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A Simple Logic of Concepts

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In Pietroski (2018) a simple representation language called SMPL is introduced, construed as a hypothesis about core conceptual structure. The present work is a study of this system from a logical perspective. In addition to establishing a completeness result and a complexity characterization for reasoning in the system, we also pinpoint its expressive limits, in particular showing that the fourth corner in the square of opposition (“Some_not”) eludes expression. We then study a seemingly small extension, called SMPL+, which allows for a minimal predicate-binding operator. Perhaps surprisingly, the resulting system is shown to encode precisely the concepts expressible in first-order logic. However, unlike the latter class, the class of SMPL+ expressions admits a simple procedural (context-free) characterization. Our contribution brings together research strands in logic—including natural logic, modal logic, description logic, and hybrid logic—with recent advances in semantics and philosophy of language.

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We would like to thank audiences at Indiana University and Stanford University for useful comments on this material. Thanks especially to Paul Pietroski for many discussions and very helpful comments and contributions on an earlier draft.


Supported by grant #586136 from the Simons Foundation.

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In this technical appendix we give detailed proofs of all results in the main text.

1.1 Proof of Theorem 2

To establish Theorem 2 we need to show how any question about satisfiability of a propositional formula α can be reduced to a question about whether tr(α) has a model in which it is non-empty. Let us abbreviate the latter property by saying that tr(α) “has a model.” The first point to notice is that tr(α) is always built up using monadic predicates, ∧, and ; in particular it never includes any subformula ∃R.φ. Such formulas have a special feature:

Lemma 14

Suppose \(\varphi \in {\mathscr{L}}\) is built up using only monadic predicates in \(\mathcal {A}\), ∧, and . Then if φ has a model, it has a model with just one point.


Let (M, 〚 〛) be a model such that \([\![\varphi ]\!] \neq \varnothing \). Pick a point a ∈ 〚φ〛 and consider a new model ({a}, 〚 〛a) where 〚Aa = 〚A〛 ∩{a} for all \(A\in \mathcal {A}\). We then show that 〚φa = {a} by induction on φ. □

To finish the proof of Theorem 2, consider \(\alpha \in {\mathscr{L}}^{prop}\). We know that α is satisfiable iff tr(α) has a model: if α is satisfiable, then by (1) we know tr(α) has a (one-point) model; and if tr(α) has a model, by Lemma 14 it has a one-point model, which by (1) again means that α is satisfiable. Because the length of tr(α) involves only a polynomial increase in length, we have achieved a polynomial reduction of propositional satisfiability to the problem whether a formula in \({\mathscr{L}}\) has a model.

Finally, to check whether two concepts are distinct is at least as difficult as checking whether a given concept has a model. In particular, to check that \(\varphi \in {\mathscr{L}}\) has a model it suffices to know that φ≠⊥, i.e., there is a model in which \([\![{\varphi }]\!] \neq [\![\bot ]\!] = \varnothing \).

1.2 Soundness Proof (Theorem 6)

For every model \({\mathscr{M}}\), if \({\mathscr{M}}\) satisfies the hypotheses of one of our rules (the sentence(s) above the line in Fig. 2), then \({\mathscr{M}}\) also satisfies the conclusion (below). For the rules pertaining to ∧, this amounts to basic facts about set intersections. We shall check this soundness fact for the rest of the rules.

Fix the model \({\mathscr{M}}\). We write M for the domain of this model.

First, assume that \({\mathscr{M}}\models \psi \leq \phi \). This means that \([\![\psi ]\!] \subseteq [\![\phi ]\!]\). To see that \([\![{\Downarrow }\phi ]\!] \subseteq [\![{\Downarrow }\psi ]\!] \), we may assume that 〚 ϕ〛≠. Thus, there is some xM such that 〚ϕ〛 = . For this same x (or indeed for any x), 〚ψ〛 = as well. Thus 〚ψ〛 = M. So we are done.

Second, we check that \([\![\phi ]\!]\subseteq [\![\top ]\!]\cap [\![{\Uparrow }\phi ]\!]\). Again, we may assume that 〚ϕ〛≠. In this case, 〚 ϕ〛 = M, and 〚⊤〛 = M always. So 〚⊤〛 ∩ 〚 ϕ〛 = M, and this is a superset of every set.

