Decidability of split equivalence Y. Abramson and A. Rabinovich Department of Computer Science The Sackler School of Exact Sciences Tel Aviv University e-mails: yotam@math.tau.ac.il, rabino@math.tau.ac.il Abstract We show that the following problem is decidable: given expressions E 1 and E 2 constructed from variables by the regular operations and shuffle, is the identity E 1 = E 2 true for all instantiations of its variables by strings? Our proof uses the notations developed in the causal approach to concurrency. As a byproduct we obtain decidability of similar equivalence for other formalisms. In particular, we prove decidability of split equivalence for Petri nets. Our paper also provides an alternative proof for a characterization of split equivalence recently given by W.Vogler. 1 Introduction 1.1 Equivalence between open expressions We deal with some problem inspired by the following classical problem: Problem 1 Equivalence of open regular expressions under instantiation of variables by languages. Input: two expressions E 1 and E 2 , constructed from variables, constants and regular op- erations - concatenation (;), union (+) and iteration (∗). Question: is the identity E 1 = E 2 valid for every instantiation of the variables by lan- guages? For example, the identity (X + Y ) ∗ =(X ∗ Y ∗ ) ∗ is true because for all languages L 1 and L 2 , the languages (L 1 + L 2 ) ∗ and (L ∗ 1 L ∗ 2 ) ∗ are the same. In this case we say that E 1 and E 2 are language equivalent (notation E 1 ∼ lang E 2 ). Problem 1 is decidable; a “folk” theorem (see [11], Chap. 3, Ex. 14) says that such an identity is valid iff it is valid when variables are instantiated by single letters. Applying this theorem to the example given above, we can see that the identity (X+Y ) ∗ =(X ∗ Y ∗ ) ∗ 1