Automated Theorem Proving Lecture 4

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Automated Theorem Proving Lecture 4.  Formula := A |  |    A  Atom := b | t = 0 | t < 0 | t  0 t  Term := c | x | t + t | t – t | ct | Select(m,t) m  MemTerm := f | Update(m,t,t) f  Field b  SymBoolConst x  SymIntConst c  {…,-1,0,1,…}. Memory axiom.

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### Automated Theorem ProvingLecture 4

 Formula := A |  |   

A  Atom := b | t = 0 | t < 0 | t  0

t  Term := c | x | t + t | t – t | ct | Select(m,t)

m  MemTerm := f | Update(m,t,t)

f  Field

b  SymBoolConst

x  SymIntConst

c  {…,-1,0,1,…}

Memory axiom

for all objects o and o’, and memories m:

 o = o’  Select(Update(m,o,v),o’) = v

o  o’  Select(Update(m,o,v),o’) = Select(m,o’)

{ b.f = 5 } a.f = 5 { a.f + b.f = 10 }

iff

Select(f,b) = 5 

Select(Update(f,a,5),a) + Select(Update(f,a,5),b)  10

is unsatisfiable

theory of arithmetic: 5, 10, +

theory of arrays: Select, Update, f

Constraints that arise in program verification are mixed!

x = w, y = w

z = z’

Theories communicating via equality and variables

Select(f,b) = 5 

Select(Update(f,a,5),a) + Select(Update(f,a,5),b)  10

Introduce:

variable w to represent Select(f,b)

variable x to represent Select(Update(f,a,w),a)

variable y to represent Select(Updatef,a,w),b)

variables z and z’ to eliminate the arithmetic disequality

Theory of arithmetic

Theory of arrays

w = 5

x + y = z

z’ = 10

w = Select(f,b)

x = Select(Update(f,a,w),a)

y = Select(Update(f,a,w),b)

z  z’

Theory of arrays
•  Formula := A |   

A  Atom := t = t | t  t

t  Term := c | Select(m,t)

m  MemTerm := f | Update(m,t,t)

c  SymConst

for all objects o and o’, and memories m:

 o = o’  Select(Update(m,o,v),o’) = v

o  o’  Select(Update(m,o,v),o’) = Select(m,o’)

Theory of Equality with Uninterpreted Functions
•  Formula := A |   

A  Atom := t = t | t  t

t  Term := c | f(t,…,t)

c  SymConst

f  Function

for all constants a and b and functions f:

- a = a

- a = b  b = a

- a = b  b = c  a = c

- a = b  f(a) = f(b)

f(f(f(f(f(a))))) = a

f(f(f(a))) = a

f(a,b) = a

f(f(a,b),b) = b

f(a,b) = b

f(f(a)) = a

a = b

f(a) = a

f(f(f(f(a)))) = a

f(f(f(f(f(a))))) = a

f(f(f(a))) = a

f

f

f

f

f

f

a

b

f

a

f(a,b) = a

f(f(a,b),b) = b

f

Congruence closure algorithm

f

f

f

f

f

a

b

f

e-graph

a

Use union-find algorithm to maintain equivalence

classes on terms.

Decision procedure for EUF

1. Construct initial e-graph for all terms appearing in

equalities and disequalities.

2. Apply congruence closure ignoring disequalities.

3. If there is a disequality t1 t2 and an equivalence

class containing both t1 and t2, return unsatisfiable.

4. Otherwise, return satisfiable.

Soundness

Theorem: If the algorithm returns unsatisfiable,

the constraints are unsatisfiable.

Lemma: At every step of the congruence closure

algorithm, each equality in the e-graph is implied

by the original set of equalities.

Proof: By induction on the number of steps.

Completeness

Theorem: If the algorithm returns satisfiable,

there is a model satisfying the constraints.

Model
• A (finite or infinite) universe U
• An interpretation I
• maps each constant symbol u to an
• element I(u)  U
• maps each function symbol f to a
• function I(f)  (UU)
Completeness

Theorem: If the algorithm returns satisfiable,

there is a model satisfying the constraints.

How do we construct the model?

f

f(a,b) = a

f(f(a,b),b) = b

f

a

b

For any term t in the e-graph, let EC(t) be the equivalence

class containing t.

U = set of equivalence classes + new element 

I(c) = EC(c)

I(f)() = EC(f(u)), if u. f(u) is a term in the e-graph

I(f)() = , otherwise

Convexity

A conjunction of facts is convex if whenever it entails a

disjunction of equalities, it also entails at least one equality

by itself.

If C  a1 = b1  …  an = bn

Then there is i  [1,n] such that C  ai = bi

A theory is convex if ever conjunction of facts in

the theory is convex.

