analyzing relational logic

Daniel Jackson, MITWG 2.3 · NewcastleApril 2000

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language assumptions

language·first-order logic·set & relation operators·uninterpreted types

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analysis desired

simulationfind a state that satisfies invariant J… and additionally condition C·find an execution of operation O… resulting in a state satisfying P… from a state satisfying P… that changes the component x… that is not an execution of operation Ov

checking·does invariant J imply invariant Jv?·does operation O preserve invariant J ?·does operation Oc refine Oa under abstraction A?

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analyses not possible

refinement·does Oc refines Oa for some abstraction?·are all executions of O also executions of O1;O2?

spec by minimization·make smallest change to connections that satisfies C …

precondition checks·does O have an execution from every state satisfying C?

temporal checks·do reachable states satisfy J ?

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semantics: formulas

M : formula env booleanX : expr env valueenv = (var + type) valuevalue = (atom atom) + (atom value)

M [a in b] e = X[a]e X[b]eM [! F] e = M[F]eM [F && G] e = M[F]e M[G]eM [all v: t | F] e = {M[F] (e v x) | x e(t)}

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semantics: expressions

X : expr env valueenv = (var + type) valuevalue = (atom atom) + (atom value)

X [a + b] e = X[a]e X[b]eX [a . b] e = {y | x. x X[a]e (x,y) X[b]e}X [~a] e = {(x,y) | (y,x) X[a]e}X [+a] e = the smallest r such that r ; r x X[a]e xX [{v: t | F}] e = {x e(t) | M[F] (e v x)}X [v] e = e(v)X [a[v]] e = e(a)(v)

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models

models are well-formed environments for which formula holdsM : formula env booleanModels (F) = {e | M[f]e}

environment e is well formed iff·tight: only bind variables declared along with formula·type correct: if expression a has type T, X[a]e X[T]e

e is within scope k iff·for all basic types T, #X[T]e = k·write Modelsk (F) for models within scope k

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small scope hypothesis

% bugscaught

scope

90

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most bugs can be caught by considering only small instances

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example

problema, b : Sp : S -> T! (a – b).p in (a.p – b.p)

a model in a scope of 2S = {S0, S1}T = {T0, T1}p = {(S0, T0), (S1, T0)}a = {S0}b = {S1}

S0

S1

T0

T1

a

b

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what Alcoa does

alcoa : formula, scope env·does not always succeed (ie, may return nothing)

properties·termination: always, with deterministic solvers·soundness: alcoa (F, k) Modelsk (F)·relative completeness: Modelsk (F) {} alcoa (F, k)

succeeds

non-properties·minimality: alcoa (F, k) not the smallest model of F in k·completeness: Models (F) {} alcoa (F, k) succeeds

so counterexamples are real, but can’t prove theorems

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scope monotonicity

Alcoa is scope monotonic·alcoa (F, k) succeeds alcoa (F, k+1) succeeds· if scope of 7 fails, no need to try 6, 5, …

because models are scope monotonic·Modelsk (F) Modelsk+1 (F)

·property of Alloy, not kernel

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every analysis is model finding

does operation O preserve invariant J ?alcoa (O && J && !J’ , 3)

show me how O1 and O2 differalcoa ((O1 && !O2) || (O2 && !O1) , 3)

show me an execution of O that changes xalcoa (O && !x = x’ , 3)

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alcoa architecture

TRANSLATEPROBLEM

TRANSLATESOLUTION

MAPPING

BOOLEANFORMULA

BOOLEANASSIGNMENT

SATSOLVER

DESIGNPROBLEM

DESIGNANALYSIS

alcoa

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overview of method

from alloy formula F and scope k generateboolean formula BFmapping : BoolAssignment Environmentsuch that

maps every solution of BFn Modelsk (F) n Models (BF)

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SAT solvers

in theory·3-SAT is NP-complete

in practice·solvers work well for <1000 variables and <100,000 clauses

·usually give small models

kinds of solver· local search (eg, WalkSAT)·Davis-Putnam (eg, RelSAT, SATO)·non-clausal (eg, Prover)

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example

problema, b : Sp : S -> T! (a – b).p in (a.p – b.p)

translation in scope of 2· formula becomes((a0b0 p00) (a1b1 p10) ((a0 p00) (a1 p10)) ((b0 p00) (b1 p10))) …

