Abstract
I develop and defend a truthmaker semantics for the relevant logic R. The approach begins with a simple philosophical idea and develops it in various directions, so as to build a technically adequate relevant semantics. The central philosophical idea is that truths are true in virtue of specific states. Developing the idea formally results in a semantics on which truthmakers are relevant to what they make true. A very natural notion of conditionality is added, giving us relevant implication. I then investigate ways to add conjunction, disjunction, and negation; and I discuss how to justify contraposition and excluded middle within a truthmaker semantics.
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Appendices
Appendix A: The Logic R
\(\textbf {R}^{\rightarrow }\) is the implicational fragment of R, axiomatised by:
 (Identity):

\(A \rightarrow A\)
 (Prefixing):

\((A \rightarrow B) \rightarrow ((C \rightarrow A) \rightarrow (C \rightarrow B))\)
 (Contraction):

\((A \rightarrow (A \rightarrow B)) \rightarrow (A \rightarrow B)\)
 (Permutation):

\((A \rightarrow (B \rightarrow C)) \rightarrow (B \rightarrow (A \rightarrow C))\)
with modus ponens, \(A, A \rightarrow B / B\) as the only rule. For ∧ and ∨, we add:
 (∧introduction):

\(((A \rightarrow B) \wedge (A \rightarrow C)) \rightarrow (A \rightarrow B \wedge C) \)
 (∧elimination):

\(A \wedge B \rightarrow A\) \(A \wedge B \rightarrow B\)
 (∨introduction):

\(A \rightarrow A \vee B\) \(B \rightarrow A \vee B\)
 (∨elimination):

\(((A \rightarrow C) \wedge (B \rightarrow C)) \rightarrow (A \vee B \rightarrow C)\)
 (Distribution):

\((A \wedge (B\vee C)) \rightarrow ((A \wedge B)\vee (A \wedge C))\)
plus adjunction, A,B/A ∧ B, as a rule. This gives the positive fragment, R^{+}. For negation, we add:
 (Contraposition):

\((A \rightarrow B) \rightarrow (\neg B \rightarrow \neg A)\)
 (Reductio):

\((A \rightarrow \neg A) \rightarrow \neg A\)
 (Double negation):

\(\neg \neg A \rightarrow A\)
The resulting logic is R. A ∘ B can be defined in R as \(\neg (A \rightarrow \neg B)\). In R^{+}, we may treat it is a primitive and add an axiom: \(A \circ B \rightarrow C \leftrightarrow (A \rightarrow (B \rightarrow C))\).
Appendix B: Semantics
We now define formal models, based on the philosophical principles for which I argued through Sections 3–7. To keep proofs brief, I will treat ∘ and ∨ as defined connectives, with A ∨ B =_{df}¬(¬A ∧¬B) and \(A \circ B =_{\text {df}} \neg (A \rightarrow \neg B)\).
Definition 1 (Models)
Models are tuples , where S and \(P \subseteq S\) are sets of states; \(\sqsubseteq \) and ≼ are partial orders on S; M is an operation on P; and V^{+} and V^{−} are functions from S to sets of primitive sentence letters. Models must satisfy the following conditions (where ‘s ≼_{P}u’ abbreviates ‘s ∈ P and s ≼ u’):
 (C1):

\(\sqsubseteq \) is a partial order with a minimal element such that any pair s,u ∈ S has a \(\sqsubseteq \)least upper bound, s ⊔ u ∈ S.
 (C2):

≼ is a partial order which distributes over ⊔: if s ≼ u then ∀t : s ⊔ t ≼ u ⊔ t.
 (C3):

If s ≼_{P}t ⊔ u, then there is a \(t^{\prime } \preceq _{P} t\) such that \(s \preceq _{P} t^{\prime } \sqcup u\).
 (C4):

If s ≼_{P}t ⊔ u and u ∈ P then u^{M} ≼ t ⊔ s^{M}.
 (C5):

p ∈ V^{+}s iff ∀u ≼_{P}s: p ∈ V^{+}u; and p ∈ V^{−}s iff ∀u ≼_{P}s: p ∈ V^{+}u.
 (C6):

