Abstract
Natural language does not express all connectives definable in classical logic as simple lexical items. Coordination in English is expressed by conjunction and, disjunction or, and negated disjunction nor. Other languages pattern similarly. Non-lexicalized connectives are typically expressed compositionally: in English, negated conjunction is typically expressed by combining negation and conjunction (not both). This is surprising: if \(\wedge \) and \(\vee \) are duals, and the negation of the latter can be expressed lexically (nor), why not the negation of the former? I present a two-tiered model of the semantics of the binary connectives. The first tier captures the expressive power of the lexicon: it is a bilateral state-based semantics that, under a restriction, can express all and only the distinctions that can be expressed by the lexicon of natural language (and, or, nor). This first tier is characterized by rejection as non-assertion and a Neglect Zero assumption. The second tier is obtained by dropping the Neglect Zero assumption and enforcing a stronger notion of rejection, thereby recovering classical logic and thus definitions for all Boolean connectives. On the two-tiered model, we distinguish the limited expressive resources of the lexicon and the greater combinatorial expressive power of the language as a whole. This gives us a logic-based account of compositionality for the Boolean fragment of the language.
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Funding
Open Access funding enabled and organized by Projekt DEAL. This work has received funding under the EU Horizon 2020 Program within the project EXPRESS: From the Expression of Disagreement to New Foundations for Expressivist Semantics (ERC Grant Agreement No. 758540), and under the Horizon Europe Program within the project Evolution of Logic (MSCA Grant Agreement No. 101064835).
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Special thanks to Luca Incurvati and Maria Aloni for extended feedback and discussion. Thanks to audiences at LORI 2021 (Xi’an Jiaotong University), the Express Conference (University of Amsterdam), the Semantic Workshop (University of Massachussetts at Amherst), and to two anonymous reviewers for this journal.
Appendix
Appendix
Proof of Lemma 1. lexical rejection. Let \(\phi \) be positive. If
![figure a](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figa_HTML.png)
then for all
![figure b](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figb_HTML.png)
.
Recall Definition 17: a sentence \(\phi \) of \(\mathscr {L}_{\textsc {llc}}\) is positive iff every atom in \(\phi \) is in the scope of an even number of \(\lnot \)s; a sentence \(\phi \) of \(\mathscr {L}_{\textsc {llc}}\) is negative iff every atom in \(\phi \) is in the scope of an odd number of \(\lnot \)s. In order to establish the Lemma we prove something stronger, namely the following generalization alongside the Lemma, for the purposes of induction. Let \(\phi \) be negative: If \(t\vDash \phi ^{\star }\) then for all \(s\supseteq t: s\vDash \phi ^{\star }\). We run two inductions in parallel for positive and negative sentences. The basis for the positive case are positive literals: these sentences are in the scope of the least even number of \(\lnot \)s, namely 0. Suppose that
![figure c](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figc_HTML.png)
. Then
![figure d](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figd_HTML.png)
and \(t\not =\varnothing \) by Definition 6. Then there is a world \(w\in t: V(w,p)=0\). Let \(s\supseteq t\). Then \(w\in s\), hence
![figure e](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Fige_HTML.png)
and \(s\not =\varnothing \). Therefore
![figure f](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figf_HTML.png)
. The basis for the negative case are negative literals. Suppose that \(t\vDash [\lnot p]^{\star }\). Then \(t\vDash \lnot p^{\star }\) and so
![figure g](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figg_HTML.png)
hence
![figure h](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figh_HTML.png)
and \(t\not =\varnothing \). Therefore
![figure i](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figi_HTML.png)
by the previous reasoning. Hence \(s\vDash \lnot p^{\star }\) and so \(s\vDash [\lnot p]^{\star }\).
Suppose that \(\lnot \phi \) is positive and
![figure j](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figj_HTML.png)
. Then
![figure k](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figk_HTML.png)
, hence \(t\vDash \phi ^{\star }\) and \(\phi \) is negative since it has one wide-scoping negation less. Assume the IH about \(\phi \). Then for all \(s\supseteq t:s\vDash \phi ^{\star }\). Hence
![figure l](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figl_HTML.png)
and so
![figure m](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figm_HTML.png)
. Suppose that \(\lnot \phi \) is negative and \(t\vDash [\lnot \phi ]^{\star }\). Then \(t\vDash \lnot \phi ^{\star }\). Hence
![figure n](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Fign_HTML.png)
and \(\phi \) is positive. Assume the IH about \(\phi \). Then for all
![figure o](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figo_HTML.png)
. Hence \(s\vDash \lnot \phi ^{\star }\) and so \(s\vDash [\lnot \phi ]^{\star }\).
Suppose that \(\phi \vee \psi \) is positive and
![figure p](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figp_HTML.png)
. Then both \(\phi \) and \(\psi \) are positive and
![figure q](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figq_HTML.png)
and
![figure r](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figr_HTML.png)
. Assume the IH about \(\phi \) and \(\psi \). Then for all
![figure s](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figs_HTML.png)
and . Hence
![figure t](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figt_HTML.png)
and so
![figure u](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figu_HTML.png)
. Suppose that \(\phi \vee \psi \) is negative and \(t\vDash [\phi \vee \psi ]^{\star }\). Then both \(\phi \) and \(\psi \) are negative and there are \(t'\) and \(t''\) such that \(t=t'\cup t''\) and \(t'\vDash \phi ^{\star }\) and \(t''\vDash \psi ^{\star }\). Assume the IH about \(\phi \) and \(\psi \). Then for all \(s\supseteq t': s\vDash \phi ^{\star }\) and for all \(s\supseteq t'': s\vDash \psi ^{\star }\). Since all superstates of t are superstates of both \(t'\) and \(t''\), for all \(s \supseteq t\) both \(s\vDash \phi ^{\star }\) and \(s\vDash \psi ^{\star }\). Since \(s=s\cup s\) then \(s\vDash \phi ^{\star }\vee \psi ^{\star }\) for all \(s\supseteq t\). Hence s \(\vDash [\phi \vee \psi ]^{\star }\) for all such states.