Third, we check that 〚ϕϕ〛 is independent of ϕ; indeed, it is always . For if 〚ϕ〛 = , then this is clear. And if 〚ϕ〛≠, them 〚 ϕ〛 = . So in this case, 〚ϕ ϕ〛 = 〚ϕ〛 ∩ = .

Fourth, we check that 〚ϕ〛 and 〚ϕ〛 are the same set. If 〚ϕ〛 = , then 〚ϕ〛 = M. So 〚 ϕ〛 = (since M), and thus 〚ϕ〛 = M. In the other case, 〚ϕ〛≠. This time, 〚ϕ〛 = . So 〚ϕ〛 = M, and 〚ϕ〛 = . Either way, we have shown what we want.

In the fifth rule, we assume that \([\![\phi ]\!] \subseteq [\![\psi ]\!]\), and we prove that \([\![\exists R.\phi ]\!] \subseteq [\![\exists R.\psi ]\!]\). For this, let x ∈ 〚∃R.ϕ〛. Then there is some y such that 〈x,y〉∈ 〚R〛 and y ∈ 〚ϕ〛. Since y ∈ 〚ϕ〛, we also have y ∈ 〚ψ〛. And thus y is a witness showing that x ∈ 〚∃R.ψ〛.

In the final rule, suppose that x ∈ 〚∃R.ϕ〛. Let y ∈ 〚ϕ〛 be such that 〈x,y〉∈ 〚R〛. Then y shows that 〚ϕ〛≠. Thus, 〚 ϕ〛 = M. And from this, we know that \([\![\exists R.\phi ]\!] \subseteq [\![{\Uparrow }\psi ]\!]\).

At this point, we have checked the soundness of the individual rules. We show by induction on formal proofs that if Γ ⊩ ϕ = ψ, then for all models \({\mathscr{M}}\), such that \({\mathscr{M}}\models {\Gamma }\), 〚ϕ〛 = 〚ψ〛. The base case of the induction is when Γ itself contains ϕ = ψ; this is immediate. We have induction steps for the rules in Fig. 2—these follow from our work just above—and for the (cases) rule. For this, assume that we have a derivation Γ ⊩ ϕ = ψ justified by (cases) at the root. We have Γ ∪{χ = ⊥}⊩ ϕ and also Γ ∪{χ≠⊥}⊩ ϕ = ψ. Let \({\mathscr{M}}\models {\Gamma }\). We show that 〚ϕ〛 = 〚ψ〛 in \({\mathscr{M}}\). There are two cases: 〚χ〛 = , and 〚χ〛≠. In the first case, we recall our assumption that Γ ∪{χ = ⊥}⊩ ϕ = ψ by a proof shorter than the given proof. So by induction hypothesis, Γ ∪{χ = ⊥}⊧ϕ = ψ. Our model \({\mathscr{M}}\) satisfies all sentences in Γ ∪{χ = ⊥}, hence it does satisfy ϕ = ψ. In the other case, when 〚χ〛≠, we have 〚 χ〛 = M. So \({\mathscr{M}}\) will satisfy all sentences in Γ ∪{χ≠⊥} in this case. And again, the induction hypothesis implies that \({\mathscr{M}}\models \phi = \psi \).

1.3 Completeness Proof (Theorem 6)

We first record some useful facts about the proof calculus.

Lemma 15

The following are all guaranteed by our calculus:

  1. 1.

    ⊥≤ φ

  2. 2.

    If φψ and ψχ, then φψ.

  3. 3.

    If φψ1 and φψ2, then φ ≤ (ψ1ψ2).

  4. 4.

    ⊤∧ ϕ = ϕ.

  5. 5.

    If ϕψ, then ϕψ.

  6. 6.

    ϕ = ϕ = ϕ.

  7. 7.

    ϕ = ϕ.

  8. 8.

    ⊤ = ⊤.

  9. 9.

    ⊤ = ⊥.

  10. 10.

    ⊥ = ⊥.