For all s ∈ P: s^{MM} = s; p ∈ V^{+}s iff p∉V^{−}s^{M}; and p ∈ V^{−}s iff p∉V^{+}s^{M}.
Definition 2 (Truthmaking and falsitymaking relations)
These models synthesise ideas from [8] and [26], and many of the subsequent proofs borrow from theirs. (Given the interrelation of these moving parts, we cannot assume that their results hold without proof.) The formal treatment of ≼ using (C2)–(C5) derives from Fine, whereas the formal treatment of negation and mates is due to Routley. Fine analyses negation via ≼, whereas here it is analysed directly in terms of falsitymaking. The philosophical interpretation here differs both from Fine’s and Routley’s. Neither author mentions truthmakers in those works and [8] understands ≼ in terms of extension of a theory, rather than determinacy of states.
Lemma 1 (Disjunction)
Disjunction has the following (derived) clauses:
Proof
\(s \Vdash A \vee B\) iff \(s \Vdash \neg (\neg A \wedge \neg B)\) iff iff ∀u ≼_{P}s: or iff ∀u ≼_{P}s: \(u \Vdash A\) or \(u \Vdash B\). iff iff \(s \Vdash \neg A \wedge \neg B\) iff \(s \Vdash \neg A\) & HCode \(s \Vdash \neg A\) iff . □
Lemma 2 (Heredity)

(a)
\(s \Vdash A\) iff for all \(u \preceq _{P} s: u \Vdash A\)

(b)
iff for all
Proof
By double induction on A. The base cases follow immediately from (C5), so suppose both (a) and (b) hold for all B less complex than A and consider these cases.

A := ¬B. For (a): \(s \Vdash \neg B\) iff iff (by IH) iff \(\forall u \preceq _{P} s: u \Vdash \neg B\). For (b): iff \(s \Vdash B\) iff \(\forall u \preceq _{P} s: u \Vdash B\) (by IH) iff .

A := B ∧ C. For (a): \(s \Vdash B \wedge C\) iff \(s \Vdash B\) and \(s \Vdash C\) iff ∀u ≼_{P}s: \(u \Vdash B\) and \(u \Vdash C\) (by IH) iff \(\forall u \preceq _{P} s: u \Vdash B \wedge C\). For (b): iff or iff ∀u ≼_{P}s∀t ≼_{P}u: or (if by transitivity of ≼; only if by IH) iff .

\(A := B \rightarrow C\). For (a) lefttoright: assume \(s \Vdash B \rightarrow C\), \(t \Vdash B\), and u ≼_{P}s. Then \(s \sqcup t \Vdash C\) and u ⊔ t ≼ s ⊔ t (by C2). Now assume v ≼_{P}u ⊔ t. Then v ≼_{P}s ⊔ t (by transitivity of ≼), so \(v \Vdash C\) (by IH only if). Since v was arbitrary, we infer ∀v ≼ u ⊔ t : C, hence \(u \sqcup t \Vdash C\) (by IH if), and so \(u \Vdash B \rightarrow C\).
Righttoleft is by contraposition. Assume . So for some \(t \Vdash B\), . By hypothesis, there is some v ≼_{P}s ⊔ t such that . By (C3), there is some u ≼_{P}s such that v ≼ u ⊔ t. By hypothesis, and so .
For (b), lefttoright: assume , u ≼_{P}s, and v ≼_{P}u. Then v ≼_{P}s, hence for any v ≼_{P}u, hence for any u ≼_{P}s. Righttoleft: assume . Then since u ≼_{P}u, by definition for any u ≼_{P}s and so .
□
Corollary 1

(a) If \(s \Vdash A\) and s ≼ u then \(u \Vdash A\); and if and s ≼ u then .

(b) If s ∈ P, then the , and clauses simplify to:

(c) \(s \Vdash A\) implies \(s \Vdash A \vee B\)
Proof
For (a), assume \(s \Vdash A\), u ≼ s, and t ≼_{P}u. By transitivity of ≼, t ≼_{P}s. So by lemma 2, \(t \Vdash A\) for any t ≼_{P}u, and hence \(u \Vdash A\). The case is similar. For (b), the lefttoright directions follow by the reflexivity of ≼ (given s ∈ P) and the righttoleft directions follow directly from lemma 2. For (c), \(s \Vdash A\) iff ∀u ≼_{P}A (2) only if ∀u ≼_{P}A ∨ B (b) iff \(s \Vdash A \vee B\) (2). □
Lemma 3 (Crossover)
For all prime states s:

(a) \(s \Vdash A\)

(b) iff
Proof
By double induction on A. Assume s is prime throughout. The base cases follow immediately from (C5), so suppose both (a) and (b) hold for all B less complex than A and consider these cases.

