introduction to satisfiability modulo...

Introduction to Satisfiability Modulo Theories

Dejan Jovanović

SRI International

SAT/SMT/AR Summer School, Lisbon, 2016

Outline

1 Introduction

2 DPLL(T) Framework

3 Decision Procedures

4 MCSAT Framework

5 Finish

Outline

1 Introduction

2 DPLL(T) Framework

4 MCSAT Framework

5 Finish

Satisfiability Modulo Theorieswhat we’re trying to solve

ProblemCheck if a given (quantifier-free) formula is satisfiable modulo theunion of background theories.

Example (QF_UFLRA)

(z = 1 ∨ z = 0) ∧ (x− y+ z = 1) ∧ (f(x) > f(y))

1 Linear real arithmetic (LRA).2 Uninterpreted functions (UF).3 Satisfiable with z 7→ 0, x 7→ 1, y 7→ 0, f(1) 7→ 1, f(0) 7→ 0

Example (QF_UFLRA)

(z = 1 ∨ z = 0) ∧ (x− y+ z = 1) ∧ (f(x) > f(y))

1 Linear real arithmetic (LRA).

2 Uninterpreted functions (UF).3 Satisfiable with z 7→ 0, x 7→ 1, y 7→ 0, f(1) 7→ 1, f(0) 7→ 0

Example (QF_UFLRA)

(z = 1 ∨ z = 0) ∧ (x− y+ z = 1) ∧ (f(x) > f(y))

1 Linear real arithmetic (LRA).2 Uninterpreted functions (UF).

3 Satisfiable with z 7→ 0, x 7→ 1, y 7→ 0, f(1) 7→ 1, f(0) 7→ 0

Example (QF_UFLRA)

(z = 1 ∨ z = 0) ∧ (x− y+ z = 1) ∧ (f(x) > f(y))

1 Linear real arithmetic (LRA).2 Uninterpreted functions (UF).3 Satisfiable with z 7→ 0, x 7→ 1, y 7→ 0, f(1) 7→ 1, f(0) 7→ 0

Satisfiability Modulo Theoriesmany applications

ExampleSchedule n jobs, each composed ofm consecutive tasks, onmmachines.

Schedule in 8 time slots.

di,j Machine 1 Machine 2Job 1 2 1Job 2 3 1Job 3 2 3

(t1,1 ≥ 0) ∧ (t1,2 ≥ t1,1 + 2) ∧ (t1,2 + 1 ≤ 8)

(t2,1 ≥ 0) ∧ (t2,2 ≥ t2,1 + 3) ∧ (t2,2 + 1 ≤ 8)

(t3,1 ≥ 0) ∧ (t3,2 ≥ t3,1 + 2) ∧ (t3,2 + 3 ≤ 8)

((t1,1 ≥ t2,1 + 3) ∨ (t2,1 ≥ t1,1 + 2))

((t1,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t1,1 + 2))

((t2,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t2,1 + 3))

((t1,2 ≥ t2,2 + 1) ∨ (t2,2 ≥ t1,2 + 1))

((t1,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t1,2 + 1))

((t2,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t2,2 + 1))

Example from [DMB11]

(t1,1 ≥ 0) ∧ (t1,2 ≥ t1,1 + 2) ∧ (t1,2 + 1 ≤ 8)

(t2,1 ≥ 0) ∧ (t2,2 ≥ t2,1 + 3) ∧ (t2,2 + 1 ≤ 8)

(t3,1 ≥ 0) ∧ (t3,2 ≥ t3,1 + 2) ∧ (t3,2 + 3 ≤ 8)

((t1,1 ≥ t2,1 + 3) ∨ (t2,1 ≥ t1,1 + 2))

((t1,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t1,1 + 2))

((t2,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t2,1 + 3))

((t1,2 ≥ t2,2 + 1) ∨ (t2,2 ≥ t1,2 + 1))

((t1,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t1,2 + 1))

((t2,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t2,2 + 1))

(t1,1 ≥ 0) ∧ (t1,2 ≥ t1,1 + 2) ∧ (t1,2 + 1 ≤ 8)

(t2,1 ≥ 0) ∧ (t2,2 ≥ t2,1 + 3) ∧ (t2,2 + 1 ≤ 8)

(t3,1 ≥ 0) ∧ (t3,2 ≥ t3,1 + 2) ∧ (t3,2 + 3 ≤ 8)

((t1,1 ≥ t2,1 + 3) ∨ (t2,1 ≥ t1,1 + 2))

((t1,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t1,1 + 2))

((t2,1 ≥ t3,1 + 2) ∨ (t3,1 ≥ t2,1 + 3))

((t1,2 ≥ t2,2 + 1) ∨ (t2,2 ≥ t1,2 + 1))

((t1,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t1,2 + 1))

((t2,2 ≥ t3,2 + 3) ∨ (t3,2 ≥ t2,2 + 1))

Run SMT solver (QF_IDL)

t1,1 7→ 5, t1,2 7→ 7

t2,1 7→ 2, t2,2 7→ 6

t3,1 7→ 0, t3,2 7→ 3

39 000

39 500

40 000

Tx1(t) = −3.2484 + 270.7t + 433.12t2 − 324.83999t3

Ty1(t) = 15.1592 + 108.28t + 121.2736t2 − 649.67999t3

Tz1(t) = 38980.8 + 5414t − 21656t2 + 32484t3

Tx2(t) = 1.0828 − 135.35t + 234.9676t22 + 3248.4t3

Ty2(t) = 18.40759 − 230.6364t − 121.2736t2 − 649.67999t3

Tz2(t) = 40280.15999 − 10828t + 24061.9816t2 − 32484t3

D = 5 H = 1000 0 ≤ t ≤1

|Tz1(t) − Tz2(t)| ≤ H (Tx1(t) − Tx2(t))2+ (Ty1(t) − Ty2(t))

2 ≤ D2

Example from [NM12]