For (1), we use all three rules for ∧, and the rule that allows us to replace ⊥ with ψψ for any ψ whatsoever (including φ):

$$ \begin{array}{@{}rcl@{}} \bot \wedge \varphi & = & (\varphi \wedge \Downarrow \varphi) \wedge \varphi \\ & = & ({\Downarrow} \varphi \wedge \varphi) \wedge \varphi \\ & = & {\Downarrow} \varphi \wedge (\varphi \wedge \varphi) \\ & = & {\Downarrow} \varphi \wedge \varphi \\ & = & \varphi \wedge {\Downarrow} \varphi \\ & = & \bot \end{array} $$

For (2), ϕχ = (ϕψ) ∧ χ = ϕ ∧ (ψχ) = ϕψ = ϕ.

For (3), let φψ1 = φ and φψ2 = φ. Then φ ∧ (ψ1ψ2) = (φψ1) ∧ ψ2 = φψ2 = φ.

For (4), ⊤∧ ϕϕ because (⊤∧ ϕ) ∧ ϕ = ⊤∧ (ϕϕ) = ⊤∧ ϕ. And in the other direction ϕ ≤⊤ and ϕϕ, so ϕ ≤⊤∧ ϕ by part (3).

For (5), assume that ϕψ. We know that ψϕ. So ϕψ. Since ϕ = ϕ, and similarly for ψ, we see that ϕ ψ.

(6) is an immediate consequence of the definition of ϕ as ϕ.

For (7), ϕ = (ϕ) = (ϕ) = ϕ = ϕ.

For (8), use (6) to calculate: ⊤ = ⊥ = ⊥ = ⊤. Another proof: ⊤≤⊤≤⊤.

For (9), first note that ⊤∧⊤ = ⊥, by definition of ⊥ and also using the law ϕϕ = ψψ. In addition, by (4), ⊤∧⊤ =⊤. Thus ⊤ = ⊤∧⊤ = ⊥.

For (10), ⊥ = ⊥ = ⊤ = ⊥. We used the definition of ⊤ as ⊥ and also part (9). Another proof: ⊥≤⊤, and so ⊤ =⊤≤⊥, using part (8). But then ⊥ =⊥∧⊤ = ⊤, using part (4) and the definition of ≤. □

Proof of Theorem 6 and Corollary 7

At this point we fix a set Γ and an equation ϕ = ψ such that Γ⊯ϕ = ψ. We show that there is a model \(\mathcal {N}\) of Γ where 〚ϕ〛≠〚ψ〛. Going forward, we shall assume that Γ is a finite set, so that we can establish the polynomial-time decidability of the consequence relation. That is, we shall find a model \(\mathcal {N}\) whose size is polynomial in the size of Γ ∪{ϕ,ψ}. The same proof, minus all of the finiteness considerations, proves the completeness of the logic. Since the completeness is easier, we omit the details.

Definition 5

Fix Γ, ϕ and ψ. We take the set of relevant predicates to be those χ which occur in Γ, ϕ, or ψ. On the assumption that Γ is finite, there are only finitely many relevant predicates, and indeed the number of them is polynomial in the size of Γ ∪{ϕ = ψ}. Every sub-concept of a relevant predicate is itself relevant. List the relevant predicates as χ1,…,χK− 1.

Expanding Γ

We expand Γ to a larger set of equations which also does not derive ϕ = ψ, and which has the following additional property: for all relevant predicates χ, either Γ contains χ = ⊥, or Γ contains χ≠⊥. We do this by a step-by-step construction, building sets Γn for 0 ≤ nK. We start with Γ0 = Γ. For each n, let χn be the n th relevant predicate. If Γn ∪{χn = ⊥}⊯ϕ = ψ, we set Γn+ 1 to be Γn ∪{χn = ⊥}. Otherwise, Γn ∪{χn = ⊥}⊩ ϕ = ψ, and we must have Γn ∪{χn≠⊥}⊯ϕ = ψ. [For if not, Γnϕ = ψ by (cases).] We set Γn+ 1 to be Γn ∪{χn≠⊥} in this case. This builds sets \({\Gamma } = {\Gamma }_{0} \subseteq {\Gamma }_{1} \subseteq {\cdots } \subseteq {\Gamma }_{K}\).

Replace Γ by ΓK

Note that Γ and ΓK have the same predicates. So the notion of a relevant predicate is the same if we expand Γ to ΓK. Moreover, ΓKϕ = ψ. The rest of this proof produces a model \(\mathcal {N}\) of ΓK where 〚ϕ〛≠〚ψ〛, and such a model \(\mathcal {N}\) is a model of Γ as well. To save on notation, we simply replace our original set Γ with ΓK. In other words, we assume that the original set Γ came to us expanded as in the last paragraph.

