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

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A propositional world

Ofer Strichman

School of Computer Science, Carnegie Mellon University

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Integrated decision procedures in Theorem-Provers

Deciding a combination of theories is the key for automationin Theorem Provers:

Boolean operators, Bit-vector, Sets, Linear-Arithmetic,Uninterpreted functions, More …

f(f(x)-f(y)) != f(z) & y <=x + 2 | b & 3 > 10

Uninterpreted functions

Linear Arithmetic

Bit-Vectoroperators

Normally, each theory is solved with its own decision procedure andthe results are combined (Shostak, Nelson..).

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Integrated decision procedures in Theorem-Provers

All of these theories, except linear arithmetic, have knownefficient 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.

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Linear Arithmetic and its sub-theories},,,,{

iii cxa

2x –3y +5z < 05x + 2w 2

Some useful methods for solving a conjunction of lineararithmetic expressions: 1. Simplex, Elliptic curve2. Variable Elimination Methods (Hodes, Fourier-Motzkin,..)3. Shostak’s loop residues4. Separation theory: Bellman / Pratt ...5. ...

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A decision procedure for separation theory

Separation predicates have the form x > y + cwhere 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)

x

y z

31

-1

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Handling disjunctions through case splitting

All previously mentioned algorithms handle disjunctionsby splitting the formula.

This can be thought of as a two stage process:

1. Convert formula to Disjunctive Normal Form (DNF)2. Solve each clause separately, until satisfying one of them.

(A common improvement: split ‘when needed’)

Case splitting is frequently the bottleneck of the procedure

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So what can be done against case-splitting ?

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

Answer: Split the domain, not the formula.

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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 ?

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SAT vs. infinite-state decision procedures

1. Pruning.

2. Learning.

3. Guidance (prioritizing internal steps)

Three mechanisms, crucial for efficient decision making:

SAT has a significant advantage in all three.

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SAT vs. infinite-state decision procedures (1/4)

1. Pruning

SAT: each clause c prunes up to 2|v|-|c| states.

Others: ? (stops when finds a satisfiable clause)

y

x0

01

1

Backtrack

Pruned!

.(x y) . .

|v|=1000, |c| =2Pruning 2998 states

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2. Learning

SAT: Partial assignments that lead to a conflict are recorded andhence not repeated.

Others: (depends on decision procedure) - Adding proved sub-goals as antecedents to new sub-goals

- …

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3. Guidance (prioritizing internal steps)

Guidance requires efficient estimation:

Consider 1 2, where 1 is unsat and hard, and 2 is sat and easy.

With proper guidance, a theorem prover should start from 2.

- How hard it is to solve each sub-formula?

- To what extent will it simplify the rest of the proof?

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3. Guidance (cont’d)

“..To what extent will it simplify the rest of the proof?”

SAT: Guidance through decision heuristics (e.g. DLIS).

Others: Expression ordering, ...

(x y z)(x v)(~x ~z)

Estimating simplification by counting literals

in each phase

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Example: Equality Logic with Uninterpreted Functions (1/3)

Equality Logic with Uninterpreted Functions:

))(),((),()()( 2121 yfxfgzuugzyfuxfu

(Uninterpreted functions are reducible to equality logic. Thus, we can concentrate on equality logic)

Traditional infinite-state decision procedure:Congruence Closure with case splitting.

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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

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Example: Equality Logic (3/3)

Let (x=y, y=z, x=z) be the equality predicates in .

x

y z

exyexz

eyz

2. Impose transitivity on cycles: exy + eyz + exz 2

1. Construct the equality graph.

The resulting formula is propositional BDDs , SAT, etc.

Bryant et. al. (CAV’00): Add transitivity constraints to the formula.

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This work

1. Separation predicates:

2. Separation predicates for integers:

3. Linear arithmetic:

4. Integer linear arithmetic:

czyxyx 232;real:,

cyxyx ;real:,

cyxyx int;:,

czyxyx 232int;:,

Extends the results of Bryant et.al. to a Boolean combination of:

Done

},{

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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.

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Reducing separation predicates to propositional logic (1/6)

: f(x) > f(y+1)

: (x=y+1 f1=f2) (f1>f2)

A. Normalize (example):

: (x>y+1 y>x-1 (f1 f2 f2 f1)) (f1>f2)

1. Uninterpreted functions equality logic

xy+1 f1=f2

Now has no negations and only the ‘>’ and ‘’ predicate symbols.

2. Normal form

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x

y z

31

-1


: ( x > z +3 (z > y –1 y x+1))

e yz ,1

, e xy,1,’:

Transitivity constraintse zx

,3, ( ))(

B. Encode + construct graph (example):

x

y z

-3-1

1

Separationgraph:

and itsdual:

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x

y z

31

-1


e yz ,1

, e xy,1,’:

Transitivity constraintse zx

,3, ( ))(

C. Add transitivity constraints for each simple cycle (example):

’: (( ))e zx,3, e yz

,1, e xy

,1, e yz

,1, e xy

,1,e zx

,3, ( ))(

x

y z

-3-1

1

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Compact representation of constraints (1/4)

.....

In most cases - yes.

e.g. If the diamonds are ‘balanced’ (c1 + c2 = c3 + c4) O(n) constraints

.....c1c2

c1+ c2

n diamonds 2n simple cycles.

Can we do better than that ?

c3c4

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Chordal graphs: each cycle of size greater than 3, has a ‘chord’.

In the equality predicates case:Let C be a cycle in GLet 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.

G:

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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.

c1c2

c3 c4

c1+ c2

c5

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Complexity of making the graph chordal:

1. If the diamonds are ‘balanced’ O(n) constraints

3. Worst case O(2n)

.....c1c1c1c1

c2c2c2c2

2. If there are uniform weights c1 and c2, c1 c2 on top and bottom

paths O(n2) constraints

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Extension to integer variables (1/2)cyxyx int;:,

Given with integer separation predicates, derive R:

• Declare all variables as real.

Theorem: is satisfiable iff R is satisfiable

(c is an integer)

• For each predicate x > y + c, add a constraint x > y + c x y + c + 1

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Experimental results (1/3)

.....

n diamonds

Each diamond has 2d edges

Top and bottom paths in each diamond are disjuncted.

There are 2n conjuncted cycles.

By adjusting the weights, we ensured that there is a single

satisfying assignment.

d=2

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n d ICS PVS CoqGraphanalysis Chaff

3 2 < 1 < 1 < 14 2 5.9 < 1 < 15 2 95.1 < 1 < 1

7 4 > 104 > 104 < 1 < 1

100 5 > 104 32 < 1

250 5 > 104 754 1.6

500 5 > 104 > 104

To be continued...

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M odel Steps ICS Graphanalysis Chaff

Load - 1 < 1 < 1 < 1store 2 87.1 < 1 < 1

unit 3 > 104 90 1Out-of- 2 < 1 < 1 < 1

order-unit 3 > 104 2.9 < 1Cache- 1 < 1 < 1 < 1Protocol 2 1.8 < 1 < 1

To be continued...

The procedure has recently been integrated into SyMP and Euclid.We currently experiment with real software verification problems.

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Next: Linear Arithmetic (1/2)

x > y + c x yc

c1c3

c2

Adding constraints according to accumulated cycle weight:

The test c1 + c2 + c3 > 0 results in a yes/no answer

Separation predicates:

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Next: Linear Arithmetic (2/2)

x > y + 2z + c x y2z + c

2z + c 3

2

x

yThe 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.

Linear Arithmetic:

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

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