/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.pl /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- Graph construction failed Graph construction failed Graph construction failed MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.pl # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern q(a,a) w.r.t. the given Prolog program could not be shown: (0) Prolog (1) CutEliminatorProof [SOUND, 0 ms] (2) Prolog (3) UndefinedPredicateHandlerProof [SOUND, 0 ms] (4) Prolog (5) PrologToPiTRSProof [SOUND, 0 ms] (6) PiTRS (7) DependencyPairsProof [EQUIVALENT, 0 ms] (8) PiDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) PiDP (11) UsableRulesProof [EQUIVALENT, 0 ms] (12) PiDP (13) PiDPToQDPProof [SOUND, 0 ms] (14) QDP (15) NonTerminationLoopProof [COMPLETE, 0 ms] (16) NO (17) PrologToPiTRSProof [SOUND, 0 ms] (18) PiTRS (19) DependencyPairsProof [EQUIVALENT, 0 ms] (20) PiDP (21) DependencyGraphProof [EQUIVALENT, 0 ms] (22) PiDP (23) UsableRulesProof [EQUIVALENT, 0 ms] (24) PiDP (25) PiDPToQDPProof [SOUND, 0 ms] (26) QDP (27) NonTerminationLoopProof [COMPLETE, 0 ms] (28) NO ---------------------------------------- (0) Obligation: Clauses: q(X, Y) :- ','(==(X, Y), p(X, Y)). p(X, X) :- !. p(X, Y) :- p(X, Y). Query: q(a,a) ---------------------------------------- (1) CutEliminatorProof (SOUND) Eliminated all cuts by simply ignoring them[PROLOG]. ---------------------------------------- (2) Obligation: Clauses: q(X, Y) :- ','(==(X, Y), p(X, Y)). p(X, X). p(X, Y) :- p(X, Y). Query: q(a,a) ---------------------------------------- (3) UndefinedPredicateHandlerProof (SOUND) Added facts for all undefined predicates [PROLOG]. ---------------------------------------- (4) Obligation: Clauses: q(X, Y) :- ','(==(X, Y), p(X, Y)). p(X, X). p(X, Y) :- p(X, Y). ==(X0, X1). Query: q(a,a) ---------------------------------------- (5) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: q_in_2: (f,f) p_in_2: (f,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (6) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa ---------------------------------------- (7) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: Q_IN_AA(X, Y) -> U1_AA(X, Y, ==_in_aa(X, Y)) Q_IN_AA(X, Y) -> ==_IN_AA(X, Y) U1_AA(X, Y, ==_out_aa(X, Y)) -> U2_AA(X, Y, p_in_aa(X, Y)) U1_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, p_in_aa(X, Y)) P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Q_IN_AA(x1, x2) = Q_IN_AA U1_AA(x1, x2, x3) = U1_AA(x3) ==_IN_AA(x1, x2) = ==_IN_AA U2_AA(x1, x2, x3) = U2_AA(x3) P_IN_AA(x1, x2) = P_IN_AA U3_AA(x1, x2, x3) = U3_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: Q_IN_AA(X, Y) -> U1_AA(X, Y, ==_in_aa(X, Y)) Q_IN_AA(X, Y) -> ==_IN_AA(X, Y) U1_AA(X, Y, ==_out_aa(X, Y)) -> U2_AA(X, Y, p_in_aa(X, Y)) U1_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, p_in_aa(X, Y)) P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Q_IN_AA(x1, x2) = Q_IN_AA U1_AA(x1, x2, x3) = U1_AA(x3) ==_IN_AA(x1, x2) = ==_IN_AA U2_AA(x1, x2, x3) = U2_AA(x3) P_IN_AA(x1, x2) = P_IN_AA U3_AA(x1, x2, x3) = U3_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes. ---------------------------------------- (10) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa P_IN_AA(x1, x2) = P_IN_AA We have to consider all (P,R,Pi)-chains ---------------------------------------- (11) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (12) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AA(X, Y) -> P_IN_AA(X, Y) R is empty. The argument filtering Pi contains the following mapping: P_IN_AA(x1, x2) = P_IN_AA We have to consider all (P,R,Pi)-chains ---------------------------------------- (13) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_AA -> P_IN_AA