A Point on on Γ

We claim that Γ⊯⊤≤⊥. For if Γ ⊩⊤≤⊥, then for all χ, Γ ⊩⊥≤ χ ≤⊤≤⊥, and so Γ ⊩ χ = ⊥. In particular, for the concepts ϕ and ψ which we fixed above, we would have Γ ⊩ ϕ = ⊥ = ψ, contrary to what we assumed at the outset.

The Model \(\mathcal {N}\)


$$ N = \{\phi: \text{$\phi$ is a relevant predicate, and }{\Gamma}\vdash \phi \neq \bot\}. $$

We interpret our language on this set N as follows: For a monadic predicate A,

$$ [\![A]\!] = \{\phi\in N : {\Gamma}\vdash \phi \leq A\}. $$

For each binary R, we set

$$ [\![R]\!] = \{\langle{\phi,\psi\rangle}\in N\times N : {\Gamma}\vdash\phi\leq \exists R. \psi\}. $$

This turns our set N into a model \(\mathcal {N}\). Please note that we need not have ∃R.ψ in N in order to have 〈ϕ,ψ〉∈ 〚R〛 in the model.

Lemma 16 (Truth Lemma for \(\mathcal {N}\))

For all relevant ψ (not necessarily in N),

$$ [\![\psi]\!]= \{\phi\in N: {\Gamma}\vdash \phi \leq \psi\}. $$


By induction on ψ. For ψ a basic predicate A, the statement in (8) is the definition in (10).

Consider a conjunction ψ, say ψ1ψ2, where ψN and therefore ψ1 and ψ2 also belong to N. We reason as follows:

$$ \begin{array}{@{}rcl@{}} [\![\psi_{1}\wedge\psi_{2}]\!] & = & [\![\psi_{1}]\!]\cap[\![\psi_{2}]\!] \\ & = & \{\phi\in N: {\Gamma}\vdash \phi \leq \psi_{1}\} \cap \{\phi\in N: {\Gamma}\vdash \phi \leq \psi_{2}\} \\ & = & \{\phi\in N: {\Gamma}\vdash \phi \leq \psi_{1}\text{ and } {\Gamma}\vdash \phi \leq \psi_{2}\} \\ & = & \{\phi\in N: {\Gamma}\vdash \phi \leq \psi_{1}\wedge \psi_{2}\} \end{array} $$

where the last line follows from Lemma 15, part (3).

The next induction step is for a relevant predicate ψ. Assume (8) for ψ. Since ψ is relevant, so is ψ. This time we prove that

$$ [\![{\Downarrow}\psi]\!]= \{\phi\in N: {\Gamma}\vdash \phi \leq {\Downarrow}\psi\}. $$

The two cases are: Γ ⊩ ψ = ⊥, and Γ ⊩ ψ≠⊥. (Notice that we used the fact that Γ contains either ψ = ⊥ or ψ≠⊥. We arranged this by expanding Γ.)

In the first case, Γ ⊩ ψ = ⊥. We claim that 〚ψ〛 = . [Here is the proof: If χ ∈ 〚ψ〛, then we use the induction hypothesis to see that Γ ⊩⊤≤χψ⊥ = ⊥. And this contradicts our earlier assumption that Γ⊯⊤≤⊥.] Since 〚ψ〛 = , 〚ψ〛 = N. We evaluate the set on the right in (9). Let ϕN. Then Γ ⊩ ϕ ≤⊤ = ⊥ = ψ. This shows that the set on the right in (9) is all of N, as desired.

In the second case, ψN, and indeed ψ ∈ 〚ψ〛 by induction hypothesis. And so 〚ψ〛 = . We claim that the set on the right in (9) is empty. For assume not, and let ϕN have Γ ⊩ ϕψ. Since ϕN, Γ ⊩⊤≤ϕ. We also have Γ ⊩⊤≤ψ, and therefore ψ⊤ = ⊥. Suppose towards a contradiction that Γ ⊩ ϕψ. Then

$${\Gamma}\vdash \top\leq {\Uparrow}\phi\leq {\Uparrow}{\Downarrow}\psi = {\Downarrow}{\Uparrow}\psi\leq\bot. $$

But recall that we are assuming about Γ that Γ⊯⊤≤⊥. This contradiction confirms that the set on the right in (9) is empty in this case.

