Technion 1 (Yet another) decision procedure for Equality Logic Ofer Strichman and Orly Meir Technion

2 Equality Logic  E : ( x 1 = x 2 Æ ( x 2  x 3 Ç x 1  x 3 )) Domain: x 1, x 2, x 3 2 N The satisfiability problem: is there an assignment to x 1, x 2, x 3 that satisfies  E ? Q: When is Equality Logic useful ?

Technion3 Equality Logic  E :( x 1 = x 2 Æ ( x 2  x 3 Ç x 1  x 3 )) A: Mainly when combined with Uninterpreted Functions f ( x, y ), g ( z ),… Mainly used in proving equivalences, but not only

Technion4 Basic notions  E : x = y Æ y = z Æ z  x x y z (non-polar) Equality Graph: Gives an abstract view of  E

Technion5 From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method  E : x 1 = x 2 Æ x 2 = x 3 Æ x 1  x 3 B : e 1,2 Æ e 2,3 Æ :e 1,3 Encode all edges with Boolean variables  This is an abstraction  Transitivity of equality is lost!  Must add transitivity constraints! e 1,3 e 1,2 e 2,3 x1x1 x2x2 x3x3

Technion6 From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method  E : x 1 = x 2 Æ x 2 = x 3 Æ x 1  x 3 B : e 1,2 Æ e 2,3 Æ :e 1,3 Transitivity Constraints: For each cycle of size n, forbid a true assignment to n -1 edges T S = ( e 1,2 Æ e 2,3 ! e 1,3 ) Æ ( e 1,2 Æ e 1,3 ! e 2,3 ) Æ ( e 1,3 Æ e 2,3 ! e 1,2 ) Check: B Æ T S e 1,3 e 1,2 e 2,3 x1x1 x2x2 x3x3

Technion7 Thm-1: It is sufficient to constrain simple cycles only e1e1 e2e2 e3e3 e4e4 e5e5 e6e6 T TT TT F From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method

Technion8 Thm-2: It is sufficient to constrain chord-free simple cycles e1e1 e2e2 e3e3 e4e4 e5e5 T T T F T F From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method

Technion9 Still, there can be an exponential number of chord- free simple cycles… Solution: make the graph ‘chordal’! …. From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method

Technion10 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. From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method

Technion11 In a chordal graph, it is sufficient to constrain only triangles. Polynomial # of edges and constraints. # constraints = 3 £ #triangles T T T T F TT Contradiction! From Equality to Propositional Logic Bryant & Velev CAV’00 – the Sparse method

Technion12 An improvement Reduced Transitivity Constraints (RTC) So far we did not consider the polarity of the edges. Assuming  E is in Negation Normal Form  E : x = y Æ y = z Æ z  x x y z (polar) Equality Graph: = = 

Technion13 Monotonicity of NNF Thm-3: NNF formulas are monotonically satisfied (in CNF this is simply the pure literal rule) Let  be in NNF and satisfiable. Thm-3 implies:  Let  ²   Derive  ’ from  by switching the value of a ‘mis-assigned’ pure literal in   Now  ’ ² 

Technion14 Claim: in the following graph T R = e 3 Æ e 2 ! e 1 is sufficient This is only true because of monotonicity of NNF (an extension of the pure literal rule) An improvement Reduced Transitivity Constraints (RTC) e1e1 e2e2 e3e3 x z y = =  Allowing e.g. e 1 = e 2 = T, e 3 = F

Technion15 Basic notions  E : x = y Æ y = z Æ z  x The Equality predicates: { x = y, y = z, z  x } which we can break to two sets: E = ={ x = y, y = z }, E  = { z  x } The Equality Graph G E (  E ) = hV, E =, E  i (a.k.a “E-graph”) x y z

Technion16 Basic notions   E : x = y Æ y = z Æ z  x unsatisfiable  2 E : x = y Æ y = z Ç z  x satisfiable The graph G E (  E ) represents an abstraction of  E It ignores the Boolean structure of  E x y z

Technion17 Basic notions Equality Path: a path made of equalities. we write x =* z Disequality Path: a path made of equalities and exactly one disequality. We write x  * y Contradictory Cycle: two nodes x and y, s.t. x= * y and x  * y form a contradictory cycle x y z

Technion18 Basic notions Thm-4: Every contradictory cycle is either simple or contains a simple contradictory cycle

Technion19 Definitions Dfn: A contradictory Cycle C is constrained under T if T does not allow this assignment C = F T T T T

Technion20 Definitions Let C = ( e s, e 1,…, e n ) where e s is solid and e 1,…, e n are dashed be a simple (contradictory) cycle. Let T  be a formula over the Boolean variables encoding C We say that C is constrained in T with respect to e s iff every assignment  s.t.   ( e s ) = F and   ( e 1 ) = …=  ( e n ) = T contradicts T

Technion21 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

Technion22 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

Technion23 Definitions for the proof… A Violating cycle under an assignment  R  This assignment violates T S but not necessarily T R eFeF e T2 e T1 T T F Either dashed or solid

