Leonardo de Moura Microsoft Research
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- Harold Gilbert
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1 Leonardo de Moura Microsoft Research
2 Is formula F satisfiable modulo theory T? SMT solvers have specialized algorithms for T
3 b + 2 = c and f(read(write(a,b,3), c-2)) f(c-b+1)
4 b + 2 = c and f(read(write(a,b,3), c-2)) f(c-b+1)
5 b + 2 = c and f(read(write(a,b,3), c-2)) f(c-b+1)
6 b + 2 = c and f(read(write(a,b,3), c-2)) f(c-b+1)
7 b + 2 = c and f(read(write(a,b,3), c-2)) f(c-b+1) Substituting c by b+2
8 b + 2 = c and f(read(write(a,b,3), b+2-2)) f(b+2-b+1) Simplifying
9 b + 2 = c and f(read(write(a,b,3), b)) f(3)
10 b + 2 = c and f(read(write(a,b,3), b)) f(3) Applying array theory axiom forall a,i,v: read(write(a,i,v), i) = v
11 b + 2 = c and f(3) f(3) Inconsistent
12 Repository of Benchmarks Benchmarks are divided in logics : QF_UF: unquantified formulas built over a signature of uninterpreted sort, function and predicate symbols. QF_UFLIA: unquantified linear integer arithmetic with uninterpreted sort, function, and predicate symbols. AUFLIA: closed linear formulas over the theory of integer arrays with free sort, function and predicate symbols.
13 For most SMT solvers: F is a set of ground formulas Many Applications Bounded Model Checking Test-Case Generation
14 An SMT Solver is a collection of Little Engines of Proof
15 An SMT Solver is a collection of Little Engines of Proof Examples: SAT Solver Equality solver
16 a = b, b = c, d = e, b = s, d = t, a e, a s a b c d e s t
17 a = b, b = c, d = e, b = s, d = t, a e, a s a b c d e s t
18 a = b, b = c, d = e, b = s, d = t, a e, a s a,b c d e s t
19 a = b, b = c, d = e, b = s, d = t, a e, a s a,b c d e s t
20 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c d e s t
21 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c d e s t
22 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c d,e s t
23 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c d,e s t
24 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c,s d,e t
25 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c,s d,e t
26 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c,s d,e,t
27 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c,s d,e,t
28 a = b, b = c, d = e, b = s, d = t, a e, a s a,b,c,s d,e,t Unsatisfiable
29 a = b, b = c, d = e, b = s, d = t, a e a,b,c,s d,e,t Model construction
30 a = b, b = c, d = e, b = s, d = t, a e 1 2 a,b,c,s d,e,t Model construction M = { 1, 2 } (universe, aka domain)
31 a = b, b = c, d = e, b = s, d = t, a e 1 2 a,b,c,s d,e,t Model construction M = { 1, 2 } (universe, aka domain) M(a) = 1 (assignment)
32 a = b, b = c, d = e, b = s, d = t, a e Alternative notation: a M = a,b,c,s d,e,t Model construction M = { 1, 2 } (universe, aka domain) M(a) = 1 (assignment)
33 a = b, b = c, d = e, b = s, d = t, a e 1 2 a,b,c,s d,e,t Model construction M = { 1, 2 } (universe, aka domain) M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2
34 a = b, b = c, d = e, b = s, d = t, a e 1 2 a,b,c,s d,e,t Model construction M = { 1, 2 } (universe, aka domain) M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2
35 Termination: easy Soundness Invariant: all constants in a ball are known to be equal. The ball merge operation is justified by: Transitivity and Symmetry rules. Completeness We can build a model if an inconsistency was not detected. Proof template (by contradiction): Build a candidate model. Assume a literal was not satisfied. Find contradiction.
