalent to a maximal atomcontradiction 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 ⇔_{FDE}ψ was defined as: for all fourvalued valuations v, v(φ) = v(ψ).
Lemma 3
Letφandψbe two sentences in disjunctive normal form such thatφ ⇔_{FDE}ψ. Then:

(i)
For all disjunctsφ_{0}ofφ 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_{j}andψ_{k}such that\(L(\varphi _{j}) = L(\psi _{k}) \subseteq L(\varphi _{i})\)andφ_{j}is minimal inφ (i.e. no disjunct ofφis properly contained inφ_{j}).
Proof
Ad (i). Assume there is a disjunct φ_{0} of φ such that for all disjuncts ψ_{∗} of ψ we have \(L(\psi _{*}) \not \subseteq L(\varphi _{0})\). We show φ⇎_{FDE}ψ.
Indeed, consider the fourvalued valuation v defined by
(Cf. the correspondence between valuations and sets of literals introduced in Section 5.) Then v(φ_{0}) ∈{1,b} since each literal in φ_{0} is either 1 or b under v. Hence v(φ) ∈{1,b} since φ is a disjunction with disjunct φ_{0}.
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 φ_{0}. If l = q for an atomic sentence q, then q∉L(φ_{0}). So if ¬q ∈ L(φ_{0}), then v(q) = 0, and if ¬q∉L(φ_{0}), 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
where C(ψ) is the set of L(ψ_{0})’s of disjuncts ψ_{0} of ψ. Since φ ⇔_{FDE}ψ, 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,
Since ψ ⇔_{FDE}φ, 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
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 ofSFnormal form)
Let φ and ψ be two sentences in standard minimal disjunctive normal form. Then
Proof
For the nontrivial direction, write
where φ_{1},…,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 ⇔_{FDE}ψ and \(\varphi _{j}^{\prime }\) is not in ψ, we have by Lemma 3(ii) that there are disjuncts φ_{0} and ψ_{∗} (primed or unprimed) such that \(L(\varphi _{0}) = L(\psi _{*}) \subseteq L(\varphi _{j}^{\prime })\). Hence φ_{0} 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:
and recall that the φ_{i}’s also occur as disjuncts in ψ and vice versa. Hence {L(φ_{1}),…,L(φ_{r})} = {L(ψ_{1}),…,L(ψ_{u})}. Since φ and ψ are standard, their order is fixed, so φ = ψ, as wanted. □
Proposition 6 (Identity ofACnormal form)
Let φ and ψ be two sentences in standard maximal disjunctive normal form. Then
Proof
For the nontrivial direction, write φ = φ_{1} ∨… ∨ φ_{n} and ψ = ψ_{1} ∨… ∨ ψ_{m}. By standardness, it suffices to show that
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 ⇔_{FDE}ψ, 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 ofSFAnormal form)
Let φ and ψ be two sentences in standard maximal positive disjunctive normal form. Then
Proof
For the nontrivial direction, write φ = φ_{1} ∨… ∨ φ_{n} and ψ = ψ_{1} ∨… ∨ 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} = φ_{0} ∧ A for a minimal disjunct φ_{0} of φ and an \(A \subseteq At(\varphi )\).
We claim that L(φ_{0}) ∈ 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 φ_{0}, L(φ_{0}) = L(φ_{j}) = L(ψ_{k}) ∈ C(ψ).
Since φ_{i} = φ_{0} ∧ A with φ_{0} 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 ofSCLnormal form)
Let φ and ψ be two sentences in standard maximal literalcontradiction closed disjunctive normal form. Then
Proof
For the nontrivial direction, assume L(φ) = L(ψ) and φ ⇔_{C}ψ. As before, given a disjunct φ_{0} of φ = φ_{1} ∨… ∨ φ_{n}, we need to show that φ_{0} also is a disjunct of ψ = ψ_{1} ∨… ∨ ψ_{m}. For contradiction, assume otherwise. Note that,
because otherwise, by maximality and L(φ) = L(ψ), φ_{0} already is a disjunct of ψ. We consider two cases.
Case 1: φ_{0} is not classically satisfiable. Then, since φ_{0} is a conjunction of literals, there is p ∈ At(φ_{0}) such that p,¬p ∈ φ_{0}. Hence p,¬p ∈ L(φ) = L(ψ). Hence, since ψ is closed under literalcontradictions, p ∧¬p is a disjunct of ψ that is contained in φ_{0}, in contradiction to (3).
Case 2: φ_{0} is classically satisfiable. We construct a classical valuation v : {p_{0},p_{1},…}→{0, 1} making φ_{0} true (in signs, v⊧φ_{0}) and every ψ_{i} false (in signs, v⊮ψ_{i}), in contradiction to φ ⇔_{C}ψ.
Since φ_{0} is assumed to be satisfiable, fix a valuation v_{0} making it true. We construct v inductively: We perform m steps (corresponding to ψ_{1},…,ψ_{m}) such that at the end of step i we have a valuation v_{i} making φ_{0} true and ψ_{1} ∨… ∨ ψ_{i} false. We can then choose v := v_{m}.