Technion24 More definitions for the proof… An edge e = ( v i, v j ) is equal under an assignment  iff there is an equality path between v i and v j all assigned T under  Denote: T T F T T v1v1 v2v2 v3v3

Technion25 More definitions for the proof… An edge e = ( v i, v j ) is disequal under an assignment  iff there is a disequality path between v i and v j in which the solid edge is the only one assigned false by  Denote: T T F T T v1v1 v2v2 v3v3

Technion26 Proof… Observation 1: The combination is impossible if  =  R (recall:  R ² T R ) Observation 2: if ( v 1, v 3 ) is solid, then F T T v1v1 v2v2 v3v3

Technion27 ReConstructing  S Type 1: It is not the case that Assign  S ( e 23 ) = F Type 2: Otherwise it is not the case that Assign  ( e 13 ) = T F T T In all other cases  S =  R F T T  F  T v1v1 v2v2 v3v3 v1v1 v2v2 v3v3

Technion28 ReConstructing  S Starting from  R, repeat until convergence:   ( e T ) := F in all Type 1 cycles   ( e F ) := 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

Technion29 Proof… Invariant: contradictory cycles are not violating throughout the reconstruction. contradicts the precondition to make this assignment… F T T v1v1 v2v2 v3v3  F T T

Technion30 Proof… Invariant: contradictory cycles are not violating throughout the reconstruction. contradicts the precondition to make this assignment… F T T v1v1 v2v2 v3v3  T T F

Technion31 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?

Technion32 The RTC algorithm Bi-Connected Component (BCC) – maximal set of edges s.t. any two edges in the set lie on a common simple cycle. Find all BCC-s with one solid edge. Add solid chords from the graph, and auxiliary dashed chords to make the BCC chordal.

Technion33 Focus on Bi-connected dashed components built on top of a solid edge  Includes all contradictory cycles involving this edge

Technion34 Make the component chordal  Chordal-ity guarantees: every cycle contains a simplicial vertex, i.e. a vertex that its neighbors are connected.

Technion35 The RTC algorithm Constraints cache:  e 2 Æ e 3 ! e 1  e 4 Æ e 7 ! e 2  e 5 Æ e 8 ! e

Technion36 Constrains all contradictory cycles Constraints cache:  e 2 Æ e 3 ! e 1  e 4 Æ e 7 ! e 2  e 6 Æ e 3 ! e

Technion37 x0x0 x1x1 x2x2 x3x3 x6x6 x4x4 x5x5 The constraint e 3,6 Æ e 3,5  e 5,6 is not added Constraining simple contradictory cycles cache: … e 5,6 Æ e 4,6  e 4,5 Open problem: constrain simple contradictory cycles in P time

Technion38 x0x0 x1x1 x2x2 x3x3 x6x6 x4x4 x5x5 the constraint e 3,6 Æ e 3,5  e 5,6 is not added, though needed Suppose the graph has 3 more edges Constraining simple contradictory cycles cache: … e 5,6 Æ e 4,6  e 4,5 Here we will stop, although … Open problem: constrain simple contradictory cycles in P time

Technion39 Results

Technion40 Example: Circuit Transformations A pipeline processes data in stages Data is processed in parallel – as in an assembly line Formal Model: Stage 1 Stage 3 Stage 2

Technion41 Example: Circuit Transformations The maximum clock frequency depends on the longest path between two latches Note that the output of g is used as input to k We want to speed up the design by postponing k to the third stage

Technion42 Validating Circuit Transformations = ?

Technion43 Validating a compilation process Source program z = ( x 1 + y 1 )  ( x 2 + y 2 ); Target program u 1 = x 1 + y 1 ; u 2 = x 2 + y 2 ; z = u 1  u 2 ; Need to prove that: ( u 1 = x 1 + y 1  u 2 = x 2 + y 2  z = u 1  u 2 ) $ z = ( x 1 + y 1 )  ( x 2 + y 2 ) Compilation Target Source

Technion44 Validating a compilation process Need to prove that: ( u 1 = x 1 + y 1  u 2 = x 2 + y 2  z = u 1  u 2 ) $ z = ( x 1 + y 1 )  ( x 2 + y 2 ) f1f1 f2f2 g1g1 g2g2 f1f1 f2f2 Source program z = ( x 1 + y 1 )  ( x 2 + y 2 ); Target program u 1 = x 1 + y 1 ; u 2 = x 2 + y 2 ; z = u 1  u 2 ; Compilation

Technion45 Need to prove that: ( u 1 = x 1 + y 1  u 2 = x 2 + y 2  z = u 1  u 2 ) $ z = ( x 1 + y 1 )  ( x 2 + y 2 ) f1f1 f2f2 g1g1 g2g2 f1f1 f2f2 Instead, prove: under functional consistency: for every uninterpreted function f x = y ! f ( x ) = f ( y ) Which translates to (via Ackermann’s reduction): Validating a compilation process