39 000

39 500

40 000

Tx1(t) = −3.2484 + 270.7t + 433.12t2 − 324.83999t3

Ty1(t) = 15.1592 + 108.28t + 121.2736t2 − 649.67999t3

Tz1(t) = 38980.8 + 5414t − 21656t2 + 32484t3

Tx2(t) = 1.0828 − 135.35t + 234.9676t22 + 3248.4t3

Ty2(t) = 18.40759 − 230.6364t − 121.2736t2 − 649.67999t3

Tz2(t) = 40280.15999 − 10828t + 24061.9816t2 − 32484t3

D = 5 H = 1000 0 ≤ t ≤1

|Tz1(t) − Tz2(t)| ≤ H (Tx1(t) − Tx2(t))2+ (Ty1(t) − Ty2(t))

2 ≤ D2

Example from [NM12]

Run SMT solver (QF_NRA)

t 7→ 319

16384≈ 0.019470215

Examplevoid swap(int* a, int* b) {*a = *a + *b;*b = *a - *b;*a = *a - *b;

Check if the swap is correct:Heap: array BV32 7→ BV32Update heap line by lineCheck if correct

Incorrect: aliasing

h1 = store(h0, a, h0[a] +32 h0[b])h2 = store(h1, b, h1[a]−32 h1[b])h3 = store(h2, a, h2[a]−32 h2[b])¬(h3[a] = h0[b] ∧ h3[b] = h0[a])

Check if the swap is correct:Heap: array BV32 7→ BV32Update heap line by lineCheck if correct

Incorrect: aliasing

Check if the swap is correct:Heap: array BV32 7→ BV32Update heap line by lineCheck if correctIncorrect: aliasing

Run SMT solver (QF_ABV)

a 7→ 0, b 7→ 0

h0[0] 7→ 1, h1[0] 7→ 2

h2[0] 7→ 0, h3[0] 7→ 0

Satisfiability Modulo Theoriesmodeling and solving

ModelingDepending on the problem domain, select a fitting theory.Consider expressivity vs solving complexity.

SolvingGet an SMT solver that supports the theory.Hope for the best.

Satisfiability Modulo Theoriescommon theories of interest

Uninterpreted Functions (QF_UF)Simplest first-order theory, with equality, applications ofuninterpreted functions, and variables of uninterpreted sorts.

Reflexivity: x = xSymmetry: x = y ⇒ y = xTransitivity: x = y ∧ y = z ⇒ x = zCongruence: x = y ⇒ f(x) = f(y)

Example

f(f(f(x))) = x f(f(f(f(f(x))))) = x f(x) ̸= x

Theory of Arrays [McC93]Operates over sorts array, index, element and function symbols

_[_] : array× index 7→ elementstore : array× index× element 7→ array .

Read-Over-Write-1: store(a, i, e)[i] = eRead-Over-Write-2: i ̸= j ⇒ store(a, i, e)[j] = a[j]Extensionality: a ̸= b ⇒ ∃i : a[i] ̸= b[i]

Example

store(store(a, i, a[j]), j, a[i]) = store(store(a, j, a[i]), i, a[j])

ArithmeticArithmetic constraints (inequalities, equalities) over arithmetic (realor integer) variables.

Difference logic (QF_RDL, QF_IDL):

x− y ≤ 1 , x− y > 10 .

Linear arithmetic (QF_LRA, QF_LIA):

2x− 3y+ 4z ≤ 5 .

Non-linear arithmetic (QF_NRA, QF_NIA):

x2 + 3xy+ y2 > 0 .

Bitvectors (QF_BV)Operates over fixed-size bit-vectors, with bit-vector operations:

concatenation a ◦ b, extraction a[i : j]bit-wise operators∼a, a|b, a&b, …shifts a ≪ k, b ≫ k (logical, arithmetic)arithmetic a+ b, a− b, a ∗ b, a/b, …predicates=,<,≤, … (signed and unsigned)

Semantics similar to programming languages.

Example (a is 32-bits)

(∼a & (a+ 1)) >u a

Satisfiability Modulo Theoriessome other interesting theories

Some other theoriesFloating point [BDG+14, ZWR14]Inductive data-types [BST07]Strings and regular expressions [LRT+14, KGG+09]Quantifiers [DMB07, RTG+13]Differential Equations [GKC13]...

Satisfiability Modulo Theoriesmore information

Books and chapters [BSST09, BM07, KS08]

OnlineSMT-LIB at http://smtlib.cs.uiowa.eduSMT-COMP at http://smtcomp.sourceforge.net

Outline

1 Introduction

2 DPLL(T) Framework

4 MCSAT Framework

5 Finish

SMT Problemhow to solve it?

Check T-satisfiability of a T-formula F

1 Convert to DNF

F ⇔D∨

(Li1 ∧ Li2 ∧ · · · ∧ Lini) .

2 If any of disjuncts is T-satisfiable, return SAT, else UNSAT.

Theory solver/Decision procedure for TProcedure to decide satisfiability of a conjunction of T-literals.

Check T-satisfiability of a T-formula F1 Convert to DNF

F ⇔D∨

(Li1 ∧ Li2 ∧ · · · ∧ Lini) .

F ⇔D∨

(Li1 ∧ Li2 ∧ · · · ∧ Lini) .

F ⇔D∨

(Li1 ∧ Li2 ∧ · · · ∧ Lini) .

SMT Problemsat solver instead of dnf?

Use a SAT solverInstead of DNF: Apply a SAT solver.Check the literals selected by the SAT solver.If not T-satisfiable, add a blocking clause.