EUF is convex

Suppose C  u1 = t1  u2 = t2

Then C  u1  t1  u2  t2 is unsatisfiable

The congruence closure algorithm demonstrates that

there is some i such that even C  ui  ti is unsatisfiable

Uninterpreted theory

Function symbols: f1, f2, … (each with an arity  {0,1,…})

Relation symbols: R1, R2, … (each with an arity  {0,1,…})

Special relation: equality (arity 2)

Variables: x1, x2, …

Boolean facts: x1 = x2,x1 x2,R(x1, x2), R(x1, x2), x. R(x,y)

A conjunction of facts is consistent iff there is a

model (U,I) that satisfies each fact in the conjunction.

e.g., EUF, arrays, lists

Interpreted theory

Function symbols: f1, f2, … (each with an arity  {0,1,…})

Relation symbols: R1, R2, … (each with an arity  {0,1,…})

Special relation: equality (arity 2)

Variables: x1, x2, …

Boolean facts: x1 = x2,x1 x2,R(x1, x2), R(x1, x2), x. R(x,y)

Fixed model (U,I) providing an interpretation for

the function and relation symbols.

A conjunction of facts is consistent iff I can be extended

to the free variables of the conjunction so that each fact

in the conjunction is satisfied.

e.g., arithmetic over rationals, arithmetic over integers

Communicating theories
• Suppose the only shared symbols between two theories T1 and T2 are equality and variables
• C1 is conjunction of facts in theory T1
• C2 is conjunction of facts in theory T2
• Suppose C1 is consistent by itself and C2 is consistent by itself
• Is C1  C2 consistent?

x = y

C2

C1

x  y

y + z  x

z  0

g1 = g2 – g3

f(g1)  f(z)

g2 = f(x)

g3 = f(y)

g2 = g3

f(f(x) – f(y))  f(z)  x  y  y + z  x  z  0

g1 = z

C1 is consistent

C2 is consistent

But C1  C2 is not consistent!

For any conjunction C1 of facts in the theory of rationals

and any conjunction C2 of facts in the theory of EUF,

it suffices to communicate equalities over shared variables.

What if C1 is a conjunction of facts in the theory of

arithmetic over integers?

C2

C1

1  x

x  2

a = 1

b = 2

f(x)  f(a)

f(x)  f(b)

C1  x = a  x = b  f(x) = f(a)  f(x) = f(b) = C2

The equality sharing procedure does not work

because the theory of integers is non-convex

(although the theory of rationals is convex)!

Fix: Communicate disjunctions of equalities!

1  x

x  2

a = 1

b = 2

f(x)  f(a)

f(x)  f(b)

 x = a  x = b

1  x

x  2

a = 1

b = 2

x = a

f(x)  f(a)

f(x)  f(b)

x = a

4, 2, x = b

Unsatisfiable

1  x

x  2

a = 1

b = 2

x = b

f(x)  f(a)

f(x)  f(b)

x = b

Unsatisfiable

1  x

x  2

a = 1

b = 2

f(x) = a

f(a) = b

f(b) = b

 x = a  x = b

a = b

a = b

1  x

x  2

a = 1

b = 2

x = a

f(x) = a

f(a) = b

f(b) = b

x = a

4, 3, x = b

Unsatisfiable

a = b

a = b

1  x

x  2

a = 1

b = 2

x = b

f(x) = a

f(a) = b

f(b) = b

x = b

Unsatisfiable

The procedure returns satisfiable only when

• C1 is consistent
• C2 is consistent
• C1 is convex
• C2 is convex
• C1 entails (x = y) iff C2 entails (x = y)

Theorem: If the procedure returns satisfiable, then

there is a model of C1  C2.

• Technical side conditions:
• Every consistent formula in T1 has a countably
• infinite model
• (2) Every consistent formula in T2 has a countably
• infinite model
Proof

Partition variables into equivalence classes Q1, …, Qn such

that for all i  [1,n], if x,y  Qi then C1 entails x = y.

Lemma: For all i  [1,n], if x,y  Qi then C2 entails x = y.

For each i  [1,n], pick representative wi Qi.

Lemma: C1 1  i < j  n(wi  wj) is consistent.

Lemma: C2 1  i < j  n(wi  wj) is consistent.

Proof continued

D1 = C1 1  i < j  n(wi  wj)

D2 = C2 1  i < j  n(wi  wj)

D1 has a countably infinite model (U1, I1)

D2 has a countably infinite model (U2, I2)

Pick an isomorphism K from U1 to U2 that is consistent

with variable assignments, i.e., for all x, K(I1(x)) = I2(x).

The interpretations of function and relation symbols can

be mapped easily using K.