A := ¬B. For (a): \(s \Vdash \neg B\) iff iff (by IH) iff . (b) is similar.

A := B ∧ C. For (a): \(s \Vdash B \wedge C\) iff \(s \Vdash B\) and \(s \Vdash C\) iff and (by IH) iff (1(b)). For (b): iff or iff or (by IH) iff .

\(A := B \rightarrow C\). For (a): \(s \Vdash B \rightarrow C\) iff \(\forall u \Vdash B: s \sqcup u \Vdash C\) iff \(\neg \exists u \Vdash B:\)(given s^{MM} = u) iff (1b, given s ∈ P). For (b): iff (1b) iff .
□
Corollary 2 (Star)
For any s ∈ P: \(s \Vdash \neg A\) iff and iff .
Corollary 3 (Negation)

(a) \(s \Vdash \neg A\)

(b) iff
Corollary 4 (Equivalent Models)
Let \(\Vvdash \) be the relation defined recursively by the \(\Vdash \)clauses for p, A ∧ B, and \(A \rightarrow B\), and 3(a) for ¬A. Then for any model M: \(s \Vvdash A\) iff \(s \Vdash A\).
Proof
2 is immediate from definition 1 and crossover (3). 3 follows from 2 and heredity (2). 4 follows from 3. □
2 shows that, for prime states at least, matehood can do the work of the Routley star. 3 shows that we can effectively do away with the falsitymaking relation altogether and work only with truthmaking (as reflected in 4). This fact will simplify the completeness proof.
We can now derive the expected clauses for intensional conjunction from its definition \(A \circ B =_{\text {df}} \neg (A \rightarrow \neg B)\):
Lemma 4 (Intensional conjunction)
Intensional conjunction has the following (derived) clauses:
Proof 6
For the truthmaking clause, lefttoright: suppose \(s \Vdash \neg (A \rightarrow \neg B)\) and let s^{+} ≼_{P}s. Then by definition, for some \(t \Vdash A\). By lemma 2, for some u ≼_{P}s^{+M} ⊔ t, hence by 2, \(u^{M} \Vdash B\) and by (C4), s^{+} ≼_{P}t ⊔ u^{M}. This holds for all s^{+} ≼_{P}s and so s ≼ t ⊔ u^{M}, \(t \Vdash A\), and \(u \Vdash B\). Righttoleft: suppose \(t \Vdash A\), \(u \Vdash B\), and s ≼ t ⊔ u and assume s^{+} ≼_{P}s. By (C3), s^{+} ≼ t ⊔ u^{+} for some u^{+} ≼_{P}u and so, by (C4), u^{+M} ≼_{P}s^{+M} ⊔ t. By 2, \(u^{+} \Vdash B\), so by 2, and so by 2, . Since s^{+} is prime, by definition and so \(s \Vdash \neg (A \rightarrow \neg B)\).
For the falsitymaking clause: iff \(s \Vdash A \rightarrow \neg B\) iff \(\forall u \Vdash A: s \sqcup u \Vdash \neg B\) iff . □
As expected, a truthmaker for both conjuncts truthmakes their intensional conjunction, and any determinate of a falsemaker for the intensional conjunction is a falsemaker for at least one of its disjuncts:
Corollary 5
(i) If \(s \Vdash A\) and \(s \Vdash B\) then \(s \Vdash A \circ B\); and (ii) if then, for any u ≼_{P}s, either .
Proof 7
For (i): since s ≼ s ⊔ s, \(s \Vdash A\) & HCode \(s \Vdash B\) implies \(s \Vdash A \circ B\). For (ii): suppose and u ≼_{P}s but (to show ). Then by 2, \(u \Vdash A \circ B\), by 3, \(u^{M} \Vdash A\), and so by 4, . Since u^{M} ≼_{P}u^{M} ⊔ u^{M}, (C4) implies u^{MM} ≼_{P}u^{MM} ⊔ u^{M}, hence u ≼_{P}u ⊔ u^{M} and, by 2, . □
As is usual in models for relevant logics, we define validity in terms of truth(making) at a designated state in all models. Since our models feature both truthmaking and falsitymaking relations, we might define entailment in terms of preservation of truthmakers (from premises to conclusion), or in terms of falsitymaker preservation (from conclusion to premises), or both. As it happens, it doesn’t matter which definition we adopt.
Definition 3 (Validity)
A is true on M, \(M \Vdash A\), iff in M. A is valid, ⊧A, iff true on all models. A set Γ is true on M when all its members are.
Definition 4 (Entailment)
Γ entails A on M, Γ⊧_{M}A, iff for all states s in M, \(s \Vdash A\) if \(s \Vdash {\Gamma }\). Γ 2entails A on M, \({\Gamma } {\models _{M}^{2}} A\), iff for all states s in M: (i) \(s \Vdash {\Gamma }\) only if \(s \Vdash A\) and (ii) only if . Entailment (and 2entailment) simpliciter is entailment (2entailment) on all models.
Lemma 5 (Entailment)