Very Lazy SMTexample

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y

(x = y ∨ ¬x = a ∨ ¬y = a)

Check with SAT solver

ViewTheory

( B2 ∨ B3 )

( B4 ∨ B5 )

( B6 ∨ B7 )

( B8 ∨ ¬ B2 ∨ ¬ B4 )

ViewBoolean

( B2 ∨ B3 )

( B4 ∨ B5 )

( B6 ∨ B7 )

( B8 ∨ ¬ B2 ∨ ¬ B4 )

ViewBoolean

( B2 ∨ B3 )

( B4 ∨ B5 )

( B6 ∨ B7 )

( B8 ∨ ¬ B2 ∨ ¬ B4 )

Check with SAT solverJ ¬ B1 , ¬ B8 , ¬ B3 , B2 , ¬ B5 , B4 , ¬ B7 , B6 K

ViewBoolean

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y

(x = y ∨ ¬x = a ∨ ¬y = a)

Check with SAT solverJ ¬a = b, ¬x = y, ¬x = b, x = a, ¬y = b, y = a, ¬z = b, z = a K

ViewTheory

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y

(x = y ∨ ¬x = a ∨ ¬y = a)

ViewTheory

Check with T-solverx = a ∧ y = a ⇒ x = y

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y(x = y ∨ ¬x = a ∨ ¬y = a)

ViewTheory

BlockAdd clause

( B2 ∨ B3 )

( B4 ∨ B5 )

( B6 ∨ B7 )

( B8 ∨ ¬ B2 ∨ ¬ B4 )

ViewBoolean

( B2 ∨ B3 )

( B4 ∨ B5 )

( B6 ∨ B7 )

( B8 ∨ ¬ B2 ∨ ¬ B4 )

Check with SAT solverJ ¬ B1 , ¬ B8 , ¬ B3 , B2 , ¬ B4 , B5 , ¬ B7 , B6 K

ViewBoolean

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y(x = y ∨ ¬x = a ∨ ¬y = a)

Check with SAT solverJ ¬a = b, ¬x = y, ¬x = b, x = a, ¬y = a, y = b, ¬z = b, z = a K

ViewTheory

¬a = b(x = a ∨ x = b)(y = a ∨ y = b)(z = a ∨ z = b)

¬x = y(x = y ∨ ¬x = a ∨ ¬y = a)

Check with SAT solverJ ¬a = b, ¬x = y, ¬x = b, x = a, ¬y = a, y = b, ¬z = b, z = a K

ViewTheory

Check with T-solverSatisfiable

a, x, z 7→ c1b, y 7→ c2

Very Lazy SMTdiscussion

PropertiesSAT and T-solver only communicate via existing literals.Number of possible conflicts finite⇒ termination.Reuse the improvements in SAT solving.SAT solver is “blind” and can get lost :(.

Integrate closely with the SAT solver: DPLL(T) [DMR02, NOT05]Incremental: Check T-satisfiability along the SAT solver.Backtrack: Backtrack with SAT solver and keep context.

Propagation: If existing literals are implied tell SAT solverConflict: Small conflict explanations.

Very Lazy SMTdiscussion

PropertiesSAT and T-solver only communicate via existing literals.Number of possible conflicts finite⇒ termination.Reuse the improvements in SAT solving.SAT solver is “blind” and can get lost :(.

Integrate closely with the SAT solver: DPLL(T) [DMR02, NOT05]Incremental: Check T-satisfiability along the SAT solver.Backtrack: Backtrack with SAT solver and keep context.

Propagation: If existing literals are implied tell SAT solverConflict: Small conflict explanations.

DPLL(T) Frameworktypical architecture

SAT Solver

Arithmetic

Bit-Vectors

UninterpretedFunctions

Theory Literals Explanations

SAT Solver

Arithmetic

Bit-Vectors

SAT SolverStandard “off-the shelf” SAT solverBuild Booleanmodel and notify theories

SAT Solver

Arithmetic

Bit-Vectors

Theory Literals ExplanationsTheory Decision ProceduresCheck conjunctions of literalsIncrementalBacktrackableProducing explanations

SAT Solver

Arithmetic

Bit-Vectors

Combination FrameworkEnsure modular reasoning is correct

DPLL(T) Frameworkgreat but not perfect

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

((ak < bk ∧ bk < ak+1) ∨ (ak < ck ∧ ck < ak+1))

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

And so on...Exponential enumeration of paths.

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

a0<a1 a1<a2 a2<a3

Example (Diamonds)

a0 > an ∧n−1∧k=0

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

a0<a1 a1<a2 a2<a3

Example (Diamonds)

a0 > an ∧n−1∧k=0

Feature/FlawCan only use existing literals!

Outline

1 Introduction

2 DPLL(T) Framework

4 MCSAT Framework

5 Finish

Uninterpreted Functionsthe theory

Literals are of the form x = y, x ̸= y, x = f(x, f(y, z)).Can be decided in O(n log(n)) based on congruence closure.Efficient theory propagation for equalities.Can generate:

small explanations [DNS05],minimal explanations [NO07],smallest explanations NP-hard [FFHP].

Typically the core of the SMT solver and used in other theories.

Uninterpreted Functionscongruence closure

ExampleJ f(x, y) = x, h(y) = g(y), f(f(x, y), y) = z, g(x) ̸= g(z) K

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:

1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:1 g(x) ̸= g(z)

2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:1 g(x) ̸= g(z)2 f(f(x, y), y) = z

3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Conflict:1 g(x) ̸= g(z)2 f(f(x, y), y) = z3 f(x, y) = x

f(f(x,y),y)

f(x,y)

h(x) g(x) g(z)

Difference Logicthe theory

Literals are of the form x− y ◃▹ k, where◃▹∈ {≤, <,=, ̸=, >,≥},x and y are arithmetic variables (integer or real),k is a constant (integer or real).

We can rewrite x− y = k to (x− y ≤ k) ∧ (x− y ≥ k).In integers, we can rewrite x− y < k to x− y ≤ k− 1.In reals, we can rewrite x− y < k to x− y ≤ k− δ.Can assume: all literals of the form x− y ≤ k.

Difference Logicthe theory

Any solution to a set of literals can be shifted:if v is a satisfying assignment, so is v′(x) = v(x) + k.