The last induction step is for ∃R. Assume that ∃R.ψ is relevant. So ψ also is relevant. We have (8) for ψN, and prove that

$$ [\![\exists R.\psi]\!]= \{\phi\in N: {\Gamma}\vdash \phi \leq \exists R.\psi\}. $$

First, let ϕ ∈ 〚∃R.ψ〛. Then there is some χ ∈ 〚ψ〛 such that 〈ϕ,χ〉∈ 〚R〛. Then χN and Γ ⊩ ϕ ≤∃R.χ. By induction hypothesis on ψ, Γ ⊩ χψ. Hence by the logic, Γ ⊩∃R.χ ≤∃R.ψ. Thus, Γ ⊩ ϕ ≤∃R.ψ. This is half of (10).

In the other direction, let ϕN have Γ ⊩ ϕ ≤∃R.ψ. Since ϕN, Γ ⊩⊤≤ϕ. By the monotonicity law in the logic, Γ ⊩ϕR.ψ. Using the logic, Γ ⊩R.ψψ. Putting these together, Γ ⊩⊤≤ψ. Since ψ is relevant, we now have ψN. We have Γ ⊩ ψψ, and so by induction hypothesis, ψ ∈ 〚ψ〛. Since ϕ and ψ belong to N and 〚R〛 is given by (7), we have 〈ϕ,ψ〉∈ 〚R〛. Thus, ϕ ∈ 〚∃R.ψ〛, as desired.

This completes the proof. □

Lemma 17

\(\mathcal {N}\) satisfies every equation in γ = δ in Γ. That is, \(\mathcal {N}\models {\Gamma }\).


Fix such an equation, and note that γ and δ are relevant by Definition 5. Let ϕ ∈ 〚γ〛. Then ϕN, and by the Truth Lemma for \(\mathcal {N}\), Γ ⊩ ϕγ. But also Γ ⊩ γδ, and so Γ ⊩ ϕδ. Since ϕN, we use the Truth Lemma for \(\mathcal {N}\) again, this time to see that ϕ ∈ 〚δ〛. This for all ϕ shows that \([\![\gamma ]\!] \subseteq [\![{\delta }]\!]\). The converse is similar, and we conclude that \(\mathcal {N}\models \gamma = \delta \). □

Lemma 18

For all predicates γ, if Γ ⊩ γ = ⊥, then 〚γ〛 = . For all relevant predicates γ, if Γ ⊩ γ≠⊥, then 〚γ〛≠.


The first assertion just comes from the fact that 〚⊥〛 = in every model of Γ. And since we know from Lemma 17 that \(\mathcal {N}\) is a model of Γ, we see from soundness that 〚γ〛 = 〚⊥〛 = .

For the second assertion, let γ be relevant with Γ ⊩ γ≠⊥. Then γN. By the Truth Lemma for \(\mathcal {N}\) and the fact that Γ ⊩ γγ, we see that γ ∈ 〚γ〛. In particular, 〚γ〛≠. □

In the next lemma, recall that our standing assumption in this proof is that Γ⊯ϕ = ψ.

Lemma 19

\(\mathcal {N}\not \models \phi ^{*}= \psi ^{*}\).


Before we start, let us recall the definition of N in (5) and also Definition 5. Note that both ϕ and ψ are relevant. Recall our construction began by arranging that for relevant χ, either Γ contains (and thus derives) χ = ⊥ or else Γ ⊩ χ≠⊥. We thus have four cases, depending on whether Γ ⊩ ϕ = ⊥ or Γ ⊩ ϕ≠⊥, and similarly for ψ. In two of these cases, we shall show that \(\mathcal {N}\not \models \phi ^{*}= \psi ^{*}\), and in the other two we derive a contradiction to the standing assumption in this completeness theorem that Γ ⊩ ϕψ.

First, if Γ ⊩ ϕ = ⊥ and also Γ ⊩ ψ = ⊥, then we easily have our contradiction Γ ⊩ ϕ = ψ.