(a)
\(M \models A \rightarrow B\) iff A⊧_{M}B

(b)
\(A {\models ^{2}_{M}} B\) iff A⊧_{M}B
Proof
For (a): \(M \models A \rightarrow B\) iff in M, iff for all \(s \Vdash A\) in M: iff for all s in M: \(s \Vdash A\) only if \(s \Vdash B\) (since ) iff A⊧_{M}B. For (b), lefttoright is trivial. For righttoleft, assume A⊧_{M}B. Then iff ∀u ≼_{P}s: (lemma 2) iff (lemma 3) iff (since A⊧_{M}B) iff (3) iff (2), and so \(A {\models ^{2}_{M}} B\). □
Note the following corollary, which holds even in the absence of (C1):
Corollary 6 (Rule Contraposition)
\(\models A \rightarrow B\) iff \(\models \neg B \rightarrow \neg A\).
Proof
For any model M: \(\models _{M} A \rightarrow B\) iff A⊧_{M}B iff \(A {\models ^{2}_{M}}\) iff ∀s: (only if ) iff ∀s: (\(s \Vdash \neg B\) only if \(s \Vdash \neg A\)) iff ¬B⊧_{M}¬A iff \(\models _{M} \neg B \rightarrow \neg A\). □
Appendix C: Soundness and Completeness
Theorem 1 (Soundness)
If A ∈R then ⊧A
Proof
We need to show that each axiom of R is valid and that both modus ponens and adjunction preserve validity. For modus ponens, if then \(s \Vdash A\) implies , hence if then . For adjunction, if and then .
For the axioms: each is of the form \(A \rightarrow B\). So, given lemma 5, we need show only that, for any state s in any model, if \(s \Vdash A\) then \(s \Vdash B\). Identity, ∧elimination, and double negation are trivial, and ∨introduction is immediate from corollary 1(c). The remaining cases are as follows:
 Prefixing::

Suppose \(s \Vdash A \rightarrow B\) and assume \(t \Vdash C \rightarrow A\) and \(u \Vdash C\). We show \((s \sqcup t) \sqcup u \Vdash B\). By definition, we have \(t \sqcup u \Vdash A\) and so \((t \sqcup u) \sqcup s \Vdash B\). Since (t ⊔ u) ⊔ s = s ⊔ (t ⊔ u), the result follows.
 Contraction::

Suppose \(s \Vdash A \rightarrow (A \rightarrow B)\) and assume \(u \Vdash A\), to show \(s \sqcup u \Vdash B\). Then \((s \sqcup u) \sqcup u \Vdash B\) and, since (s ⊔ u) ⊔ u = s ⊔ (u ⊔ u) = s ⊔ u, the result follows.
 Permutation::

Suppose \(s \Vdash A \rightarrow (B \rightarrow C)\) and assume \(t \Vdash B\) and \(y \Vdash A\), to show \((s \sqcup t) \sqcup u \Vdash C\). Then \(s \sqcup u \Vdash B \rightarrow C\), so \((s \sqcup u) \sqcup t \Vdash C\). Since (s ⊔ u) ⊔ t = (s ⊔ t) ⊔ u, the result follows.
 ∧introduction::