We can use this to also process simple bounds x ≤ k:introduce fresh variable z (for zero),rewrite each x ≤ k to x− z ≤ k,given a solution v, shift it so that v(z) = 0.

If we allow (dis)equalities as literals, then:in reals, satisfiability is polynomial;in integers, satisfiability is NP-hard.

Difference Logicfrom literals to graph

Example J x ≤ 1, x− y ≤ 2, y− z ≤ 3, z− x ≤ −6 KConstruct a graph from literals.Check if there is a negative path.

Theoremliterals unsatisfiable⇔∃ negative path.

Conflict:(x− y ≤ 2), (y− z ≤ 3), (z− x ≤ −6).

Linear Arithmeticthe theory

Language:Atoms are of the form a1x1 + · · · anxn ≤ b.We can rewrite t = b to (t ≤ b) ∧ (t ≥ b).In integers, we can rewrite t < b to t ≤ b− 1.In reals, we can rewrite t < b to t ≤ b− δ.

Variant of simplex designed for DPLL(T) [DDM06]:IncrementalCheap backtrackingCan do theory propagationCan generate minimal explanationsWorst case exponential (fast in practice)

Linear Arithmetictableau

Rewrite each∑

aixi ≤ b to s ≤ bwith s =∑

We get tableau of equations + simple bounds on variables.Tableau is fixed.Bounds can be asserted and retracted.

Tableaus1 = a1,1 · x1 + · · ·+ a1,i · xj + · · ·+ a1,n · xn...si = ai,1 · x1 + · · ·+ ai,i · xj + · · ·+ ai,n · xn...

sm = am,1 · x1 + · · ·+ am,i · xj + · · ·+ am,n · xn

Bounds−∞ ≤ x1 ≤ u1

l2 ≤ x2 ≤ +∞...

li ≤ si ≤ uilj ≤ xj ≤ uj

l2 ≤ x2 ≤ +∞...

Variables can be basic and non-basic in the tableau.Keep an assignment v of all variables:

v satisfies the tableau,v satisfies bounds on the non-basic variables.

Initially v(x) = 0 and−∞ ≤ x ≤ +∞.

l2 ≤ x2 ≤ +∞...

Variables can be basic and non-basic in the tableau.

Keep an assignment v of all variables:v satisfies the tableau,v satisfies bounds on the non-basic variables.

l2 ≤ x2 ≤ +∞...

Case 1:v satisfies bound on the basic variables too.Satisfiable, v is the model!

l2 ≤ x2 ≤ +∞...

Case 2:v doesn’t satisfy bound on the basic variables si, andall xj’s that si depends on are at their bounds (can’t fix).Unsatisfiable, the row is the explanation.

l2 ≤ x2 ≤ +∞...

Case 3:v doesn’t satisfy bound on the basic variables si, andexists xj’s that si depends on, with slack available.Pivot, update, and continue.

Linear Arithmeticexample

-1 1 2 3 4 5 6

J 2y− x− 2 ≤ 0, −2y− x+ 4 ≤ 0 K

Tableaus1 = 2y− xs2 = −2y− x

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ +∞−∞ ≤ s2 ≤ +∞

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ +∞−∞ ≤ s2 ≤ +∞

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ +∞−∞ ≤ s2 ≤ +∞

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ +∞

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ +∞

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

s2 is inconsistent

There is slack in y.Pivot s2 and y.Update s2 value.Update basics.

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

s2 is inconsistentThere is slack in y.

Pivot s2 and y.Update s2 value.Update basics.

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 K

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

s2 is inconsistentThere is slack in y.Pivot s2 and y.

Update s2 value.Update basics.

1 1 2 3 4 5 6

J s1 ≤ 2, s2 ≤ −4 KTableau

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

s2 is inconsistentThere is slack in y.Pivot s2 and y.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ 0

s2 is inconsistentThere is slack in y.Pivot s2 and y.Update s2 value.

Update basics.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ −4

s2 is inconsistentThere is slack in y.Pivot s2 and y.Update s2 value.

Update basics.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 0

s1 7→ 0

s2 7→ −4

s2 is inconsistentThere is slack in y.Pivot s2 and y.Update s2 value.Update basics.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

s1 is inconsistent

There is slack in x.Pivot s1 and x.Update s1 value.Update basics.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

s1 is inconsistentThere is slack in x.

Pivot s1 and x.Update s1 value.Update basics.

1 1 2 3 4 5 6

s1 = −s2 − 2x

y = −1

2s2 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 4

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.Update s1 value.

Update basics.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 2

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.Update s1 value.

Update basics.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 0

y 7→ 2

s1 7→ 2

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.Update s1 value.Update basics.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 1

y 7→ 3

2s1 7→ 2

s2 7→ −4

s1 is inconsistentThere is slack in x.Pivot s1 and x.Update s1 value.Update basics.

1 1 2 3 4 5 6

x = −1

2s1 −

4s1 −

Bounds−∞ ≤ x ≤ +∞−∞ ≤ y ≤ +∞−∞ ≤ s1 ≤ 2

−∞ ≤ s2 ≤ − 4

Assignment

x 7→ 1

y 7→ 3

2s1 7→ 2

s2 7→ −4

Linear Arithmeticintegers

Classic NP-complete problem [Pap81].Admits quantifier elimination [Coo72].Common approach:

Simplex + Branch-And-Bound [DDM06, Gri12, Kin14]Use Simplex as if variables were real.If UNSAT in reals, then UNSAT in integers too.If SAT and solution is integral, then SAT (lucky).If non-integral solution v(x), then refine:

Branch-and-Bound lemmas: (x ≤ ⌊v(x)⌋) ∨ (x ≥ ⌈v(x)⌉).Cutting plane lemmas: new implied inequality refuting v.

Additionally solve integer equalities.

Sad, but not guaranteed to terminate.

Alternatives [JdM13, BSW15] not yet mature.

