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A propositional world

A propositional world. Ofer Strichman School of Computer Science, Carnegie Mellon University. Integrated decision procedures in Theorem-Provers. Deciding a combination of theories is the key for automation in Theorem Provers: Boolean operators, Bit-vector, Sets, Linear-Arithmetic,

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A propositional world

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  1. A propositional world Ofer Strichman School of Computer Science, Carnegie Mellon University

  2. Integrated decision procedures in Theorem-Provers Deciding a combination of theories is the key for automation in Theorem Provers: Boolean operators, Bit-vector, Sets, Linear-Arithmetic, Uninterpreted functions, More … Bit-Vector operators Linear Arithmetic Uninterpreted functions f(f(x)-f(y)) != f(z) & y <=x + 2 | b & 3 > 10 Normally, each theory is solved with its own decision procedure and the results are combined (Shostak, Nelson..).

  3. Integrated decision procedures in Theorem-Provers All of these theories, except linear arithmetic, have known efficient direct reductions to propositional logic. Thus, reducing linear arithmetic to propositional logic will: 1. Enable integration of theories in the propositional logic level. 2. Potentially be faster than known techniques.

  4. Linear Arithmetic and its sub-theories 2x –3y +5z < 0 5x + 2w 2 • Some useful methods for solving a conjunction of linear • arithmetic expressions: • Simplex, Elliptic curve • Variable Elimination Methods (Hodes, Fourier-Motzkin,..) • Shostak’sloop residues • Separation theory: Bellman / Pratt ... • ...

  5. x 3 1 z y -1 A decision procedure for separation theory Separation predicates have the form x > y + c where x,y are real variables, and c is a constant Pratt [73] (/Bellman[57]): Given a set of conjuncted separation predicates  1. Construct the `inequality graph’ 2.  is satisfiable iff there is no cycle with non-negative accumulated weight : (x > z +3 z>y –1 y > x+1)

  6. Handling disjunctions through case splitting • All previously mentioned algorithms handle disjunctions • by splitting the formula. • This can be thought of as a two stage process: • Convert formula to Disjunctive Normal Form (DNF) • Solve each clause separately, until satisfying one of them. (A common improvement: split ‘when needed’) Case splitting is frequently the bottleneck of the procedure

  7. So what can be done against case-splitting ? Answer: Split the domain, not the formula. Given a formula , this transformation can be done if ’ s.t. |= |=’, and ’ is decidable under a finite domain. • When is this possible? •  enjoys the ‘Small model property’, or • Tailor-made reduction

  8. SAT vs. infinite-state decision procedures With finite instantiation (e.g. SAT), we split the domain. Infinite state decision procedures split the formula. So what’s the big difference ?

  9. 1. Pruning. 2. Learning. 3. Guidance (prioritizing internal steps) SAT vs. infinite-state decision procedures Three mechanisms, crucial for efficient decision making: SAT has a significant advantage in all three.

  10. x 0 1 . (x  y) . . Backtrack y 1 0 Pruned! SAT vs. infinite-state decision procedures (1/4) 1. Pruning SAT: each clause c prunes up to 2|v|-|c| states. |v|=1000, |c| =2 Pruning 2998states Others: ? (stops when finds a satisfiable clause)

  11. SAT vs. infinite-state decision procedures (2/4) 2. Learning SAT: Partial assignments that lead to a conflict are recorded and hence not repeated. Others: (depends on decision procedure) - Adding proved sub-goals as antecedents to new sub-goals - …

  12. SAT vs. infinite-state decision procedures (3/4) 3. Guidance (prioritizing internal steps) Consider 1 2, where 1is unsat and hard, and 2is sat and easy. With proper guidance, a theorem prover should start from 2. Guidance requires efficient estimation: - How hard it is to solve each sub-formula? - To what extent will it simplify the rest of the proof?

  13. SAT vs. infinite-state decision procedures (4/4) 3. Guidance (cont’d) “..To what extent will it simplify the rest of the proof?” SAT: Guidance through decision heuristics (e.g. DLIS). (x  y  z) (x  v) (~x  ~z) Estimating simplification by counting literals in each phase Others: Expression ordering, ...

  14. Example: Equality Logic with Uninterpreted Functions (1/3) Equality Logic with Uninterpreted Functions: (Uninterpreted functions are reducible to equality logic. Thus, we can concentrate on equality logic) Traditional infinite-state decision procedure: Congruence Closurewith case splitting.