Suppose \(s \Vdash (A \rightarrow B) \wedge (A \rightarrow C)\) and assume \(u \Vdash A\), to show \(s \sqcup u \Vdash B \wedge C\). Then \(s \Vdash (A \rightarrow B)\) and \(s \Vdash (A \rightarrow C)\), hence \(s \sqcup u \Vdash B\) and \(s \sqcup u \Vdash C\), and so \(s \sqcup u \Vdash B \wedge C\).
 ∨elimination::

Assume \(s \Vdash (A \rightarrow C) \wedge (B \rightarrow C)\) and \(u \Vdash A \vee B\). Then \(s \Vdash A \rightarrow C\), \(s \Vdash B \rightarrow C\). Now consider any t ≼_{P}s ⊔ u. By (C1), there is some v ≼_{P}u such that t ≼_{P}s ⊔ v. So \(v \Vdash A \vee B\), hence \(v \Vdash A\) or \(v \Vdash B\) and, either way, \(s \sqcup v \Vdash C\). Then by lemma 2, \(t \Vdash C\), for all t ≼_{P}s ⊔ u, and hence \(s \sqcup u \Vdash C\).
 Distribution::

Assume \(s \Vdash A \wedge (B\vee C)\), so that \(s \Vdash A\) and \(s \Vdash B \vee C\). Assume u ∈ P and u ≼ s. Then \(u \Vdash B\) or \(u \Vdash C\); and, given lemma 2, \(u \Vdash A\). So either \(u \Vdash A \wedge B\) or \(u \Vdash A \wedge C\), and so \(s \Vdash (A \wedge B) \wedge (A \wedge C)\).
 Contraposition::

Assume \(s \Vdash A \rightarrow \neg B\) but . Then \(\exists u \Vdash B:\). Then ∃t ≼_{P}s ⊔ u st , so \(t^{M} \Vdash A\). So \(s \sqcup t^{M} \Vdash \neg B\). By (C1), there is a v ≼_{P}u such that t ≼_{P}s ⊔ v, so by (C1), v^{M} ≼_{P}s ⊔ t^{M}, so \(v^{M} \Vdash \neg B\), so , so . Contradiction.
 Reductio::

Assume \(s \Vdash A \rightarrow \neg A\) but . Then \(s^{M} \Vdash A\), so \(s \sqcup s^{M} \Vdash \neg A\). We also have s^{M} ≼ s^{M} ⊔ s^{M} (reflexivity and idempotence of ⊔), hence s ≼ s ⊔ s^{M} (by C1 and s^{MM} = s), and so (by 2) \(s \Vdash \neg A\): contradiction.
□
Definition 5 (Deduction and Theories)
Γ ⊩_{R}A just in case there is a finite sequence of sentences A_{1},…,A_{n} where A = A_{n} and each A_{i≤n} is either in Γ, or the conjunction of some previous A_{j},A_{k} (i.e. j,k < i), or for some A_{k<i}, we have \(A_{k} \rightarrow A_{i} \in \textbf {R}\). A theoryT is any set of sentences closed under deduction: if A ∈ T and Γ ⊩_{R}A then A ∈ T. A theory is prime iff either A ∈ T or B ∈ T whenever A ∨ B ∈ T. T_{R} and P_{R} are the sets of all theories and all prime theories, respectively.
This definition of deduction is not the classical one, for we must ensure that theorems are not automatically derivable from any premises. We do not have \(B \vdash _{\textbf {R}} A \rightarrow A\), for example, even though \(A \rightarrow A\) is valid.
Lemma 6 (Prime theories)
Assume that Γ is closed under disjunction and T is a theory which does not intersect Γ. Then there is a prime theory \(T^{+} \supseteq T\) which does not intersect Γ.
Proof
A standard syntactic result for theories in R; see, e.g., Fine [8]. □
Definition 6 (Canonical model)
The canonical model \({\mathcal{M}}^{R}\) is a tuple
where T_{R}, P_{R}, and R are as above and:

\(S \sqcup _{\textbf {R}} U = \{ B \mid \exists A \in U : A \rightarrow B \in S \}\); set \(S \sqsubseteq _{\textbf {R}} U\) iff S ⊔_{R}U = U

For S ∈ P_{R}, \(S^{M_{\textbf {R}}} = \{ A \mid \neg A \notin S \}\)

\(V_{\textbf {R}}^{+}S = \{ p \mid p \in S\}\)