Additionally solve integer equalities.Sad, but not guaranteed to terminate.

Arraysthe theory

∀a, i, e : store(a, i, e)[i] = e∀a, i, j, e : i ̸= j ⇒ store(a, i, e)[j] = a[j]

∀a, b : a ̸= b ⇒ ∃i : a[i] ̸= b[i]

Common approach:UF + lemmas on demand [BB09, DMB09].Use UF as if store and _[_]were uninterpreted.If UNSAT in UF, then UNSAT in arrays too.If SAT and solution respects array axioms, then SAT (lucky).If not, then refine by instantiating violated axioms.

Bit-Vectorsthe theory

Common approach:1 Heavy preprocessing2 Encode into SAT (bit-blasting)3 Run a SAT solver

Alternatives [HBJ+14, ZWR16] not yet mature.

Bit-Vectorsbit-blasting

XOR si

OR ci+1

Translation to CNFEach node a new variablesXOR introduces 4 clausesAND introduces 3 clausesOR introduces 3 clauses17 new clauses5 new variables

Bit-Vectorsbit-blasting

XOR si

OR ci+1

Translation to CNFEach node a new variablesXOR introduces 4 clausesAND introduces 3 clausesOR introduces 3 clauses17 new clauses5 new variables

Bit-Blasting Addition/Multiplicationx[32] + y[32] 544 new clauses, 160 new

variablesx[32] × y[32] 10016 new clauses, 3008 new

variables

Outline

1 Introduction

2 DPLL(T) Framework

4 MCSAT Framework

5 Finish

DPLL(T)architecture

SAT Solver

Arithmetic

Bit-Vectors

DPLL(T)pros and cons

GoodSimple interfaceOnly conjunctions of constraints

What can be improved?Simple interface can be restrictiveArbitrary conjunctions of constraints

DPLL(T)pros and cons

GoodSimple interfaceOnly conjunctions of constraints

What can be improved?Simple interface can be restrictiveArbitrary conjunctions of constraints

DPLL(T)simple interface issues

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

a0<a1 a1<a2 a2<a3

Example (Diamonds)

a0 > an ∧n−1∧k=0

DPPL(T)simple interface issues

How to fix this?Extensions of DPLL(T) can add new literals [BNOT06].Magic needed to discover these literals [BDdM08, HAMM14].More pragmatic approach would be desirable.

Rethink the Architecture!Why is SAT solver special?Why the restriction on the interface?Let’s dig deeper into how SAT solvers work.

Boolean Satisfiabilityhistory

xn ∨ · · · ∨ x1 ∨ ym ∨ · · · ∨ y1

Resolution procedure by Davis, Putnam [DP60]Search procedure by Davis, Logemann, Loveland [DLL62]

Resolution (DP)Find a proofSaturationExponential

Search (DLL)Find amodelSearch and backtrackingExponential

Boolean Satisfiabilitymodern

Marques-Silva, Sakallah[SS97] GRASP: A new search algorithm for satisfiabiliy

Moskewicz, Madigan, Zhao, Zhang, Malik[MMZ+01] CHAFF: Engineering an efficient SAT solver

Conflict-Directed Clause LearningUse the search to guide resolutionUse resolution to guide the search

Resolve

Boolean Satisfiabilitycdcl mechanics

Model ConstructionBuild partial model by assigning variables to valuesJ. . . , x, . . . , y, . . . , z, . . .K .

Unit ReasoningReason about unit constraints

(x ∨ y ∨ z ∨ w) .

Explain ConflictsExplain conflicts using clausal reasons

(x ∨ y ∨ z) .

Linear Real Arithmetic

Linear Arithmetica1x1 + · · ·+ anxn ≥ b a1x1 + · · ·+ anxn = b

DPLL(T): SimplexAmodel builder for a conjunction of linear constraints.

Search for a modelEscape conflicts through pivotingBuilt for the DPLL(T) framework

[DDM06] A fast linear-arithmetic solver for DPLL(T)

Linear Arithmetica1x1 + · · ·+ anxn ≥ b a1x1 + · · ·+ anxn = b

Fourier-Motzkin Resolution2x+ 3y− z ≥ −1

6x+ 9y− 3z ≥ −3

−3x− 2y+ 4z ≥ 2

−6x− 4y+ 8z ≥ 4

5y+ 5z ≥ 1

Feels like Boolean resolution (elimination).Behaves like Boolean resolution (exponential).

Model ConstructionBuild partial model by assigning variables to valuesJ. . . , C1, C2, . . . , x 7→ 1/2, . . . , y 7→ 1/2, . . . , z 7→ −1, . . .K .

C1 ≡ (x+ y+ z+ w ≥ 0) C2 ≡ (x+ y+ z− w > 0) .

Explain ConflictsExplain conflicts using valid clausal reasons

(C1 ∨ C2 ∨ x+ y+ z > 0) .

Linear Real Arithmeticexample

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

Explanation C1 ∧ C2 ⇒ x ̸= 0.5

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2K

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5K

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5K

Unit Constraint Reasoning

2y− x− 2 < 0 ⇒ (y < 1.25)

−2y− x+ 4 < 0 ⇒ (y > 1.75)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5KExplanation C1 ∧ C2 ⇒ x ̸= 0.5

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5KExplanation C1 ∧ C2 ⇒

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5KExplanation C1 ∧ C2 ⇒

Fourier-Motzkin

2y− x− 2 < 0 −2y− x+ 4 < 0

−2x+ 2 < 0

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5KExplanation C1 ∧ C2 ⇒ x > 1

Fourier-Motzkin

2y− x− 2 < 0 −2y− x+ 4 < 0

−2x+ 2 < 0

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x 7→ 0.5KExplanation C1 ∨ C2 ∨ (x > 1)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2KExplanation C1 ∨ C2 ∨ (x > 1)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x > 1KExplanation C1 ∨ C2 ∨ (x > 1)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x > 1, x 7→ 2KExplanation C1 ∨ C2 ∨ (x > 1)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

