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Fischer-Servi logic does not have interpolation

Rodrigo Nicolau Almeida, Nick Bezhanishvili, Simon Lemal

Abstract

We prove that the Fischer-Servi logic $\mathsf{IK}$ does not have the (Craig) interpolation property. This is obtained by showing that the corresponding class of modal Heyting algebras lacks the amalgamation property. We also generalize this result to some extensions of the Fischer-Servi logic such as $\mathsf{IT}$, $\mathsf{IK4}$, $\mathsf{IS4}$, and $\mathsf{IGL}$.

Fischer-Servi logic does not have interpolation

Abstract

We prove that the Fischer-Servi logic does not have the (Craig) interpolation property. This is obtained by showing that the corresponding class of modal Heyting algebras lacks the amalgamation property. We also generalize this result to some extensions of the Fischer-Servi logic such as , , , and .

Paper Structure

This paper contains 11 sections, 19 theorems, 8 equations, 3 figures.

Table of Contents

  1. Introduction
  2. Super-Fischer-Servi logics, FS-algebras and IK-frames
  3. FS-algebras
  4. IK frames and spaces
  5. Craig interpolation and superamalgamation
  6. Failure of interpolation for $\mathsf{IK}$
  7. Conclusions
  8. Acknowledgments
  9. Appendix: Some folklore proofs
  10. Logics, Algebras and Duality
  11. Craig interpolation from Superamalgamation

Key Result

theorem 1

Let $L$ be a super FS-logic. For each formula $\phi(\overline{p})\in \mathcal{L}_{\boxempty,\Diamond}$ the following are equivalent: $\blacktriangleleft$$\blacktriangleleft$

Figures (3)

  • Figure 1: Confluence conditions for Fischer-Servi
  • Figure 2: Amalgamation of lattices
  • Figure 3: The co-V formation $(Z,X,Y)$

Theorems & Definitions (45)

  • definition 1
  • definition 2
  • theorem 1
  • definition 3
  • definition 4
  • definition 5
  • theorem 2
  • lemma 1
  • definition 6
  • definition 7
  • ...and 35 more