·a model isa0 , a1 , b0 , b1 , p00 , p01 , p10 ,

p11

mapping function ·set to vector of bool var

a [a0 a1]b [b0 b1]

· relation to matrixp [p00 p01 , p10 p11]

final resultS = {S0, S1}T = {T0, T1}p = {(S0, T0), (S1, T0)}

a = {S0}b = {S1}

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compositional translation

translating relation r: S -> TXT [r]ij boolean var, true when r contains (Si, Tj)

translating expression e: TXT [a]i boolean formula, true when a contains Ti

translating formulasMT [F] boolean formula, true for models of F

sample rulesMT [F && G] = MT[F] MT[F] XT [a - b]i = XT [a]i XT [b]iXT [a . b]i = j. XT [a]j XT [b]ji

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example

a [a0 a1]b [b0 b1]p [p00 p01 , p10 p11]a – b [a0b0 a1b1](a – b).p [(a0b0 p00) (a1b1 p10) …]a.p [(a0 p00) (a1 p10) (a0 p01) (a1 p11)]b.p [(b0 p00) (b1 p10) (b0 p01) (b1 p11)]a.p – b.p [((a0 p00) (a1 p10)) ((b0 p00) (b1 p10)) …]

! (a – b).p in (a.p – b.p) (((a0b0 p00) (a1b1 p10)

((a0 p00) (a1 p10)) ((b0 p00) (b1 p10))))

…

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quantifiers

example!((all x | x in x.p) -> (all x | x in x.p.p))

put in negation normal form(all x | x in x.p) && (some x | ! x in x.p.p)

skolemize(all x | x in x.p) && ! (xc in xc.p.p)

how to translate remaining universal quantifiers?

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environments & trees

semanticsM : formula env booleanX : expr env valueenv = (var + type) valuevalue = (atom atom) + (atom value)

translationMT : formula boolFormula treeXT : expr boolValue tree tree = (var (index tree) + boolValue = booleanFormulaMatrix + (index boolValue)

env becomes (tree, boolean encoding of relations)

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examples

x

[1 0] [0 1]

0 1x : T

x

[p00 p01]

0 1

[p10 p11]

x.p

x

p00

0 1

p11

x in x.p

p00 p11

all x | x in x.p

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compositional rules

MT [a in b] = merge (MT[a], MT[b], u,v. i (ui vj) )

MT [all x | F] = fold (MT[F], )

merging·subexpressions may have different variables·so interpose layers as necessary, then merge·maintain consistent ordering from root to leaf

x

[1 0] [0 1]

0 1y

[1 0] [0 1]

0 1

x0 1

y0 1

y0 1

0 11

x yx in y

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symmetry

observation·types are uninterpreted·permuting elements of a type preserves modelhood

examplea = {S0} , b = {S1}, p = {(S0, T0), (S1, T0)}a = {S1} , b = {S0}, p = {(S0, T0), (S1, T0)}both models of !(a – b).p in (a.p – b.p)

exploitation·environments form equivalence classes·avoid considering all elements of a class

environments

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symmetry in boolean formula

preserved!·express symmetry as permutation on boolean vars·then is a symmetry of the boolean formula too·want to rule out one of A, A

Crawford’s idea·order boolean vars into sequence V·view assignments as binary numbers & pick smaller· for each , add constraint V V

example·A = 0123, A = 0123, = (02)·V V is 0123 < 2103 = 0 < 2

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symmetry constraint for a relation

suppose we translated relation p: S -> T to the matrix0 1 23 4 56 7 8

symmetry (S0 S1) exchanges top two rows3 4 50 1 26 7 8

constraint obtained is0 1 2 3 4 5 6 7 8 < 3 4 5 0 1 2 6 7 8= 0 1 2 < 3 4 5= 03 (03 14) (03 14 25)

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generalizing symmetry

extend to·multiple relations and sets·multiple types

but·diminishing returns·pick ordering of vars carefully·homogeneous relations tricky

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results

observations·solver time dominates·RelSAT dominates other solvers·symmetry gives 100x speedup for ‘proofs’·bugs in boolean code not translation

so end-to-end check

where are we?· interactive analysis up to 200 bits·small but real specs (longest so far is 400 lines)·30loc list-processing procedures

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challenges

symmetry·why do a few symmetries seem to work so well?·what symmetries should be used?

progress bar·will symmetry spoil our heuristics?

visualization of models·key for novice use

language extensions·numbers (easy?)·sequences (hard?)

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allex

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allox

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allix