JC1, C2, x > 1, x 7→ 2, y 7→ 1.5KExplanation C1 ∨ C2 ∨ (x > 1)

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

-1 1 2 3 4 5 6

C1︷︸︸︷2y− x− 2 < 0 ∧

C2︷︸︸︷−2y− x+ 4 < 0

Linear Real Arithmeticcomparison to dpll(t)

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

Example (Diamonds)

a0 > an ∧n−1∧k=0

a 1<b 1b

c 1<a 2

a 2<b 2 b

a2 <c2

a0<a1 a1<a2 a2<a3

Example (Diamonds)

a0 > an ∧n−1∧k=0

mcsat cvc4 z3 mathsat5 yicesset solved time (s) solved time (s) solved time (s) solved time (s) solved time (s)clocksynchro (36) 36 123.11 36 1166.55 36 1828.74 36 1732.59 36 1093.80DTPScheduling (91) 91 31.33 91 72.92 91 100.55 89 1980.96 91 926.22miplib (42) 8 97.16 27 3359.40 23 3307.92 19 5447.46 23 466.44sal (107) 107 12.68 107 13.46 107 6.37 107 7.99 107 2.45sc (144) 144 1655.06 144 1389.72 144 954.42 144 880.27 144 401.64spiderbenchmarks (42) 42 2.38 42 2.47 42 1.66 42 1.22 42 0.44TM (25) 25 1125.21 25 82.12 25 51.64 25 1142.98 25 55.32ttastartup (72) 70 4443.72 72 1305.93 72 1647.94 72 2607.49 72 1218.68uart (73) 73 5244.70 73 1439.89 73 1379.90 73 1481.86 73 679.54

596 12735.35 617 8832.46 613 9279.14 607 15282.82 613 4844.53

DPLL(T) Simplex (CVC4)Total Physical Source Lines of Code (SLOC) = 22,597Development Effort Estimate, Person-Years (Person-Months) = 5.28 (63.38)(Basic COCOMO model, Person-Months = 2.4 * (KSLOC**1.05))

Schedule Estimate, Years (Months) = 1.01 (12.10)(Basic COCOMO model, Months = 2.5 * (person-months**0.38))

Estimated Average Number of Developers (Effort/Schedule) = 5.24Total Estimated Cost to Develop = $ 713,502(average salary = $56,286/year, overhead = 2.40).

MCSAT Fourier-Motzkin (CVC4)Total Physical Source Lines of Code (SLOC) = 1,966Development Effort Estimate, Person-Years (Person-Months) = 0.41 (4.88)(Basic COCOMO model, Person-Months = 2.4 * (KSLOC**1.05))

Schedule Estimate, Years (Months) = 0.38 (4.57)(Basic COCOMO model, Months = 2.5 * (person-months**0.38))

Estimated Average Number of Developers (Effort/Schedule) = 1.07Total Estimated Cost to Develop = $ 54,942(average salary = $56,286/year, overhead = 2.40).

Generated using David A. Wheeler’s SLOCCount.

Non-Linear Arithmetic

f(⃗y, x) = am · xdm + am−1 · xdm−1 + · · ·+ a1 · xd1 + a0

f is in Z[⃗y, x], ai are in Z[⃗y]

Examples

f(x, y) = (x2 − 1)y2 + (x+ 1)y− 1 ∈ Z[x, y]g(x) = 16x3 − 8x2 + x+ 16 ∈ Z[x]

Polynomial Constraints

f(x, y) > 0 ∧ g(x) < 0

Non-Linear Arithmetichistory

Tarski [Tar48]Quantifier elimination

Decidable, non-elementary Collins [Col75]Cylindrical Algebraic Decomposition

Doubly-exponential

Non-Linear Real Arithmeticcomplexity

PSPACE

Canny [Can88], Grigor’ev [Gri88]

Non-Linear Real Arithmeticcylindrical algebraic decomposition

p1 > 0 ∨ (p2 = 0 ∧ p3 < 0) p1, p2, p3 ∈ Z[x1, . . . , xn]

Projection (Saturation)Project polynomials using a projection P

{p1, p2, p3} 7→ {p1, p2, p3, p4, . . . , pn} .

Lifting (Model construction)For each variable xk

1 Isolate roots of pi(α, xk).2 Choose a cell C and assign xk 7→ αk ∈ C, continue.3 If no more cells, backtrack.

Non-Linear Real Arithmetic

Model ConstructionBuild partial model by assigning variables to valuesJ. . . , C1, C2, . . . , x 7→ √

2/2, . . .K .

C1 ≡ (x2 + y2 < 1) C2 ≡ (xy > 1) .

(C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1) .

2/2, . . .K .

C1 ≡ (x2 + y2 < 1) C2 ≡ (xy > 1) .

(C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1) .

Non-Linear Real Arithmeticunit constraints

x3 − 2x2 + 1 > 0

− 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0

− 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0

− 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0

− 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0 − 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0 − 3x3 + 8x2 − 4x > 0

-1 1 2 3

x3 − 2x2 + 1 > 0 − 3x3 + 8x2 − 4x > 0

2/2, . . .K .

C1 ≡ (x2 + y2 < 1) C2 ≡ (xy > 1) .

(C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1) .

2/2, . . .K .

C1 ≡ (x2 + y2 < 1) C2 ≡ (xy > 1) .

(C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1) .

