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

Daniel Kroening and Ofer Strichman 1

Decision Procedures in First Order Logic

Decision Procedures forEquality Logic

Decision Procedures An algorithmic point of view 2

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


We will first investigate methods that solve Equality Logic. Uninterpreted functions are eliminated with one of the reduction schemes.

Our starting point: the E-Graph GE(E)

Recall: GE(E) represents an abstraction of E:

It represents ALL equality formulas with the same set of equality predicates as E

Decision Procedures for Equality Logic


From Equality to Propositional LogicBryant & Velev 2000: the Sparse method

E = x1 = x2 Æ x2 = x3 Æ x1 x3

enc = e1 Æ e2 Æ :e3

Encode all edges with Boolean variables (note: for now, ignore polarity) This is an abstraction Transitivity of equality is lost! Must add transitivity constraints!

e 3

e2

e1


From Equality to Propositional Logic

E = x1 = x2 Æ x2 = x3 Æ x1 x3

enc = e1 Æ e2 Æ :e3

For each cycle add a transitivity constrainttrans = (e1 Æ e2 ! e3) Æ

(e1 Æ e3 ! e2) Æ

(e3 Æ e2 ! e1)

Check: enc Æ trans

e 3

e2

e1



There can be an exponential number of cycles, so let’s try to make it better.

Thm: it is sufficient to constrain simple cycles only

e1

e2 e3

e4

e5e6

T

T T

TT

F



Still, there is an exponential number of simple cycles. Thm [Bryant & Velev]: It is sufficient to constrain

chord-free simple cycles

e1

e2

e3

e4

e5

T

TF

T

F

T


Still, there can be an exponential number of chord-free simple cycles…

Solution: make the graph ‘chordal’ by adding edges.

….




Dfn: A graph is chordal iff every cycle of size 4 or more has a chord.

How to make a graph chordal ? eliminate vertices one at a time, and connect their neighbors.



Once the graph is chordal, we can constrain only the triangles.

Note that this procedure adds not more than a polynomial # of edges, and results in a polynomial no. of constraints.

T

T

TT

F

TTContradiction!


Improvement

So far we did not consider the polarity of the edges.

Claim: in the following graph trans = e3 Æ e2 ! e1 is sufficient

This is only true because of monotonicity of NNF

e1

e2

e3


Definitions Dfn: A contradictory Cycle C is constrained under

T if T does not allow this assignment

C =

F

T

T T

T


Main theorem

If

T R constrains all simple contradictory cycles,

and

For every assignment S, S ² T S ! S ² T R

then E is satisfiable iff B Æ T R is satisfiable

The Equality Formula

From the Sparse method


Transitivity: 5 constraintsRTC: 0 constraints

Transitivity: 5 constraintsRTC: 1 constraint

F

T

T

T

T


Proof of the main theorem

() E is satisfiable BÆT S is satisfiable BÆT R is satisfiable

() Proof strategy: Let R be a satisfying assignment to B Æ T R We will construct S that satisfies B Æ T S

From this we will conclude that E is satisfiable

Skip proof


Definitions for the proof…

A Violating cycle under an assignment R

This assignment violates T S but not necessarily T R

eF

eT2

eT1

T

TF

Either dashed or

solid


More definitions for the proof… An edge e = (vi,vj) is equal under an assignment iff

there is an equality path between vi and vj all assigned T under Denote:

T

TF

TTv1 v2

v3


More definitions for the proof… An edge e = (vi,vj) is disequal under an assignment iff

there is a disequality path between vi and vj in which the solid edge is the only one assigned false by Denote:

T

TF

TTv1 v2

v3


Proof… Observation 1:

The combinationis impossible if = R

(recall: R ² T R)

Observation 2: if (v1,v3) is solid, then

FT

Tv1 v2

v3


ReConstructing S

Type 1:

It is not the case that

Assign S (e23) = F

Type 2:

Otherwise it is not the case that

Assign (e13) = T

FT

T

In all other cases S = R

FT

T

F T

v1 v2

v3

v1 v2

v3


ReConstructing S

Starting from R, repeat until convergence: (eT) := F in all Type 1 cycles

(eF) := T in all Type 2 cycles

All Type 1 and Type 2 triangles now satisfy T S B is still satisfied (monotonicity of NNF) Left to prove: all contradictory cycles are still

satisfied


Proof…

Invariant: contradictory cycles are not violating throughout the reconstruction.

contradicts the precondition to make this assignment…

FT

Tv1 v2

v3

F

T

T


Proof…

Invariant: contradictory cycles are not violating throughout the reconstruction.

contradicts the precondition to make this assignment…

FT

Tv1 v2

v3

TT

F


Applying RTC

How can we use the theorem without enumerating contradictory cycles ?

Answer: Consider the chordal graph. Constrain triangles if they are part of a (simple)

contradictory cycle How?


•Should we constrain this triangle?•In which direction ?

•Is this constraint necessary ?


Decomposing the graph

Focus on Bi-connected dashed components built on top of a solid edge Includes all contradictory cycles involving this edge


Make the component chordal Chordal-ity guarantees: every cycle contains a simplicial

vertex, i.e. a vertex that its neighbors are connected.


The RTC algorithm Constraints cache:

e2 Æ e3 ! e1

e4 Æ e7 ! e2

e5 Æ e8 ! e4

1

23

4

5

6

8

9

1211

7


Constrains all contradictory cycles Constraints cache:

e2 Æ e3 ! e1

e4 Æ e7 ! e2

e6Æ e3 ! e4

1

23

4

5

6

8

9

1211

7


Results – random graphs

0

50000

100000

150000

200000

250000

300000

350000

400000

01:1001:0501:0201:0102:0105:0110:01

Solid : Dashed

# c

on

str

ain

ts

Sparse

RTC

V=200, E=800, 16 random topologies


Random graphs (Satisfiable)

Documents

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