36 Completeness We can build a model if an inconsistency was not detected. Instantiating the template for our procedure: Assume some literal c = d is not satisfied by our model. That is, M(c) M(d). This is impossible, c and d must be in the same ball. i c,d, M(c) = M(d) = i
37 Completeness We can build a model if an inconsistency was not detected. Instantiating the template for our procedure: Assume some literal c d is not satisfied by our model. That is, M(c) = M(d). Key property: we only check the disequalities after we processed all equalities. This is impossible, c and d must be in the different balls i c, j d, M(c) = i M(d) = j
38 a = b, b = c, d = e, b = s, d = t, f(a, g(d)) f(b, g(e)) Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
39 a = b, b = c, d = e, b = s, d = t, f(a, g(d)) f(b, g(e)) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
40 a = b, b = c, d = e, b = s, d = t, f(a, v 1 ) f(b, g(e)) v 1 g(d) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
41 a = b, b = c, d = e, b = s, d = t, f(a, v 1 ) f(b, g(e)) v 1 g(d) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
42 a = b, b = c, d = e, b = s, d = t, f(a, v 1 ) f(b, v 2 ) v 1 g(d), v 2 g(e) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
43 a = b, b = c, d = e, b = s, d = t, f(a, v 1 ) f(b, v 2 ) v 1 g(d), v 2 g(e) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
44 a = b, b = c, d = e, b = s, d = t, v 3 f(b, v 2 ) v 1 g(d), v 2 g(e), v 3 f(a, v 1 ) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
45 a = b, b = c, d = e, b = s, d = t, v 3 f(b, v 2 ) v 1 g(d), v 2 g(e), v 3 f(a, v 1 ) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
46 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) First Step: Naming subterms Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
47 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1 v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
48 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1 v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) d = e implies g(d) = g(e)
49 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1 v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) d = e implies v 1 = v 2
50 We say: v 1 and v 2 are congruent. a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) d = e implies v 1 = v 2
51 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) a = b, v 1 = v 2 implies f(a, v 1 ) = f(b, v 2 )
52 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3 v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) a = b, v 1 = v 2 implies v 3 = v 4
53 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n ) a = b, v 1 = v 2 implies v 3 = v 4
54 a = b, b = c, d = e, b = s, d = t, v 3 v 4 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Unsatisfiable Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
55 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Changing the problem a,b,c,s d,e,t v 1,v 2 v 3, v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
56 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
57 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Congruence Rule: x 1 = y 1,, x n = y n implies f(x 1,, x n ) = f(y 1,, y n )
58 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4
59 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 Missing: Interpretation for f and g.
60 Building the interpretation for function symbols M(g) is a mapping from M to M Defined as: M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise ( k is an arbitrary element) Is M(g) well-defined?
61 Building the interpretation for function symbols M(g) is a mapping from M to M Defined as: M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise ( k is an arbitrary element) Is M(g) well-defined? Problem: we may have v g(a) and w g(b) s.t. M(a) = M(b) = 1 and M(v) = 2 3 = M(w) So, is M(g)( 1 ) = 2 or M(g)( 1 ) = 3?
62 Building the interpretation for function symbols M(g) is a mapping from M to M Defined as: M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j This is impossible because of the congruence rule! a and b are in the same ball, then so are v and w = k, otherwise ( k is an arbitrary element) Is M(g) well-defined? Problem: we may have v g(a) and w g(b) s.t. M(a) = M(b) = 1 and M(v) = 2 3 = M(w) So, is M(g)( 1 ) = 2 or M(g)( 1 ) = 3?
63 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) a,b,c,s d,e,t v 1,v 2 v 3, v 4 Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4
64 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise
65 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 M(g) = { 2 3 } M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise
66 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 M(g) = { 2 3 } M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise
67 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 M(g) = { 2 3, else 1 } M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise
68 a = b, b = c, d = e, b = s, d = t, a v 4, v 2 v 3 v 1 g(d), v 2 g(e), v 3 f(a, v 1 ), v 4 f(b, v 2 ) Model construction: M = { 1, 2, 3, 4 } M(a) = M(b) = M(c) = M(s) = 1 M(d) = M(e) = M(t) = 2 M(v 1 ) = M(v 2 ) = 3 M(v 3 ) = M(v 4 ) = 4 M(g) = { 2 3, else 1 } M(f) = { ( 1, 3 ) 4, else 1 } M(g)( i ) = j if there is v g(a) s.t. M(a) = i M(v) = j = k, otherwise
69 It is possible to implement our procedure in O(n log n)
70 d,e,t Sets (equivalence classes) d,e t = d,e,t Union a,b,c,s a s Membership
71 d,e,t Sets (equivalence classes) Key observation: The sets are disjoint! d,e t = d,e,t Union a,b,c,s a s Membership
72 Union-Find data-structure Every set (equivalence class) has a root element (representative). a,b,c,s,r b s r root a c We say: find(c) is b
73 Union-Find data-structure a,b,c b s,r = s a,b,c,s,r s a c r b r a c
74 Tracking the equivalence classes size is important! a 1 a 2 a 3 = a 1 a 2 a 3 a 1 a 2 a 3 a 4 = a 1 a 2 a 3 a 4 a 1 a 2 a 3 a n-1 a n = a 1 a 2 a 3 a n-1 a n
75 Tracking the equivalence classes size is important! a 1 a 2 a 3 = a 1 a 2 a 3 a 1 a 2 a 3 a 4 = a 1 a 2 a 3 a 2 a n = a 2 a 1 a an-1 3 a 1 a 3 a n an-1 a 4
76 Tracking the equivalence classes size is important! a 1 a 2 a 3 = a 1 a 2 a 3 We can do n merges in O(n log n) a 1 a 2 a 3 a 4 = a 1 a 2 a 3 a 2 a n = a 2 a 1 a an-1 3 a 1 a 3 a n an-1 a 4 Each constant has two fields: find and size.