Non-Linear Real Arithmeticexample

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

Explanation C1 ∧ C2 ⇒ x ̸=√2/2

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2K

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

Unit Constraint Reasoning

x2 + y2 < 1 ⇒ −√3/2 < y <

√3/2

−2y− x+ 4 < 0 ⇒ y >√2

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

Explanation C1 ∧ C2 ⇒

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

CAD Projection

P = {x,−4 + 4x2, 1− x2 + x4}

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

CAD Projection

P = {x,−4 + 4x2, 1− x2 + x4}

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

Explanation C1 ∧ C2 ⇒ x ≤ 0 ∨ x ≥ 1

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x 7→ √2/2K

Explanation C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2KExplanation C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x ≥ 1KExplanation C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1

- 2 -1 0 1 2

C1︷︸︸︷x2 + y2 < 1 ∧

C2︷︸︸︷xy > 1

JC1, C2, x ≥ 1, x 7→ 3/2KExplanation C1 ∨ C2 ∨ x ≤ 0 ∨ x ≥ 1

Non-Linear Real Arithmeticsmt-comp 2015

100000

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

tive t

problems solved

Yices2Z3

SMT-RATraSATCVC3

CVC4aCVC4b

Model-Based Procedures

Linear Real ArithmeticGeneralizing DPLL to Richer Logics [MKS09]Conflict Resolution [KTV09]Natural Domain SMT [Cot10]

Linear Integer ArithmeticCutting to the Chase: Solving Linear Integer Arithmetic [JdM13]

Non-Linear Real ArithmeticSolving Non-Linear Arithmetic [JDM12, Jov12]

General FrameworkModel-Constructing Satisfiability Calculus [DMJ13, JBdM13]

MCSATsimple architecture

BCP FM

Each plugins reasons about their domain:1 Track when a constraint becomes unit [MMZ+01].2 Unit constraints imply feasible sets of individual variables.3 Propagate any constraints/variables whose value is implied.4 Explain any unit conflicts with clausal explanations.5 When asked, decide unassigned variable to feasible value.

MCSATimplementations

NRA + NIA + UF in Yices2 Link

QF_NRA in Z3 Link

QF_UFLRA in CVC4 Link

Outline

1 Introduction

2 DPLL(T) Framework

4 MCSAT Framework

5 Finish

Research Challengestodo

Extensible and simple SMT solver ala MiniSAT.Integer arithmetic: a complete and practical procedure.Bit-vectors: other than bit-blasting [HBJ+14, ZWR16].Proofs: can we have themwithout toomuch trouble.Quantifiers: push-button, with model generation.MCSAT: arrays, bit-vectors.

Homeworkfor the practical

Download and install the following solversCVC4 Link

MathSAT5 Link

Yices2 Link

Z3 Link

THE END

Homeworkfor the practical

Download and install the following solversCVC4 Link

MathSAT5 Link

Yices2 Link

Z3 Link

THE END

References I

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[BDdM08] Nikolaj Bjørner, Bruno Dutertre, and Leonardo de Moura.Accelerating lemma learning using joins-dppl (join).Proceedings of short papers at LPAR, 8, 2008.

[BDG+14] Martin Brain, Vijay D’Silva, Alberto Griggio, Leopold Haller, and Daniel Kroening.Deciding floating-point logic with abstract conflict driven clause learning.Formal Methods in System Design, 45(2):213–245, 2014.

[BM07] Aaron R Bradley and Zohar Manna.The calculus of computation: decision procedures with applications to verification.Springer Science & Business Media, 2007.

[BNOT06] Clark Barrett, Robert Nieuwenhuis, Albert Oliveras, and Cesare Tinelli.Splitting on demand in sat modulo theories.In Logic for Programming, Artificial Intelligence, and Reasoning, pages 512–526,2006.

[BSST09] Clark W Barrett, Roberto Sebastiani, Sanjit A Seshia, and Cesare Tinelli.Satisfiability modulo theories.Handbook of satisfiability, 185:825–885, 2009.

References II

[BST07] Clark Barrett, Igor Shikanian, and Cesare Tinelli.An abstract decision procedure for satisfiability in the theory of recursive datatypes.Electronic Notes in Theoretical Computer Science, 174(8):23–37, 2007.

[BSW15] Martin Bromberger, Thomas Sturm, and Christoph Weidenbach.Linear integer arithmetic revisited.In International Conference on Automated Deduction, pages 623–637. Springer,2015.

[Can88] John Canny.Some algebraic and geometric computations in pspace.In Proceedings of the twentieth annual ACM symposium on Theory of computing,pages 460–467, 1988.

[Col75] George E Collins.Quantifier elimination for real closed fields by cylindrical algebraicdecomposition.In Automata Theory and Formal Languages, pages 134–183, 1975.

[Coo72] David C Cooper.Theorem proving in arithmetic without multiplication.Machine Intelligence, 7(91-99):300, 1972.

References III

[Cot10] Scott Cotton.Natural domain SMT: A preliminary assessment.Formal Modeling and Analysis of Timed Systems, pages 77–91, 2010.

[DDM06] Bruno Dutertre and Leonardo De Moura.A fast linear-arithmetic solver for DPLL(T).In International Conference on Computer Aided Verification, pages 81–94.Springer, 2006.

[DLL62] Martin Davis, George Logemann, and Donald Loveland.A machine program for theorem-proving.Communications of the ACM, 5(7):394–397, 1962.

[DMB07] Leonardo De Moura and Nikolaj Bjørner.Efficient e-matching for smt solvers.In International Conference on Automated Deduction, pages 183–198. Springer,2007.

[DMB09] Leonardo De Moura and Nikolaj Bjørner.Generalized, efficient array decision procedures.In Formal Methods in Computer-Aided Design, pages 45–52. IEEE, 2009.

[DMB11] Leonardo De Moura and Nikolaj Bjørner.Satisfiability modulo theories: introduction and applications.Communications of the ACM, 54(9):69–77, 2011.

References IV[DMJ13] Leonardo De Moura and Dejan Jovanović.

A model-constructing satisfiability calculus.In Verification, Model Checking, and Abstract Interpretation, pages 1–12, 2013.

[DMR02] Leonardo De Moura and Harald Rueß.Lemmas on demand for satisfiability solvers.2002.

[DNS05] David Detlefs, Greg Nelson, and James B Saxe.Simplify: a theorem prover for program checking.Journal of the ACM (JACM), 52(3):365–473, 2005.