R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (15) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = P_IN_AA evaluates to t =P_IN_AA Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from P_IN_AA to P_IN_AA. ---------------------------------------- (16) NO ---------------------------------------- (17) PrologToPiTRSProof (SOUND) We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes: q_in_2: (f,f) p_in_2: (f,f) Transforming Prolog into the following Term Rewriting System: Pi-finite rewrite system: The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (18) Obligation: Pi-finite rewrite system: The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa ---------------------------------------- (19) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem: Pi DP problem: The TRS P consists of the following rules: Q_IN_AA(X, Y) -> U1_AA(X, Y, ==_in_aa(X, Y)) Q_IN_AA(X, Y) -> ==_IN_AA(X, Y) U1_AA(X, Y, ==_out_aa(X, Y)) -> U2_AA(X, Y, p_in_aa(X, Y)) U1_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, p_in_aa(X, Y)) P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Q_IN_AA(x1, x2) = Q_IN_AA U1_AA(x1, x2, x3) = U1_AA(x3) ==_IN_AA(x1, x2) = ==_IN_AA U2_AA(x1, x2, x3) = U2_AA(x3) P_IN_AA(x1, x2) = P_IN_AA U3_AA(x1, x2, x3) = U3_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (20) Obligation: Pi DP problem: The TRS P consists of the following rules: Q_IN_AA(X, Y) -> U1_AA(X, Y, ==_in_aa(X, Y)) Q_IN_AA(X, Y) -> ==_IN_AA(X, Y) U1_AA(X, Y, ==_out_aa(X, Y)) -> U2_AA(X, Y, p_in_aa(X, Y)) U1_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, p_in_aa(X, Y)) P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa Q_IN_AA(x1, x2) = Q_IN_AA U1_AA(x1, x2, x3) = U1_AA(x3) ==_IN_AA(x1, x2) = ==_IN_AA U2_AA(x1, x2, x3) = U2_AA(x3) P_IN_AA(x1, x2) = P_IN_AA U3_AA(x1, x2, x3) = U3_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (21) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 5 less nodes. ---------------------------------------- (22) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AA(X, Y) -> P_IN_AA(X, Y) The TRS R consists of the following rules: q_in_aa(X, Y) -> U1_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U1_aa(X, Y, ==_out_aa(X, Y)) -> U2_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, X) -> p_out_aa(X, X) p_in_aa(X, Y) -> U3_aa(X, Y, p_in_aa(X, Y)) U3_aa(X, Y, p_out_aa(X, Y)) -> p_out_aa(X, Y) U2_aa(X, Y, p_out_aa(X, Y)) -> q_out_aa(X, Y) The argument filtering Pi contains the following mapping: q_in_aa(x1, x2) = q_in_aa U1_aa(x1, x2, x3) = U1_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U2_aa(x1, x2, x3) = U2_aa(x3) p_in_aa(x1, x2) = p_in_aa p_out_aa(x1, x2) = p_out_aa U3_aa(x1, x2, x3) = U3_aa(x3) q_out_aa(x1, x2) = q_out_aa P_IN_AA(x1, x2) = P_IN_AA We have to consider all (P,R,Pi)-chains ---------------------------------------- (23) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (24) Obligation: Pi DP problem: The TRS P consists of the following rules: P_IN_AA(X, Y) -> P_IN_AA(X, Y) R is empty. The argument filtering Pi contains the following mapping: P_IN_AA(x1, x2) = P_IN_AA We have to consider all (P,R,Pi)-chains ---------------------------------------- (25) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (26) Obligation: Q DP problem: The TRS P consists of the following rules: P_IN_AA -> P_IN_AA R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (27) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by semiunifying a rule from P directly. s = P_IN_AA evaluates to t =P_IN_AA Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [ ] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence The DP semiunifies directly so there is only one rewrite step from P_IN_AA to P_IN_AA. ---------------------------------------- (28) NO