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Using First-Order Theorem Provers in Data Structure Verification Charles Bouillaguet Viktor Kuncak Martin Rinard Ecole Normale Supérieure, Cachan, France MIT CSAIL Implementing Data Structures is Hard Often small, but complex code Lots of pointers Unbounded, dynamic allocation Complex shape invariants Dag Properties involving arithmetic (ordering…) Need strong invariants to guarantee correctness e.g. lookup in ordered tree needs sortedness How to obtain reliable data structure implementations? Approach Prove that the program is correct For all program executions (sound) Verified properties: Data structure operations do not crash Data structure invariants are preserved Data structure content is correctly updated Infrastructure Jahob system for verifying data structure implementation Kuncak, Wies, Zee, Rinard, Nguyen, Bouillaguet, Schmitt, Marnette, Bugrara Analyzed programs: subset of Java Specification : subset of Isabelle’s language Summary of Verified Data Structures Implementations of relations Add a binding Remove all bindings for a given key Test key membership Retrieve data bound to a key Test emptiness Verified implementations Linked list Ordered tree Hash table An Example : Ordered Trees Implementation of a finite map Operations key value right left insert lookup remove Representation invariants: tree shaped (acyclicity, unique parent) ordering constraints Sample code public static FuncTree update(int k, Object v, FuncTree t) { FuncTree new_left, new_right; Object new_data; int new_key; if (t==null) { new_data = v; new_key = k; new_left = null; new_right = null; } else { if (k < t.key) { new_left = update(k, v, t.left); new_right = t.right; new_key = t.key; new_data = t.data; } else if (t.key < k) { … } else { new_data = v; new_key = k; new_left = t.left; new_right = t.right; }} FuncTree r = new FuncTree(); r.left = new_left; r.right = new_right; r.data = new_data; r.key = new_key; return r; } Sample code public static FuncTree update(int k, Object v, FuncTree t) /*: requires "v ~= null” ensures "result..content = t..content - {(x,y). x=k} + {(k,v)} */ { FuncTree new_left, new_right; Object new_data; int new_key; if (t==null) { new_data = v; new_key = k; new_left = null; new_right = null; } else { if (k < t.key) { new_left = update(k, v, t.left); new_right = t.right; new_key = t.key; new_data = t.data; no null dereferences } else if (t.key < k) { … } else { new_data = v; new_key = k; new_left = t.left; new_right = t.right; }} FuncTree r = new FuncTree(); postcondition holds and r.left = new_left; r.right = new_right; invariants preserved r.data = new_data; r.key = new_key; //: "r..content" := "t..content - {(x,y). x=k} + {(k,v)}"; return r; } 3 lines spec 30 lines code Ordered tree: interface public ghost specvar content :: "(int * obj) set" = "{}"; public static FuncTree empty_set() ensures "result..content = {}" public static FuncTree add(int k, Object v, FuncTree t) requires "v ~= null & (ALL y. (k,y) ~: t..content)” ensures "result..content = t..content Un {(k,v)}” public static FuncTree update(int k, Object v, FuncTree t) requires "v ~= null” ensures "result..content = t..content - {(x,y). x=k} + {(k,v)}” public static Object lookup(int k, FuncTree t) ensures "((k, result) : t..content) | (result = null & (ALL v. (k,v) ~: t..content))” public static FuncTree remove(int k, FuncTree t) ensures "result..content = t..content - {(x,y). x=k}” Representation Invariants public final class FuncTree{ private int key; abstract set-valued field private Object data; private FuncTree left, right; /*: public ghost specvar content :: "(int * obj) set"; tuples invariant ("content definition") "this ~= null -->implicit universal quantification content = {(key, data)} Un left..content Un right..content"over this invariant ("nullequality implies empty") "this = null --> content = {}" between sets invariant ("left children are smaller") "ALL k v. (k,v) : left..content --> k < key” invariant ("right children are bigger") "ALL k v. (k,v) : right..content --> k > key" */ explicit quantification arithmetic How could these properties be verified? Standard Approach eauto ; intros . intuition ; subst . apply Extensionality_Ensembles. unfold Same_set. unfold Included. unfold In. unfold In in H1. intuition. destruct H0. destruct (eq_nat_dec x1 ArraySet_size). subst. rewrite arraywrite_match in H0 ; auto. intuition. subst. apply Union_intror. auto with sets. assert (x1 < ArraySet_size). omega. clear n. apply Union_introl. rewrite arraywrite_not_same_i in H0. unfold In. exists x1. intuition.omega. Transform program into a logic formula Using weakest precondition inversion H0 ; subst ; clear H0. unfold In in H3. The program is correct iff the formula valid destruct H3. exists x1.is intuition. low efficiency 1 line per grad student-minute parallelization looks non-trivial rewrite arraywrite_not_same_i. intuition ; omega. omega. exists ArraySet_size. intuition. inversion H3. subst. rewrite arraywrite_match ; trivial. Prove