Suppose that \(\phi \wedge \psi \) is positive and
![figure v](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figv_HTML.png)
. Then both \(\phi \) and \(\psi \) are positive and there are \(t'\) and \(t''\) such that \(t=t'\cup t''\) and
![figure w](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figw_HTML.png)
and
![figure x](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figx_HTML.png)
. Assume the IH about \(\phi \) and \(\psi \). Then for all
![figure y](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figy_HTML.png)
and for all
![figure z](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figz_HTML.png)
. Since all superstates of t are superstates of both \(t'\) and \(t''\), for all \(s\supseteq t\) both
![figure aa](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaa_HTML.png)
and
![figure ab](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figab_HTML.png)
. Since \(s=s\cup s\) then
![figure ac](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figac_HTML.png)
for all \(s\supseteq t\). Hence
![figure ad](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figad_HTML.png)
for all such states. Suppose that \(\phi \wedge \psi \) is negative and \(t\vDash [\phi \wedge \psi ]^{\star }\). Then both \(\phi \) and \(\psi \) are negative and \(t\vDash \phi ^{\star }\) and \(t\vDash \psi ^{\star }\). Assume the IH about \(\phi \) and \(\psi \). Then for all \(s\supseteq t: s\vDash \phi ^{\star }\) and \(s\vDash \psi ^{\star }\). Hence \(s\vDash \phi ^{\star }\wedge \psi ^{\star }\) and so \(s\vDash [\phi \wedge \psi ]^{\star }\).
Proof of Lemma 5. If \(s\not \vDash \phi \) then
![figure ae](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figae_HTML.png)
and if
![figure af](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaf_HTML.png)
then \(s\vDash \phi \).
This is shown by induction, generalizing from the atomic case. For the atomic case, (i) suppose that \(s\not \vDash p\). Then
![figure ag](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figag_HTML.png)
by Definition 2. (ii) Suppose that
![figure ah](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figah_HTML.png)
. Then \(s\not =\varnothing \) and \(s\vDash p\) by the same definition.
Assume the Induction Hypothesis about \(\phi \). (i) Suppose that \(s\not \vDash \lnot \phi \). Then
![figure ai](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figai_HTML.png)
. Then \(s\vDash \phi \) by the IH. Then
![figure aj](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaj_HTML.png)
(ii) Suppose that
![figure ak](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figak_HTML.png)
. Then
![figure al](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figal_HTML.png)
. Then
![figure am](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figam_HTML.png)
by the IH. Then \(s\vDash \lnot \phi \).
Assume the IH about \(\phi \) and \(\psi \). (i) Suppose that \(s\not \vDash \phi \vee \psi \). Then it is not the case that there are t and \(t'\) such that \(s=t\cup t'\) and \(t\vDash \phi \) and \(t'\vDash \psi \). Let \(s=t\) and \(t'=\varnothing \). Since \(\varnothing \vDash \psi \), the assumption implies that \(s\not \vDash \phi \). Then
![figure an](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figan_HTML.png)
by the IH. Suppose for contradiction that
![figure ao](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figao_HTML.png)
. Then \(s\vDash \psi \) by IH and since \(s=s\cup \varnothing \) and \(\varnothing \vDash \phi \), we would have \(s\vDash \phi \vee \psi \), contrary to assumption. Hence
![figure ap](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figap_HTML.png)
. Hence
![figure aq](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaq_HTML.png)
. (ii) Suppose that
![figure ar](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figar_HTML.png)
. Then either
![figure as](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figas_HTML.png)
or
![figure at](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figat_HTML.png)
. Suppose the former. Then \(s\vDash \phi \) by the IH. Since \(s=s\cup \varnothing \) and \(\varnothing \vDash \psi \), then \(s\vDash \phi \vee \psi \). Likewise if you suppose the latter.
(i) Suppose that \(s\not \vDash \phi \wedge \psi \). Then either \(s\not \vDash \phi \) or \(s\not \vDash \psi \). Suppose the former. Then
![figure au](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figau_HTML.png)
by the IH. Since \(s=s\cup \varnothing \) and
![figure av](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figav_HTML.png)
, then
![figure aw](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaw_HTML.png)
. Likewise if you suppose the latter. (ii) Suppose that
![figure ax](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figax_HTML.png)
. Then it is not the case that there are t and \(t'\) such that \(s=t\cup t'\) and
![figure ay](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figay_HTML.png)
and
![figure az](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figaz_HTML.png)
. Let \(s=t\) and \(t'=\varnothing \). Since
![figure ba](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figba_HTML.png)
, the assumption implies that
![figure bb](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figbb_HTML.png)
. Then \(s\vDash \phi \) by the IH. Moreover, suppose for contradiction that \(s\not \vDash \psi \). Then
![figure bc](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figbc_HTML.png)
by IH and since \(s=s\cup \varnothing \) and
![figure bd](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figbd_HTML.png)
, we would have
![figure be](http://media.springernature.com/lw685/springer-static/image/art%3A10.1007%2Fs10992-023-09708-5/MediaObjects/10992_2023_9708_Figbe_HTML.png)
, contrary to assumption. Hence \(s\vDash \psi \). Hence \(s\vDash \phi \wedge \psi \).
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Sbardolini, G. The Logic of Lexical Connectives. J Philos Logic 52, 1327–1353 (2023). https://doi.org/10.1007/s10992-023-09708-5
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DOI: https://doi.org/10.1007/s10992-023-09708-5