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RECURSIVE AXIOMATISATIONS FROM SEPARATION PROPERTIES

Published online by Cambridge University Press:  03 February 2021

ROB EGROT*
Affiliation:
FACULTY OF INFORMATION AND COMMUNICATION TECHNOLOGY MAHIDOL UNIVERSITY 999 PHUTTHAMONTHON 4 RD, SALAYA NAKHON PATHOM 73170, THAILAND E-mail: robert.egr@mahidol.ac.th
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Abstract

We define a fragment of monadic infinitary second-order logic corresponding to an abstract separation property. We use this to define the concept of a separation subclass. We use model theoretic techniques and games to show that separation subclasses whose axiomatisations are recursively enumerable in our second-order fragment can also be recursively axiomatised in their original first-order language. We pin down the expressive power of this formalism with respect to first-order logic, and investigate some questions relating to decidability and computational complexity. As applications of these results, by showing that certain classes can be straightforwardly defined as separation subclasses, we obtain first-order axiomatisability results for these classes. In particular we apply this technique to graph colourings and a class of partial algebras arising from separation logic.

Information

Type
Article
Copyright
© Association for Symbolic Logic 2021
Figure 0

Figure 1 Class inclusions for separation subclasses.

Figure 1

Figure 2 Complexity class inclusions.

Figure 2

Figure 3 A game played on $C_5$, with different ‘colours’ being denoted by $\diamond $ and $\circ $. Here $\forall $ first asks $\exists $ to colour the top vertex, which she does with $\diamond $ (both players making arbitrary choices here). To minimise the maximum length of a chain of uncoloured elements $\forall $ then demands that one of the bottom two vertices be coloured (the left one say). To be consistent with the closest coloured vertex $\exists $ responds by colouring with $\diamond $. Now $\forall $ demands that either of the vertices between the two coloured ones (moving clockwise) be coloured (the higher one, say). To be consistent with the nearest neighbour $\exists $ must colour it with $\circ $. Now $\forall $ can force a forbidden colouring in the next round.