the formula Very difficult formulas: interactively (Coq, Isabelle) Decidable classes: automated (MONA, CVCL, Omega) This talk: difficult formulas in automated way :) Formulas in Jahob Very expressive specification language Higher-Order features How to prove formulas automatically? Convert them to something simpler Decidable classes First-Order Logic Automated reasoning in Jahob Why FOL? Existing theorem provers: SPASS, E, Vampire, Theo, Prover9, … continuously improving (yearly competition) Effective on formulas with short proofs Handle nicely formulas with quantifiers HOL FOL Ideas : avoid axiomatizing rich theories Translate what can naturally be expressed in FOL soundly approximate the rest Sound, incomplete approach Full details in long version of the paper (x,y) ∈ z.content ⇢ Content(x,y,z) w.f := y ⇢(x=y ⋀ w=v) ⋁ (x ≠ y ⋀ w=f(y) ) λx.E = λx.F⇢∀x. E=F … Arithmetic Numbers are uninterpreted constants in FOL Provers do not know that 1+1=2 ! Still need to reason about arithmetic… Our Solution Provide partial, incomplete axiomatization • Still cannot deduce 1+1=2 ! • comparison between constants in formula Satisfactory results in practice • ordering of elements in tree • array bound checks Observation Most formulas are easy to prove ie in no measurable time have very short proofs (in # of resolution step) Problem often concentrated in a small number that take very long to prove We applied two existing techniques to make them easier 1. Eliminating type/sort information 2. Filtering unnecessary assumptions Sort Information Specification language has sorts Integers Objects Boolean Translate to unsorted FOL ∀(x : Obj). P(x) ⇣ ∀x. Obj(x) ⇒P(x) Sort Information Encoding sort information bigger formulas longer proofs Formulas become harder to prove Temptation to omit sort information… Effect on hard formulas Formulas that take more than 1s to prove, from the Tree implementation (SPASS) Benchmark Time (s) with Tree.remove Tree.remove_max w/o Proof length with w/o Generated clauses with w/o 5.3 1.1 799 155 18 376 9 425 3.6 0.3 1 781 309 19 601 1 917 9.8 4.9 1 781 174 33 868 27 108 8.1 0.5 1 611 301 31 892 3 922 8.1 4.7 1 773 371 37 244 28 170 7.9 0.3 1 391 308 41 354 3 394 +∞ 0.22 78.9 6.8 2 655 1 159 177 755 19 527 34.8 0.8 4 062 597 115 713 5 305 97 1 075 Omitting Sorts (cont’d) Great speed-up (more than x10 sometimes) ! However: ∀ (x y:S). x = y ∃ (x y:T). x ≠ y Satisfiable with sorts (S={a}, T={b,c})… Unsatisfiable without! Omitting sort guards breaks soundness!!! Possible workaround: type-check generated proof When it is possible to skip type-checking ? Omitting Sorts Result We proved the following Theorem. Suppose that i. Sorts are pair-wise disjoint (no sub-sorting) ii. Sorts have the same cardinality Then omitting sort guards is sound and complete This justify this useful optimization Assumption Filtering Provers get confused by too many assumptions Lots of useless assumptions Hardest shown benchmark needs 12 out of 56 Big benchmark: on average 33% necessary Assumption filtering Try to eliminate irrelevant assumptions automatically Give a score to assumption based on relevance Experimental results Benchmark lines of lines of # of code specification methods verif. time Sets as imperative list 60 24 9 18s Relation as functional Linked list 76 26 9 12s 186 38 10 178s 69 53 10 119s Relation as functional Ordered trees Relation as hash table (using f.list) Verification effort Decreased as we improved the system functional list was easy a few days for trees two hours for simple hash table FOL : Currently most usable method for these kind of data structures Related work Interactive Provers – Isabelle, Coq, HOL, PVS, ACL2 First-Order ATP Vampire – Voronkov [04] SPASS – Weidenbach [01] E – Shultz [IJCAR04] Program Checking ESC/Java2 – Kiniry, Chalin, Hurlin Krakatoa – Marche, Paulin-Mohring, Urbain [03] Spec# – Barnett, DeLine, Jacobs, Fähndrich, Leino, Schulte, Venter [05] Hob system: verify set implementations (we verify relations) Shape analysis PALE - Møller and Schwartzbach [PLDI01] TVLA - Sagiv, Reps, and Wilheim [TOPLAS02] Roles - Kuncak, Lam, and Rinard [POPL02] Multiple Provers - Screenshot Conclusion Jahob verification system Automation by translation HOLFOL omitting sorts theorem gives speedup filtering automates selection of assumptions Promising experimental results strong properties: correct implementation • Do not crash • operations correctly update the content, clarifies behavior in case of duplicate keys, … • representation invariants preserved (ordering, treeness, each element is in appropriate bucket) relatively fast verification effort much smaller than using interactive provers Thank you Formal Methods are the Future of computer Science. Always have been… Always will be. Questions ? Converting to GCL Conditionnal statement: easy [| if cond then tbranch else fbranch |] = (Assume cond; [| tbranch|] ) (Assume !cond; [| fbranch|] ) Procedure calls: Could inline (potentially exponential blowup) Desugaring (modularity) : • [| r = CALL