Step 1. Find a literal l in ψ_{1} which is not in φ_{0}—which must exists since, by (3), \(L (\psi _{1}) \not \subseteq L(\varphi _{0})\). Define v_{1} to be like v_{0} except that it makes l false (then v_{1}⊧φ_{0} and v_{1}⊮ψ_{1}, as required).
Step i + 1. Assume v_{i} has been constructed such that v_{i}⊧φ_{0} and v_{i}⊮ψ_{1} ∨… ∨ ψ_{i}. We build v_{i+ 1}. We consider two cases.^{Footnote 2}
Case (a). There is a literal l in L(ψ_{i+ 1}) ∖(L(φ_{0}) ∪ L(ψ_{1}) ∪… ∪ 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}⊧φ_{0} and v_{i+ 1}⊮ψ_{1} ∨… ∨ 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 φ_{0}):
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 nonempty. Let j_{1},…,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}⊧φ_{0} and v_{i+ 1}⊮ψ_{1} ∨… ∨ 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}⊧φ_{0} because v_{i}⊧φ_{0} and v_{i+ 1} only differs from v_{i} on the set L which is disjoint from At(φ_{0}) (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∉φ_{0} since v_{i}⊧φ_{0}. 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 ofSCAnormal form)
Let φ and ψ be two sentences in standard maximal atomcontradiction closed disjunctive normal form. Then
Proof
For the nontrivial direction, assume At(φ) = At(ψ) and φ ⇔_{C}ψ.
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 atomcontradictions, 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, literalcontradiction 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 ⇔_{FDE}ψ).
Proof
Soundness (left to right). It is readily checked that if φ ≡ ψ is an axiom of SF, then φ ⇔_{FDE}ψ. 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 ⇔_{FDE}ψ, that then also \(\varphi ^{\prime } \Leftrightarrow _{\mathsf {FDE}} \psi ^{\prime }\).
Completeness (right to left). Assume φ ⇔_{FDE}ψ. By Proposition 1, there are normal forms \(\varphi _{{\min \limits }}\) and \(\psi _{{\min \limits }}\) of φ and ψ, respectively. By soundness,
so, by Proposition 5, \(\varphi _{{\min \limits }} = \psi _{{\min \limits }}\), so we can prove in SF that
so ⊩_{SF}φ ≡ 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
Proof
Let’s start with the lefttoright direction. We show AC ⊩ φ ≡ ψ implies ⇔_{FDE}ψ by contraposition (though it could also be shown directly by induction): If φ⇎_{FDE}ψ, 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 ACequivalent 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 ACproofs.)
For the other direction, assume \(L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})\) and φ ⇔_{FDE}ψ. 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 lefttoright direction that ACequivalence entails FDEequivalence, so we have
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 ACprovably 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 = l_{1} ∧… ∧ l_{n} we have, by induction on n, that φ_{C} = {L_{φ}} where \(L_{\varphi } = \{ s_{l_{1}} , {\ldots } , s_{l_{n}} \}\).^{Footnote 3} (b) For a disjunctive form φ = φ_{1} ∨… ∨ 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
Proof
The lefttoright direction is immediate by induction on SFAproofs: For SFAaxioms φ ≡ ψ we have that At(φ) = At(ψ) and φ ⇔_{FDE}ψ, and these two properties are preserved by the SFArules.
For the other direction, assume At(φ) = At(ψ) and φ ⇔_{FDE}ψ. 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 lefttoright direction that SFAequivalence entails having the same atoms and FDEequivalence, so we have
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
Proof
The lefttoright direction is shown by induction on SCLproofs: 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 SCLaxiom. 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 ACsynonymy implies Lidentity, \(L([\varphi \vee (p \wedge \neg p)]_{{\max \limits }}) = L(\varphi _{{\max \limits }} \vee (p \wedge \neg p) ) = L(\varphi _{{\max \limits }}) \cup \{p , \neg p \} = L(\varphi _{{\max \limits }})\). Concerning the rules, we only have the rules of AC of which we know that they preserve Lidentity.
For the other direction, assume \(L(\varphi _{{\max \limits }}) = L(\psi _{{\max \limits }})\) and φ ⇔_{C}ψ. By Proposition 3, there are φ_{lcl} and ψ_{lcl} in standard maximal literalcontradiction closed disjunctive form such that SCL ⊩ φ_{lcl} ≡ φ and SCL ⊩ ψ_{lcl} ≡ ψ. Moreover, from the lefttoright direction we know
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
Proof
The lefttoright direction is immediate by induction on SCAproofs.
For the other direction, assume At(φ) = At(ψ) and φ ⇔_{C}ψ. By Proposition 4, there are φ_{acl} and ψ_{acl} in standard maximal atomcontradiction closed disjunctive form such that SCA ⊩ φ_{acl} ≡ φ and SCA ⊩ ψ_{acl} ≡ ψ. Moreover, from the lefttoright direction we know
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 \(\varphi _{\mathcal {C}} = \varphi _{\mathcal {C}}^{\prime }\). □
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Hornischer, L. Logics of Synonymy. J Philos Logic 49, 767–805 (2020). https://doi.org/10.1007/s10992019095375
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DOI: https://doi.org/10.1007/s10992019095375