77 Implementing the congruence rule. Occurrences of a constant: we say a occurs in v iff v f(,a, ) When we merge two equivalence classes we can traverse these occurrences to find new congruences. a b c s r Occurrences(b) = { v 1 g(b), v 2 f(a) } Occurrences(s) = { v 3 f(r) }
78 Implementing the congruence rule. Occurrences of a constant: we say a occurs in v iff v f(,a, ) When we merge two equivalence classes we can traverse these occurrences to find new congruences. a b c occurrences(b) = { v 1 g(b), v 2 f(a) } occurrences(s) = { v 3 f(r) } s r Inefficient version: for each v in occurrences(b) for each w in occurrences(s) if v and w are congruent add (v,w) to todo queue A queue of pairs that need to be merged.
79 b a c r occurrences(b) = { v 1 g(b), v 2 f(a) } occurrences(s) = { v 3 f(r) } s We also need to merge occurrences(b) with occurrences(s). This can be done in constant time: Use circular lists to represent the occurrences. (More later) v 1 v 3 = v 2 v 1 v 2 v 3
80 Avoiding the nested loop: for each v in occurrences(b) for each w in occurrences(s) Avoiding the nested loop: Use a hash table to store the elements v 1 f(a 1,, a n ). Each constant has an identifier (e.g., natural number). Compute hash code using the identifier of the (equivalence class) roots of the arguments. hash(v 1 ) = hash-tuple(id(f), id(root(a 1 )),, id(root(a n )))
81 Avoiding the nested loop: for each v in occurrences(b) for each w in occurrences(s) Avoiding the nested loop: Use a hash table to store hash-tuple the can elements be the Jenkin s v 1 f(a 1,, a n ). Each constant has an hash identifier function (e.g., for strings. natural number). Just adding the ids produces a Compute hash code using the identifier of the (equivalence very bad hash-code! class) roots of the arguments. hash(v 1 ) = hash-tuple(id(f), id(root(a 1 )),, id(root(a n )))
82 Efficient implementation of the congruence rule. Merging the equivalences classes with roots: a 1 and a 2 Assume a 2 is smaller than a 1 Before merging the equivalence classes: a 1 and a 2 for each v in occurrences(a 2 ) remove v from the hash table (its hashcode will change) After merging the equivalence classes: a 1 and a 2 for each v in occurrences(a 2 ) if there is w congruent to v in the hash-table add (v,w) to todo queue else add v to hash-table
83 Efficient implementation of the congruence rule. Merging the equivalences classes with roots: a 1 and a 2 Assume a 2 is smaller than a 1 Before merging the equivalence classes: a 1 and a 2 for each v in occurrences(a 2 ) remove v from the hash table (its hashcode will change) After merging the equivalence classes: a 1 and a 2 for each v in occurrences(a 2 ) if there is w congruent to v in the hash-table add (v,w) to todo queue else add v to hash-table add v to occurrences(a 1 ) Trick: Use dynamic arrays to represent the occurrences
84 The efficient version is not optimal (in theory). Problem: we may have v = f(a 1,, a n ) with huge n. Solution: currying Use only binary functions, and represent f(a 1, a 2,a 3,a 4 ) as f(a 1, h(a 2, h(a 3, a 4 ))) This is not necessary in practice, since the n above is small.
85 Each constant has now three fields: find, size, and occurrences. We also has use a hash-table for implementing the congruence rule. We will need many more improvements!
