Term inequalities in finite algebras
Given an algebra \(\mathbf{A}\), and terms \(s(x_{1},x_{2},\dots x_{k})\) and \(t(x_{1},x_{2},\dots x_{k})\) of the language of \({\mathbf A}\), we say that \(s\) and \(t\) are {\em separated} in \({\mathbf A}\) iff for all \(a_{1},a_{2}\dots a_{k}\in A\), \(s(a_{1},a_{2},\dots a_{k})\) and \(t(a_{1...
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Veröffentlicht in: | arXiv.org 2016-01 |
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Format: | Artikel |
Sprache: | eng |
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Zusammenfassung: | Given an algebra \(\mathbf{A}\), and terms \(s(x_{1},x_{2},\dots x_{k})\) and \(t(x_{1},x_{2},\dots x_{k})\) of the language of \({\mathbf A}\), we say that \(s\) and \(t\) are {\em separated} in \({\mathbf A}\) iff for all \(a_{1},a_{2}\dots a_{k}\in A\), \(s(a_{1},a_{2},\dots a_{k})\) and \(t(a_{1},a_{2},\dots a_{k})\) are never equal. We prove that given two terms that are separated in any algebra, there exists a finite algebra in which they are separated. As a corollary, we obtain that whenever the sentence \(\sigma\) is a universally quantified conjunction of negated atomic formulas, \(\sigma\) is consistent iff it has a finite model. |
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ISSN: | 2331-8422 |