\(V_{\textbf {R}}^{}S = \{ p \mid \neg p \in S\}\)
The canonical model deviates from our philosophical characterisation in that (i) M_{R} here is none other than the standard canonical model Routley star and (ii) we model determination using the converse of set inclusion. Nevertheless, the approach meets the the formal requirements on a model and greatly simplifies the completeness proof.
Lemma 7 (Theory heredity)
A ∈ S iff, for all prime \(U \supseteq S\), A ∈ S.
Proof
Lefttoright is trivial. For righttoleft, assume A∉S, and let . Then Γ is disjunctionclosed (since ) and Γ ∩ S = ∅. So by lemma 6, there is a prime \(U \supseteq S\) such that Γ ∩ U = ∅, hence A∉U. Contraposing, if A ∈ U for all \(U \supseteq S\) then A ∈ S. □
Lemma 8
\({\mathcal{M}}^{\textbf {R}}\)is a model.
Proof
We need to demonstrate that each of (C1)–(C6) hold of \({\mathcal{M}}^{\textbf {R}}\):
 (C1):

We must show that (i) \(\sqsubseteq _{\textbf {R}}\) is a partial order on T_{R}; (ii) R is its minimal element; and (iii) S ⊔_{R}U ∈ T_{R} for every S,U ∈R.
(i): A standard proof (e.g. Dunn and Restall [7]) shows that the permutation, prefixing, and contraction axioms guarantee that ⊔_{R} is commutative, associative, and idempotent. It is then a standard result of order theory that defining \(S \sqsubseteq _{\textbf {R}} U\) iff S ⊔_{R}U = U results in a partial order.
(ii): If A ∈ S then, since \(A \rightarrow A \in \textbf {R}\), we have A ∈R ⊔_{R}S. If A ∈R ⊔_{R}S then \(B \rightarrow A \in \textbf {R}\) for some B ∈ S. Then B ⊩_{R}A (definition 5) and so, since S is a theory (definition 5), A ∈ S. So R ⊔_{R}S = S and hence \(\textbf {R} \sqsubseteq _{\textbf {R}} S\) for every state S.
(iii): This requires that each S ⊔_{R}U is a theory (by definition 5’s lights), which is a standard syntactic result (e.g. Fine [8]).
 (C2):

Clearly \(\supseteq \) is a partial order; we show it distributes over ⊔_{R}. Suppose \(S \supseteq U\) and A ∈ U ⊔ T. Then for some B ∈ U, \(B \rightarrow A \in T\) and B ∈ S, hence A ∈ S ⊔ T. So \(S \sqcup T \supseteq U \sqcup T\).
 (C3):

Suppose \(S \supseteq T \sqcup U\) and S is prime, and let \({\Gamma } = \{ A \mid \exists B \notin S : A \rightarrow B \in U \}\). To show Γ is closed under disjunction, suppose A_{1},A_{2} ∈Γ. Then there are B_{1},B_{2}∉S such that \(A_{1} \rightarrow B_{1}, A_{2} \rightarrow B_{2} \in U\). Since \(A_{1} \rightarrow B_{1}, A_{2} \rightarrow B_{2} \vdash _{\textbf {R}} A_{1} \vee A_{2} \rightarrow B_{1} \vee B_{2}\) and U is ⊩_{R}closed, we have \(A_{1} \vee A_{2} \rightarrow B_{1} \vee B_{2} \in U\). Since S is prime, B_{1} ∨ B_{2}∉S, hence A_{1} ∨ A_{2} ∈Γ. Moreover, Γ does not intersect T. For if there were any A ∈Γ∩ T then, by definition, there would be some B∉S such that \(A \rightarrow B \in U\), hence B ∈ S: contradiction. So, by lemma 6, there is a prime theory \(T^{+} \supseteq T\) which also does not intersect Γ. Now suppose B ∈ T^{+} ⊔ U. Then there is an A ∈ T^{+} such that \(A \rightarrow B \in U\), hence B ∈ S (else Γ and T^{+} would intersect) and so \(T^{+} \sqcup U \subseteq S\).
 (C4):

Suppose S,U ∈ P, \(S \supseteq T \sqcup U\) and A ∈ T ⊔ S^{M}. Then for some B ∈ S^{M}, \(B \rightarrow A \in T\) and hence, given contraposition, \(\neg A \rightarrow \neg B \in T\). B ∈ S^{M} implies ¬B∉S, hence ¬B∉T ⊔ U. Then ¬A∉U (else we get ¬B ∈ S), hence A ∈ U^{M}, and so \(U^{M} \supseteq T \sqcup S^{M}\).
 (C5):