86 Many verification/analysis problems require: case-analysis x 0, y = x + 1, (y > 2 y < 1)
87 Many verification/analysis problems require: case-analysis x 0, y = x + 1, (y > 2 y < 1) Naïve Solution: Convert to DNF (x 0, y = x + 1, y > 2) (x 0, y = x + 1, y < 1)
88 Many verification/analysis problems require: case-analysis x 0, y = x + 1, (y > 2 y < 1) Naïve Solution: Convert to DNF (x 0, y = x + 1, y > 2) (x 0, y = x + 1, y < 1) Too Inefficient! (exponential blowup)
89 Case Analysis Equality + UF Arithmetic Bit-vectors
90 M F Partial model Set of clauses
91 Guessing p p q, q r p, q p q, q r
92 Deducing p p q, p s p, s p q, p s
93 Backtracking p, s, q p q, s q, p q p, s p q, s q, p q
94 Efficient indexing (two-watch literal) Non-chronological backtracking (backjumping) Lemma learning
95 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1)
96 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver
97 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver Assignment p 1, p 2, p 3, p 4
98 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver Assignment p 1, p 2, p 3, p 4 x 0, y = x + 1, (y > 2), y < 1
99 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver Assignment p 1, p 2, p 3, p 4 x 0, y = x + 1, (y > 2), y < 1 Unsatisfiable x 0, y = x + 1, y < 1 Theory Solver
100 Basic Idea x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) p 1, p 2, (p 3 p 4 ) p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver Assignment p 1, p 2, p 3, p 4 x 0, y = x + 1, (y > 2), y < 1 New Lemma p 1 p 2 p 4 Unsatisfiable x 0, y = x + 1, y < 1 Theory Solver
101 New Lemma p 1 p 2 p 4 Unsatisfiable x 0, y = x + 1, y < 1 Theory Solver AKA Theory conflict
102 procedure SmtSolver(F) (F p, M) := Abstract(F) loop (R, A) := SAT_solver(F p ) if R = UNSAT then return UNSAT S := Concretize(A, M) (R, S ) := Theory_solver(S) if R = SAT then return SAT L := New_Lemma(S, M) Add L to F p
103 Basic Idea F: x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) F p : p 1, p 2, (p 3 p 4 ) M: p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver A: Assignment p 1, p 2, p 3, p 4 S: x 0, y = x + 1, (y > 2), y < 1 L: New Lemma p 1 p 2 p 4 S : Unsatisfiable x 0, y = x + 1, y < 1 Theory Solver
104 F: x 0, y = x + 1, (y > 2 y < 1) Abstract (aka naming atoms) F p : p 1, p 2, (p 3 p 4 ) M: p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1) SAT Solver A: Assignment p 1, p 2, p 3, p 4 S: x 0, y = x + 1, (y > 2), y < 1 L: New Lemma p 1 p 2 p 4 S : Unsatisfiable x 0, y = x + 1, y < 1 Theory Solver procedure SMT_Solver(F) (F p, M) := Abstract(F) loop (R, A) := SAT_solver(F p ) if R = UNSAT then return UNSAT S = Concretize(A, M) (R, S ) := Theory_solver(S) if R = SAT then return SAT L := New_Lemma(S, M) Add L to F p Lazy translation to DNF
105 State-of-the-art SMT solvers implement many improvements.
106 Incrementality Send the literals to the Theory solver as they are assigned by the SAT solver p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1), p 5 (x < 2), p 1, p 2, p 4 p 1, p 2, (p 3 p 4 ), (p 5 p 4 ) Partial assignment is already Theory inconsistent.
107 Efficient Backtracking We don t want to restart from scratch after each backtracking operation.
108 Efficient Lemma Generation (computing a small S ) Avoid lemmas containing redundant literals. p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1), p 5 (x < 2), p 1, p 2, p 3, p 4 p 1, p 2, (p 3 p 4 ), (p 5 p 4 ) p 1 p 2 p 3 p 4 Imprecise Lemma
109 Theory Propagation It is the SMT equivalent of unit propagation. p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1), p 5 (x < 2), p 1, p 2 p 1, p 2, (p 3 p 4 ), (p 5 p 4 ) p 1, p 2 imply p 4 by theory propagation p 1, p 2, p 4 p 1, p 2, (p 3 p 4 ), (p 5 p 4 )
110 Theory Propagation It is the SMT equivalent of unit propagation. p 1 (x 0), p 2 (y = x + 1), p 3 (y > 2), p 4 (y < 1), p 5 (x < 2), p 1, p 2 p 1, p 2, (p 3 p 4 ), (p 5 p 4 ) p 1, p 2 imply p 4 by theory propagation p 1, p 2, p 4 p 1, p 2, (p 3 p 4 ), (p 5 p 4 ) Tradeoff between precision performance.
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Propositional Logic: Models and Proofs C. R. Ramakrishnan CSE 505 1 Syntax 2 Model Theory 3 Proof Theory and Resolution Compiled at 11:51 on 2016/11/02 Computing with Logic Propositional Logic CSE 505