[DP60] Martin Davis and Hilary Putnam.A computing procedure for quantification theory.Journal of the ACM (JACM), 7(3):201–215, 1960.

[FFHP] Andreas Fellner, Pascal Fontaine, Georg Hofferek, and BrunoWoltzenlogel Paleo.Np-completeness of small conflict set generation for congruence closure.

[GKC13] Sicun Gao, Soonho Kong, and Edmund M Clarke.Satisfiability modulo ODEs.In Formal Methods in Computer-Aided Design (FMCAD), 2013, pages 105–112. IEEE,2013.

[Gri88] D Yu Grigor’ev.Complexity of deciding tarski algebra.Journal of symbolic Computation, 5(1):65–108, 1988.

References V

[Gri12] Alberto Griggio.A practical approach to satisfiability modulo linear integer arithmetic.Journal on Satisfiability, Boolean Modeling and Computation, 8:1–27, 2012.

[HAMM14] Julien Henry, Mihail Asavoae, David Monniaux, and Claire Maïza.How to compute worst-case execution time by optimization modulo theory anda clever encoding of program semantics.In Proceedings of the 2014 SIGPLAN/SIGBED conference on Languages, compilersand tools for embedded systems, pages 43–52. ACM, 2014.

[HBJ+14] Liana Hadarean, Kshitij Bansal, Dejan Jovanović, Clark Barrett, and CesareTinelli.A tale of two solvers: Eager and lazy approaches to bit-vectors.In International Conference on Computer Aided Verification, pages 680–695.Springer, 2014.

[JBdM13] Dejan Jovanović, Clark Barrett, and Leonardo de Moura.The design and implementation of themodel constructing satisfiability calculus.Formal Methods in Computer-Aided Design, 2013.

[JDM12] Dejan Jovanović and Leonardo De Moura.Solving non-linear arithmetic.In International Joint Conference on Automated Reasoning, pages 339–354.Springer, 2012.

References VI

[JdM13] Dejan Jovanović and Leonardo de Moura.Cutting to the chase.Journal of automated reasoning, 51(1):79–108, 2013.

[Jov12] Dejan Jovanović.SMT Beyond DPLL(T): A New Approach to Theory Solvers and Theory Combination.PhD thesis, Courant Institute of Mathematical Sciences New York, 2012.

[KGG+09] Adam Kiezun, Vijay Ganesh, Philip J Guo, Pieter Hooimeijer, and Michael D Ernst.HAMPI: a solver for string constraints.In Proceedings of the eighteenth international symposium on Software testing andanalysis, pages 105–116. ACM, 2009.

[Kin14] Tim King.Effective Algorithms for the Satisfiability of Quantifier-Free Formulas Over LinearReal and Integer Arithmetic.PhD thesis, Courant Institute of Mathematical Sciences New York, 2014.

[KS08] Daniel Kroening and Ofer Strichman.Decision procedures: an algorithmic point of view.Springer Science & Business Media, 2008.

[KTV09] Konstantin Korovin, Nestan Tsiskaridze, and Andrei Voronkov.Conflict resolution.In Principles and Practice of Constraint Programming, pages 509–523. 2009.

References VII[LRT+14] Tianyi Liang, Andrew Reynolds, Cesare Tinelli, Clark Barrett, and Morgan Deters.

A DPLL(T) theory solver for a theory of strings and regular expressions.In International Conference on Computer Aided Verification, pages 646–662.Springer, 2014.

[McC93] John McCarthy.Towards a mathematical science of computation.In Program Verification, pages 35–56. Springer, 1993.

[MKS09] Kenneth L McMillan, Andreas Kuehlmann, and Mooly Sagiv.Generalizing DPLL to richer logics.In Computer Aided Verification, pages 462–476, 2009.

[MMZ+01] MatthewWMoskewicz, Conor F Madigan, Ying Zhao, Lintao Zhang, and SharadMalik.Chaff: Engineering an efficient sat solver.In Proceedings of the 38th annual Design Automation Conference, pages 530–535.ACM, 2001.

[NM12] Anthony Narkawicz and César A Munoz.Formal verification of conflict detection algorithms for arbitrary trajectories.Reliable Computing, 17(2):209–237, 2012.

[NO07] Robert Nieuwenhuis and Albert Oliveras.Fast congruence closure and extensions.Information and Computation, 205(4):557–580, 2007.

References VIII

[NOT05] Robert Nieuwenhuis, Albert Oliveras, and Cesare Tinelli.Abstract dpll and abstract dpll modulo theories.In International Conference on Logic for Programming Artificial Intelligence andReasoning, pages 36–50. Springer, 2005.

[Pap81] Christos H Papadimitriou.On the complexity of integer programming.Journal of the ACM (JACM), 28(4):765–768, 1981.

[RTG+13] Andrew Reynolds, Cesare Tinelli, Amit Goel, Sava Krstić, Morgan Deters, andClark Barrett.Quantifier instantiation techniques for finite model finding in smt.In International Conference on Automated Deduction, pages 377–391. Springer,2013.

[SS97] João P Marques Silva and Karem A Sakallah.Grasp—a new search algorithm for satisfiability.In Proceedings of the 1996 IEEE/ACM international conference on Computer-aideddesign, pages 220–227. IEEE Computer Society, 1997.

[Tar48] Alfred Tarski.A decision method for elementary algebra and geometry.1948.

References IX

[ZWR14] Aleksandar Zeljić, Christoph MWintersteiger, and Philipp Rümmer.Approximations for model construction.In International Joint Conference on Automated Reasoning, pages 344–359.Springer, 2014.

[ZWR16] Aleksandar Zeljić, Christoph MWintersteiger, and Philipp Rümmer.Deciding bit-vector formulas with mcsat.In International Conference on Theory and Applications of Satisfiability Testing,pages 249–266. Springer, 2016.

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