Second, suppose that Γ ⊩ ϕ = ⊥ but that Γ ⊩ ψ≠⊥. In this case, Lemma 18 shows that in \(\mathcal {N}\), 〚ϕ〛 = and 〚ψ〛≠. This tells us that \(\mathcal {N}\models \phi ^{*}\neq \psi ^{*}\), as desired.

Mutatis mutandis, we obtain the same conclusion \(\mathcal {N}\models \phi ^{*}\neq \psi ^{*}\) in the case that Γ ⊩ ϕ≠⊥ but Γ ⊩ ψ = ⊥.

Finally, suppose that Γ ⊩ ϕ≠⊥ and also Γ ⊩ ψ≠⊥. Thus, both ϕ and ψ belong to the set N which underlies our model. Since Γ ⊩ ϕϕ, the Truth Lemma implies that ϕ∈ 〚ϕ〛. Similarly ψ∈ 〚ψ〛. Suppose towards a contradiction that \(\mathcal {N} \models \phi ^{*}= \psi ^{*}\). Since 〚ϕ〛 = 〚ψ〛, ϕ∈ 〚ψ〛 and ψ∈ 〚ϕ〛. By the Truth Lemma again, Γ ⊩ ϕψϕ. This contradicts Γ⊯ϕ = ψ. □

Concluding the Proof of Completeness

We have shown that if Γ⊯ϕ = ψ, then there is a model of Γ where ϕ = ψ is false. This is the completeness of the proof system.

On Complexity

Our foregoing work also shows that if Γ has any model whatsoever in which ϕ = ψ fails, then it has such a model whose universe is a subset of S, where S is the set of predicates that appear in Γ ∪{ϕ = ψ}. And the size of S is polynomially related to the size of Γ ∪{ϕ = ψ}. This is behind the complexity assertion that the relation “Γ⊯ϕ = ψ” is in the class np of problems decidable in non-deterministic polynomial time.

1.4 Proof of Theorem 10

Recall that we assume in our language \({\mathscr{L}}^{+}\) that we have a predicate variable X for every first-order variable \(x \in {\mathscr{L}}^{FO}\). We use this correspondence freely in what follows.

For the translations τx as defined in Section 4.2, we show Theorem 10 by induction on the structure of φ. For the first base case we simply have \({\mathscr{M}},f \models \tau _{x}(A)\) iff f(x) ∈ 〚A〛. For the second base case we have \({\mathscr{M}},f \models \tau _{x}(Y)\) iff \({\mathscr{M}},f \models x=y\) iff f(x) = f(y) iff f(x) ∈{gf(Y )} iff \(f(x) \in [\![Y]\!]_{g_{f}}\). The case of conjunction is straightforward, as are the next three cases. For the final case:

$$ \begin{array}{@{}rcl@{}} \mathcal{M} ,f \models \tau_{x}(\zeta Y.\varphi) & \text{ iff } & \mathcal{M} ,f \models \exists z \exists y \big(x=y \wedge \tau_{z}(\varphi)\big) \\ & \text{ iff } & \text{there is }a\text{ s.t. }\mathcal{M},f[z\mapsto a,y\mapsto f(x)] \models \tau_{z}(\varphi) \\ & \text{ iff } & \text{there is }a\text{ s.t. }f[z\mapsto a,y\mapsto f(x)](z) \in [\![\varphi]\!]_{g_{f[z\mapsto a,y\mapsto f(x)]}} \\ & \text{ iff } & \text{there is }a\text{ s.t. }f[z\mapsto a,y\mapsto f(x)](z) \in [\![\varphi]\!]_{g_{f}[Y\mapsto f(x)]} \\ & \text{ iff } &[\![\varphi]\!]_{g_{f}[Y\mapsto f(x)]} \neq \varnothing \\ & \text{ iff } & f(x) \in [\![\zeta Y.\varphi]\!]_{g_{f}} \end{array} $$

The third equivalence is by the induction hypothesis, and fourth is by the fact that the assignment to Z is not included in gf[Yf(x)] since the variable z was chosen fresh. That is, Z did not occur in φ, and so the assignment \([\![{\varphi }]\!]_{g_{f[z\mapsto a,y\mapsto f(x)]}} = [\![{\varphi }]\!]_{g_{f}[Z \mapsto a,Y\mapsto f(x)]}\) is the same as \([\![{\varphi }]\!]_{g_{f}[Y\mapsto f(x)]}\).

