Playing with Quantified Satisfaction Nikolaj Bjørner 1 and Mikol´ aˇ s Janota 2 1 Microsoft Research, Redmond, USA 2 Microsoft Research, Cambridge, UK Abstract We develop an algorithm for satisfiability of quantified formulas. The algorithm is based on recent progress in solving Quantified Boolean Formulas, but it generalizes beyond propositional logic to theories, such as linear arithmetic over integers (Presburger arithmetic), linear arithmetic over reals, algebraic data-types and arrays. Compared with previous algorithms for satisfiability of quantified arithmetical formulas our new implementation outperforms previous implementations in Z3 by a significant margin. 1 Introduction Evaluating a quantified formula of the form ∃x∀y∃z∀u.F [x,y,z,u] can be naturally seen as a game between two parties: the existential player that seeks to find values for x and z that satisfy F and a universal player (spoiler) that attempts to counter the moves for x with values for y and the move for z with values for u such that falsify F . The connection between games and quantifiers is used in many connections, in model-theoretical tools such as Ehrenfeucht-Fra¨ ıss ´ e games [1, 12], and also in algorithms for QBF formulas [15]. This paper takes as starting point the recent algorithm Qesto from [15] and develops generalizations for non-propositional formulas. Let us illustrate one central ingredient of Qesto on a simple example. Example 1. Consider the formula G, where: G = ∀u 1 ,u 2 ∃e 1 ,e 2 .F F = (u 1 ∧ u 2 → e 1 ) ∧ (u 1 ∧¬u 2 → e 2 ) ∧ (e 1 ∧ e 2 →¬u 1 ) We wish to evaluate the formula G and the algorithm proceeds as a game between a universal and existential player. The universal player seeks to falsify F and the existential player seeks to satisfy F . • In the first round, the ∀ player can falsify F using a model M , where M =[u 1 →⊤,u 2 → ⊤,e 1 →⊥,e 2 →⊥]. Thus, ¬F [⊤, ⊤,e 1 ,e 2 ] is satisfiable. • In its quest to falsify F , the ∀ player took control over some existential variables. The ∃ player can now strike back and change values to satisfy F . For example, F [⊤, ⊤, ⊤, ⊥] is true. • The ∀ player has no move to counter this assignment. It has to backtrack over the assignment to e 1 ,e 2 . In particular ¬F [u 1 ,u 2 , ⊤, ⊥] ∧u 2 is already unsatisfiable without regard to the assignment to u 1 . Therefore, the ∀ player learns ¬u 2 and reconsiders its original move and instead settles on the assignment M =[u 1 →⊤,u 2 →⊥,e 1 →⊥,e 2 →⊥] which satisfies ¬F . • But the ∃ player can easily counter this move by setting e 1 →⊥,e 2 →⊤. • Therefore, the ∀ player has yet again to consider what went wrong by determining the core ¬u 2 ∧¬F [u 1 ,u 2 , ⊥, ⊤], which is empty. The universal player has to give up and the existential player established that the formula is true. 1