  15. Example: Equality Logic (2/3) • Since 1998, several groups devised finite-state decision procedures • for this theory: • Goel et. al. (CAV’98) – Boolean encoding and BDDs • Bryant et. al.(CAV’99) – Positive-equality + finite instantiation • Pnueli et. al. (CAV’99) – Small domains instantiation • Bryant et. al.(CAV’00) – Boolean encoding with explicit constraints

  16. x exz exy z y eyz Example: Equality Logic (3/3) Bryant et. al. (CAV’00): Add transitivity constraints to the formula. Let (x=y, y=z, x=z) be the equality predicates in . 1. Construct the equality graph. 2. Impose transitivity on cycles: exy + eyz +exz  2 The resulting formula is propositional BDDs , SAT, etc.

  17. This work Extends the results of Bryant et.al. to a Boolean combination of: • Separation predicates: • Separation predicates for integers: • Linear arithmetic: • Integer linear arithmetic: Done

  18. Usability Separation predicates: “Most verification conditions involving inequalities are separation predicates” [Pratt, 1973]: Array bounds checks, tests on index variables, timing constraints, worst execution time analysis, etc. Linear arithmetic: All of the above + … + Linear programming, + Integer Linear programming.

  19. Reducing separation predicates to propositional logic (1/6) A. Normalize (example): : f(x) > f(y+1) 1. Uninterpreted functions  equality logic : (x=y+1f1=f2)(f1>f2) 2. Normal form xy+1 f1=f2 : (x>y+1 y>x-1(f1 f2 f2 f1)) (f1>f2) Now  has no negations and only the ‘>’ and ‘’ predicate symbols.

  20. x 3 1 z y -1 x -3 -1 z y 1 Reducing separation predicates to propositional logic (3/6) B. Encode + construct graph (example): : (x > z +3  (z>y –1 yx+1)) Transitivity constraints  )) ( ’:   ( and its dual: Separation graph:

  21. x 3 1 z y -1 x -3 -1 z y 1 Reducing separation predicates to propositional logic (5/6) C. Add transitivity constraints for each simple cycle (example): Transitivity constraints )) (   (  ’: (( )) )) (   (  ’:  

  22. ..... ..... Compact representation of constraints (1/4) n diamonds  2nsimple cycles. Can we do better than that ? In most cases - yes. e.g. If the diamonds are ‘balanced’ (c1 +c2 =c3 +c4)  O(n) constraints c2 c1 c1+ c2 c4 c3

  23. Compact representation of constraints (2/4) Chordal graphs: each cycle of size greater than 3, has a ‘chord’. G: In the equality predicates case: Let C be a cycle in G Let  be an assignment that violates C’s transitivity ( |C) Theorem: there exists a cycle c of size 3 in G s.t.  | c Conclusion: add transitivity constraints only for triangles. Now only a polynomial no. of constraints is required.

  24. c2 c1 c1+ c2 c3 c4 c5 Compact representation of constraints (3/4) • Our case is more complicated: • G is directed • G is a multi-graph • Edges have weights • There are two types of edges G is chordal iff: Every directed cycle of size greater than 3 has a chord which ‘accumulates’ the weight of the path between its ends.

  25. ..... 2. If there are uniform weights c1 and c2, c1c2 on top and bottom paths  O(n2) constraints c1 c1 c1 c1 c2 c2 c2 c2 Compact representation of constraints (4/4) Complexity of making the graph chordal: 1. If the diamonds are ‘balanced’  O(n) constraints 3. Worst case  O(2n)

  26. Extension to integer variables (1/2) (c is an integer) Given  with integer separation predicates, derive R: • Declare all variables as real. • For each predicate x>y + c, add a constraint • x > y + c  x y + c + 1 Theorem:  is satisfiable iff R is satisfiable

  27. Experimental results (1/3) d=2 ..... n diamonds Each diamond has 2d edges Top and bottom paths in each diamond are disjuncted. There are 2nconjuncted cycles. By adjusting the weights, we ensured that there is a single satisfying assignment.

  28. Experimental results (2/3) To be continued...

  29. Experimental results (3/3) To be continued... The procedure has recently been integrated into SyMP and Euclid. We currently experiment with real software verification problems.

  30. c1 c3 c2 Next: Linear Arithmetic (1/2) Separation predicates: c x > y + c y x Adding constraints according to accumulated cycle weight: The testc1 + c2 + c3 > 0 results in a yes/no answer

  31. x 2z + c 3 y 2 Next: Linear Arithmetic (2/2) Linear Arithmetic: 2z + c x > y + 2z + c y x The test1 + 2 + 3 > 0 results in a new predicate! Shostak[81]: ‘Deciding linear inequalities by computing loop residues’ - Determine a fixed variable order - Represent each predicate by its two ‘highest’ variables This procedure guarantees termination.

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