Trivial, given lemma 7.
 (C6):

We show that, for all s ∈ P: (i) S^{MM} = S; (ii) p ∈ V^{+}s iff p∉V^{−}s^{M}; and (iii) p ∈ V^{−}s iff p∉V^{+}s^{M}. We use the fact that and so, since S is a theory, ¬¬A ∈ S iff A ∈ S. For (i): \(A \in S^{M_{\textbf {R}}M_{\textbf {R}}}\) iff \(\neg A \notin S^{M_{\textbf {R}}}\) iff ¬¬A ∈ S iff A ∈ S. For (ii): \(p \in V_{\textbf {R}}^{+}S\) iff p ∈ S iff ¬¬p ∈ S iff \(\neg p \notin S^{M_{\textbf {R}}}\) iff \(p \in V_{\textbf {R}}^{}S^{M_{\textbf {R}}}\). For (iii): \(p \in V_{\textbf {R}}^{}S\) iff ¬p ∈ S iff \(p \notin S^{M_{\textbf {R}}}\) iff \(p \notin V_{\textbf {R}}^{+}S^{M_{\textbf {R}}}\).
□
Lemma 9 (Truth)
For any state S in \({\mathcal{M}}^{\textbf {R}}\): \(S \Vdash A\) iff A ∈ S.
Proof
By induction on A. The base case is given by the definition of \(V_{\textbf {R}}^{+}\); so assume the result holds for all B less complex than A and consider these cases:

A := ¬B. \(S \Vdash \neg B\) iff \(U \Vdash \neg B\) for all prime \(U \supseteq S\) (lemma 2) iff for all prime \(U \supseteq S\) (corollary 2) iff B∉U^{M} for all prime \(U \supseteq S\) (by hypothesis) iff ¬B ∈ U for all prime \(U \supseteq S\) (definition of M_{R}) iff ¬B ∈ S (lemma 6).

A := B ∧ C. \(S \Vdash B \wedge C\) iff \(S \Vdash B\) and \(S \Vdash C\) iff B ∈ S and C ∈ S (by hypothesis) iff B ∧ C ∈ S (since B,C ⊩_{R}B ∧ C).

\(A := B \rightarrow C\). \(S \Vdash B \rightarrow C\) iff \(\forall U \Vdash B: S \sqcup _{\textbf {R}} U \Vdash C\) iff ∀U ∋ B : C ∈ S ⊔_{R}U (by hypothesis) iff \(B \rightarrow C \in S\) (by definition of ⊔_{R}).
□
Theorem 2 (Completeness)
If ⊧A then A ∈R.
Proof
By contraposition. Assume A∉R. Then by lemma 9, in \({\mathcal{M}}^{\textbf {R}}\). and, by lemma 8, . □
Definition 7 (Closed models)
A model is closed when, for any of its states s,u: \(s \sqcup u \Vdash A\) if \(s \Vdash A\) and \(u \Vdash A\). A is valid on the class of closed models, ⊧_{C}A, when in every closed model.
Theorem 3
RM is sound and complete on the class of closed models: A ∈RM iff ⊧_{C}A.
Proof
For soundness: suppose \(s, u \Vdash A\) in a closed model \({\mathcal{M}}\). By closure, \(s \sqcup u \Vdash A\), hence \(s \Vdash A \rightarrow A\) and so . For completeness, we show that the canonical model formed from RM in place of R is closed. Suppose S and U are RMtheories, each containing A. Since \(A \rightarrow (A \rightarrow A)\) is a theorem, \(A \rightarrow A \in S\). By construction, \(S \sqcup _{\textbf {RM}} U = \{B \mid \exists A \in U: A \rightarrow B \in S \}\) and so A ∈ S ⊔_{RM}U. □
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Jago, M. Truthmaker Semantics for Relevant Logic. J Philos Logic 49, 681–702 (2020). https://doi.org/10.1007/s10992019095339
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DOI: https://doi.org/10.1007/s10992019095339
Keywords
 Truthmaking
 Truthmaker semantics
 Relevant logic
 Relevance
 Conditionality
 Contraposition
 Nonclassical logic
 Semantics