1. Daniel Kroening and Ofer Strichman 1 Decision Procedures in First Order Logic Decision Procedures for Equality Logic Range Allocation

2. Decision Procedures An algorithmic point of view2 Part III – Decision Procedures for Equality Logic and Uninterpreted Functions Algorithm I – From Equality to Propositional Logic  Adding transitivity constraints  Making the graph chordal  An improved procedure: consider polarity Algorithm II – Range-Allocation  What is the small-model property?  Finding a small adequate range (domain) to each variable  Reducing to Propositional Logic    

3. Decision Procedures An algorithmic point of view3 Range allocation The small model property Range Allocation

4. Decision Procedures An algorithmic point of view4 To a formula with uninterpreted functions Uninterpreted functions From a general formula:

5. Decision Procedures An algorithmic point of view5 From a formula with uninterpreted functions: To a formula in the theory of equality Ackerman’s reduction

6. Decision Procedures An algorithmic point of view6 The Small Model Property Equality Logic enjoys the Small Model Property This means that if a formula in this logic is satisfiable, then there is a finite, bounded in size, model that satisfies it. It gets better: in Equality Logic we can compute this bound, which suggests a decision procedure. What is this bound?

7. Decision Procedures An algorithmic point of view7 The Small Model Property Claim: the range 1.. n is adequate, where n is the number of variables in  Proof:  Every satisfying assignment defines a partition of the variables  Every assignment that results in the same partitioning also satisfies the formula  The range 1.. n allows all partitionings

8. Decision Procedures An algorithmic point of view8 Complexity We need log n variables to encode the range 1… n For n variables we need n log n bits. This is already better than the worst-case O( n 2 ) bits required by the Boolean encoding method …

9. Decision Procedures An algorithmic point of view9 Instead of giving the range [1..11], analyze connectivity: x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 x 1, y 1, x 2, y 2 :{0-1} u 1, f 1, f 2, u 2 : {0-3}g 1, g 2, z: {0-2} The state-space: from 11 11 to ~10 5 Finite Instantiations revisited

10. Decision Procedures An algorithmic point of view10 Or even better: x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 x 1, y 1, g 1, u 1 : {0} {0}{0}{0-1} An Upper-bound: State-space  n! x 2, y 2, g 2, f 1 : {0-1} u 2 : {0-3} f 2, z: {0-2} The state-space: from ~10 5 to 576

11. Decision Procedures An algorithmic point of view11 Choosing a minimal range for the integer variables 0. a,b,c,d,e : { 1 } ( 1 )(normal) 1.a,b,c,d,e : {1..5} (3125)(1..n) 2. a,b,c: {1..3}(connectivity d,e: {1..2} (108) analysis) 3. a: {1}, b:{1-2}, c: {1-3}(factorial d: {1}, e: {1-2} (12) reduction) 4..........  E : : (( a = b ) Ç : ( b = c )) ! (( d  e ))

12. Decision Procedures An algorithmic point of view12 Before and after, in SMV

13. Decision Procedures An algorithmic point of view13 The Range-Minimization Problem Given an Equality formula  E, find in polynomial time a small domain D* sufficient to preserve its satisfiability In other words: find D* such that  E is satisfiable under an infinite domain D !  E is satisfiable under the finite domain D*

14. Decision Procedures An algorithmic point of view14 The strategy 1. Determine a range allocation R, mapping each variable x i  E into a small set of integers, s.t.  E is satisfiable iff it is satisfiable over some R -interpretation. 2. Encode each variable x i as an enumerated type over R ( x i ), and use a standard satisfiability checker as a decision procedure.

15. Decision Procedures An algorithmic point of view15 What range is adequate ? Recall that a subgraph of an E-Graph G E (  E ) is unsatisfiable iff it contains a contradictory cycle. Dfn: A Range Allocation R is adequate for G E (  E ) if every satisfiable subgraph can be satisfied under R. Now we need an algorithm to find adequate ranges

16. Decision Procedures An algorithmic point of view16 Examples: The price of a polynomial procedure: The predicates of  E hold less information than  E.

17. Decision Procedures An algorithmic point of view17 x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 Building the E-Graph E  : E = : Note: 1. Inconsistent subsets appear as contradictory cycles 2. Some of the vertices are mixed

18. Decision Procedures An algorithmic point of view18 The Range-Allocation Algorithm A. Remove all solid edges not belonging to contradictory cycles. B. Add a single unique value to singleton vertices, and remove them from the graph. x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 {0}{0}{1}{1}{3}{3}{2}{2} Step I - pre-processing:

19. Decision Procedures An algorithmic point of view19 Step II - Set construction: A. For each mixed vertex x i : 1. Add a unique value u i to R ( x i ) 2. Broadcast u i on G = 3. Remove x i from the graph B. Add a unique value to each remaining G = component g1g1 g2g2 z {4}{4} {4}{4} {4}{4} g1g1 z {4, } g1g1 g2g2 z {4}{4} 1.2.

20. Decision Procedures An algorithmic point of view20 u1u1 f1f1 f2f2 u2u2 {6}{6}{6}{6}{6}{6} {6}{6} f1f1 f2f2 u2u2 {6,7}{6,7} {6,7}{6,7} {6,7}{6,7} u2u2 {6,7, } u1u1 f1f1 f2f2 u2u2 {6}{6}{6,7}{6,7} 1. 2. 3. f1f1 {6,7, }

21. Decision Procedures An algorithmic point of view21 x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 {3}{3}{2}{2} {4}{4} {4, } {6}{6}{6,7}{6,7} {6,7, } {1}{1} {0}{0} Is the allocated range adequate? We have to satisfy every consistent subset B :  For all x  B, assign the smallest value allocated in step A to a mixed vertex which is G = (B) connected to x.  If there isn’t any, choose the value given in step B.

22. Decision Procedures An algorithmic point of view22 Further optimizations The order in which mixed vertices are eliminated has a strong effect. Not all mixed vertices need to start from a unique value. An analysis that involves solving a coloring problem can help here… … (see lecture notes)

23. Decision Procedures An algorithmic point of view23 x1x1 x2x2 y1y1 y2y2 g1g1 g2g2 zu1u1 f1f1 f2f2 u2u2 {3}{3}{2}{2} {4}{4} {4,5}{4,5} {4,5}{4,5} {6,7}{6,7} {6}{6}{6}{6}{6,8}{6,8} {1}{1} {0}{0} A state-space story 11 11! 16 1..n1..ibasic ordercolor 4872? 576 connectivity