m(x, y, z) |] = Assert (m’s precondition); Havoc r; Havoc {vars modified by m} ; Assume (m’s postcondition) Converting to GCL (cont’d) Loops: invariant required [| while /*: invariant */ (condition) {lbody} |] = assert invariant; havoc vars(lbody); assume invariant; ((assume condition; [| lbody |]; assert invariant; assume false) (assume !condition)) invariant hold initially no assumptions on variables except that invariant hold condition hold invariant is preserved no need to verify anything more or condition do not hold and execution continues Verification condition for remove ((((fieldRead Pair_data null) = null) & ((fieldRead FuncTree_data null) = null) & ((fieldRead FuncTree_left null) = null) & ((fieldRead FuncTree_right null) = null) & (ALL (xObj::obj). (xObj : Object)) & ((Pair Int FuncTree) = {null}) & ((Array Int FuncTree) = {null}) & ((Array Int Pair) = {null}) & (null : Object_alloc) & (pointsto Pair Pair_data Object) & (pointsto FuncTree FuncTree_data Object) & (pointsto FuncTree FuncTree_left FuncTree) & (pointsto FuncTree FuncTree_right FuncTree) & comment ''unalloc_lonely'' (ALL (x::obj). ((x ~: Object_alloc) --> ((ALL (y::obj). ((fieldRead Pair_data y) ~= x)) & (ALL (y::obj). ((fieldRead FuncTree_data y) ~= x)) & (ALL (y::obj). ((fieldRead FuncTree_left y) ~= x)) & (ALL (y::obj). ((fieldRead FuncTree_right y) ~= x)) & ((fieldRead Pair_data x) = null) & ((fieldRead FuncTree_data x) = null) & ((fieldRead FuncTree_left x) = null) & ((fieldRead FuncTree_right x) = null)))) & comment ''ProcedurePrecondition'' (True & comment ''FuncTree_PrivateInv content definition'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree) & ((this :: obj) ~= null)) --> ((fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (this :: obj)) = (({((fieldRead (FuncTree_key :: (obj => int)) (this :: obj)), (fieldRead (FuncTree_data :: (obj => obj)) (this :: obj)))} Un (fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (fieldRead (FuncTree_left :: (obj => obj)) (this :: obj)))) Un (fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (fieldRead (FuncTree_right :: (obj => obj)) (this :: obj))))))) & comment ''FuncTree_PrivateInv null implies empty'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree) & ((this :: obj) = null)) --> ((fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (this :: obj)) = {}))) & comment ''FuncTree_PrivateInv no null data'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree) & ((this :: obj) ~= null)) --> ((fieldRead (FuncTree_data :: (obj => obj)) (this :: obj)) ~= null))) & comment ''FuncTree_PrivateInv left children are smaller'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree)) --> (ALL k. (ALL v. (((k, v) : (fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (fieldRead (FuncTree_left :: (obj => obj)) (this :: obj)))) --> (intless k (fieldRead (FuncTree_key :: (obj => int)) (this :: obj)))))))) & comment ''FuncTree_PrivateInv right children are bigger'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree)) --> (ALL k. (ALL v. (((k, v) : (fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (fieldRead (FuncTree_right :: (obj => obj)) (this :: obj)))) --> ((fieldRead (FuncTree_key :: (obj => int)) (this :: obj)) < k))))))) & comment ''t_type'' (((t :: obj) : (FuncTree :: obj set)) & ((t :: obj) : (Object_alloc :: obj set)))) --> ((comment ''TrueBranch'' (((t :: obj) = null) :: bool) --> (comment ''ProcedureEndPostcondition'' ((((fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (null :: obj)) = ((fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (t :: obj)) - {p. (EX x y. ((p = (x, y)) & (x = (k :: int))))})) & (ALL (framedObj::obj). (((framedObj : Object_alloc) & (framedObj : FuncTree)) --> ((fieldRead FuncTree_content framedObj) = (fieldRead FuncTree_content framedObj))))) & comment ''FuncTree_PrivateInv content definition'' (ALL (this::obj). (((this : Object_alloc) & (this : FuncTree) & ((this :: obj) ~= null)) --> ((fieldRead (FuncTree_content :: (obj => ((int * obj)) set)) (this :: obj)) = (({((fieldRead (FuncTree_key :: (obj => int)) (this :: obj)), (fieldRead (FuncTree_data :: (obj => obj)) (this :: obj)))} Un (fieldRead (FuncTree_content :: (obj => … And 200 more kilobytes… Infeasible to prove directly Splitting heuristic Verification condition is big conjunction conjunctions in postcondition proving each invariant proving each branch in program Solution: split VC into individual conjuncts Prove each conjunct separately Each conjunct has form H1 /\ … /\ Hn Gi Tree.Remove has 230 such conjuncts How do we prove them? Detupling (cont’d) Complete rules: Handling of Fields (cont’d) We dealt with field updates New function expressed in terms of old one Base case: field variables Natural encoding in FOL using functions: x = y.f ! x = f(y) Future work Verify more examples balanced trees fancy priority queues (binomial, Fibonacci, …) hash table with dynamic resizing hash function verify clients of data structures Improve assumption filtering take rarity of symbols into account check for occurring polarity …