1.5 Proof of Theorem 11

We show both simultaneously by induction on formulas in \({\mathscr{L}}^{FO}\). Consider the second base case. We have \({\mathscr{M}},f \models R(x_{i},x_{j})\) implies 〈f(xi),f(xj)〉∈ 〚R〛, which in turn implies \([\![{X_{i}}]\!]_{g_{f}} \cap [\![{\exists R.X_{j}}]\!]_{g_{f}} \neq \varnothing \), and thus \([\![{\Uparrow (X_{i} \wedge \exists R. X_{j})}]\!]_{g_{f}} = M\). For (8) we have \({\mathscr{M}},f \ |\neq R(x_{i},x_{j})\) implies 〈f(xi),f(xj)〉∉〚R〛, which gives \([\![{X_{i}}]\!]_{g_{f}} \cap [\![{\exists R.X_{j}}]\!]_{g_{f}} = \varnothing \), and thus \([\![{\Uparrow (X_{i} \wedge \exists R. X_{j})}]\!]_{g_{f}} = \varnothing \). The other two base cases are analogous.

Conjunction is straightforward, so consider negation. For (2) we have:

$$ \begin{array}{@{}rcl@{}} \mathcal{M},f \models \neg \alpha & \text{ implies } & \mathcal{M},f \ |\neq \alpha \\ & \text{ implies } & [\![\widehat{\alpha}]\!]_{g_{f}} = \emptyset \\ & \text{ implies } & [\![\zeta Y.{\Downarrow}(Y \wedge \widehat{\alpha})]\!]_{g_{f}} = M \end{array} $$

where the last implication holds because Y is chosen fresh and thus is not in dom(gf): if \([\![\widehat {\alpha }]\!]_{g_{f}} = \varnothing \) then every point a is such that \(\{a\} \cap [\![{\widehat {\alpha }}]\!]_{g_{f}} = \varnothing \). And for (3) we have:

$$ \begin{array}{@{}rcl@{}} \mathcal{M},f \ |\neq \neg \alpha & \text{ implies } & \mathcal{M},f \models \alpha \\ & \text{ implies } & [\![\widehat{\alpha}]\!]_{g_{f}} = M \\ & \text{ implies } & [\![\zeta Y.{\Downarrow}(Y \wedge \widehat{\alpha})]\!]_{g_{f}} = \emptyset \end{array} $$

where the reasoning in the last step is as in the previous case.

Consider (2) for existential quantification:

$$ \begin{array}{@{}rcl@{}} \mathcal{M},f \models \exists x_{i} \alpha & \text{ implies } & \text{there is }a\text{ s.t. }\mathcal{M},f[x_{i} \mapsto a] \models \alpha \\ & \text{ implies } & \text{there is }a\text{ s.t. }[\![\widehat{\alpha}]\!]_{g_{f}[X_{i} \mapsto a]} = M \\ & \text{ implies } & [\![\zeta X_{i} . \widehat{\alpha}]\!]_{g_{f}} \neq \varnothing \\ & \text{ implies } & [\![{\Uparrow}\zeta X_{i} . \widehat{\alpha}]\!]_{g_{f}} = M \end{array} $$

And for (3) we have:

$$ \begin{array}{@{}rcl@{}} \mathcal{M},f \ |\neq \exists x_{i} \alpha & \text{implies } & \text{for all }a\text{: }\mathcal{M},f[x_{i} \mapsto a] \ |\neq \alpha \\ & \text{ implies } & \text{for all }a\text{: }[\![\widehat{\alpha}]\!]_{g_{f}[X_{i} \mapsto a]} = \varnothing \\ & \text{ implies } & [\![\zeta X_{i} . \widehat{\alpha}]\!]_{g_{f}} = \varnothing \\ & \text{ implies } & [\![{\Uparrow}\zeta X_{i} . \widehat{\alpha}]\!]_{g_{f}} = \varnothing \ \end{array} $$

This completes the proof.

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Icard, T.F., Moss, L.S. A Simple Logic of Concepts. J Philos Logic 52, 705–730 (2023).

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