/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- Graph construction failed Graph construction failed Graph construction failed MAYBE proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Left Termination of the query pattern q(a,a) w.r.t. the given Prolog program could not be shown: (0) Prolog (1) UndefinedPredicateHandlerProof [SOUND, 0 ms] (2) Prolog (3) PrologToPiTRSProof [SOUND, 0 ms] (4) PiTRS (5) DependencyPairsProof [EQUIVALENT, 0 ms] (6) PiDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) PiDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) PiDP (11) PiDPToQDPProof [SOUND, 0 ms] (12) QDP (13) TransformationProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) NonTerminationLoopProof [COMPLETE, 0 ms] (20) NO (21) PrologToPiTRSProof [SOUND, 0 ms] (22) PiTRS (23) DependencyPairsProof [EQUIVALENT, 0 ms] (24) PiDP (25) DependencyGraphProof [EQUIVALENT, 0 ms] (26) PiDP (27) UsableRulesProof [EQUIVALENT, 0 ms] (28) PiDP (29) PiDPToQDPProof [SOUND, 0 ms] (30) QDP (31) TransformationProof [EQUIVALENT, 0 ms] (32) QDP (33) UsableRulesProof [EQUIVALENT, 0 ms] (34) QDP (35) QReductionProof [EQUIVALENT, 0 ms] (36) QDP (37) NonTerminationLoopProof [COMPLETE, 0 ms] (38) NO ---------------------------------------- (0) Obligation: Clauses: q(X, Y) :- ','(\==(X, Y), p(X, Y)). p(X, Y) :- ','(==(X, Y), p(X, Y)). p(X, Y). Query: q(a,a) ---------------------------------------- (1) UndefinedPredicateHandlerProof (SOUND) Added facts for all undefined predicates [PROLOG]. ---------------------------------------- (2) Obligation: Clauses: q(X, Y) :- ','(\==(X, Y), p(X, Y)). p(X, Y) :- ','(==(X, Y), p(X, Y)). p(X, Y). \==(X0, X1). ==(X0, X1). Query: q(a,a) ---------------------------------------- (3) 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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (4) 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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa ---------------------------------------- (5) 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, ==_in_aa(X, Y)) P_IN_AA(X, Y) -> ==_IN_AA(X, Y) U3_AA(X, Y, ==_out_aa(X, Y)) -> U4_AA(X, Y, p_in_aa(X, Y)) U3_AA(X, Y, ==_out_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa 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) ==_IN_AA(x1, x2) = ==_IN_AA U4_AA(x1, x2, x3) = U4_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (6) 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, ==_in_aa(X, Y)) P_IN_AA(X, Y) -> ==_IN_AA(X, Y) U3_AA(X, Y, ==_out_aa(X, Y)) -> U4_AA(X, Y, p_in_aa(X, Y)) U3_AA(X, Y, ==_out_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa 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) ==_IN_AA(x1, x2) = ==_IN_AA U4_AA(x1, x2, x3) = U4_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes. ---------------------------------------- (8) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, ==_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa 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) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (10) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, ==_in_aa(X, Y)) The TRS R consists of the following rules: ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) The argument filtering Pi contains the following mapping: ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa 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 ---------------------------------------- (11) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_in_aa) The TRS R consists of the following rules: ==_in_aa -> ==_out_aa The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (13) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule P_IN_AA -> U3_AA(==_in_aa) at position [0] we obtained the following new rules [LPAR04]: (P_IN_AA -> U3_AA(==_out_aa),P_IN_AA -> U3_AA(==_out_aa)) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) The TRS R consists of the following rules: ==_in_aa -> ==_out_aa The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (15) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) R is empty. The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (17) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. ==_in_aa ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (19) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: 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 P_IN_AA -> U3_AA(==_out_aa) with rule P_IN_AA -> U3_AA(==_out_aa) at position [] and matcher [ ] U3_AA(==_out_aa) -> P_IN_AA with rule U3_AA(==_out_aa) -> P_IN_AA Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (20) NO ---------------------------------------- (21) 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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog ---------------------------------------- (22) 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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa ---------------------------------------- (23) 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, ==_in_aa(X, Y)) P_IN_AA(X, Y) -> ==_IN_AA(X, Y) U3_AA(X, Y, ==_out_aa(X, Y)) -> U4_AA(X, Y, p_in_aa(X, Y)) U3_AA(X, Y, ==_out_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa 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) ==_IN_AA(x1, x2) = ==_IN_AA U4_AA(x1, x2, x3) = U4_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (24) 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, ==_in_aa(X, Y)) P_IN_AA(X, Y) -> ==_IN_AA(X, Y) U3_AA(X, Y, ==_out_aa(X, Y)) -> U4_AA(X, Y, p_in_aa(X, Y)) U3_AA(X, Y, ==_out_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa 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) ==_IN_AA(x1, x2) = ==_IN_AA U4_AA(x1, x2, x3) = U4_AA(x3) We have to consider all (P,R,Pi)-chains ---------------------------------------- (25) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 6 less nodes. ---------------------------------------- (26) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, ==_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, Y) -> U3_aa(X, Y, ==_in_aa(X, Y)) ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) U3_aa(X, Y, ==_out_aa(X, Y)) -> U4_aa(X, Y, p_in_aa(X, Y)) p_in_aa(X, Y) -> p_out_aa(X, Y) U4_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 U3_aa(x1, x2, x3) = U3_aa(x3) ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa U4_aa(x1, x2, x3) = U4_aa(x3) p_out_aa(x1, x2) = p_out_aa q_out_aa(x1, x2) = q_out_aa 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 ---------------------------------------- (27) UsableRulesProof (EQUIVALENT) For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R. ---------------------------------------- (28) Obligation: Pi DP problem: The TRS P consists of the following rules: U3_AA(X, Y, ==_out_aa(X, Y)) -> P_IN_AA(X, Y) P_IN_AA(X, Y) -> U3_AA(X, Y, ==_in_aa(X, Y)) The TRS R consists of the following rules: ==_in_aa(X0, X1) -> ==_out_aa(X0, X1) The argument filtering Pi contains the following mapping: ==_in_aa(x1, x2) = ==_in_aa ==_out_aa(x1, x2) = ==_out_aa 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 ---------------------------------------- (29) PiDPToQDPProof (SOUND) Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi. ---------------------------------------- (30) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_in_aa) The TRS R consists of the following rules: ==_in_aa -> ==_out_aa The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (31) TransformationProof (EQUIVALENT) By rewriting [LPAR04] the rule P_IN_AA -> U3_AA(==_in_aa) at position [0] we obtained the following new rules [LPAR04]: (P_IN_AA -> U3_AA(==_out_aa),P_IN_AA -> U3_AA(==_out_aa)) ---------------------------------------- (32) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) The TRS R consists of the following rules: ==_in_aa -> ==_out_aa The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (33) UsableRulesProof (EQUIVALENT) As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (34) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) R is empty. The set Q consists of the following terms: ==_in_aa We have to consider all (P,Q,R)-chains. ---------------------------------------- (35) QReductionProof (EQUIVALENT) We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN]. ==_in_aa ---------------------------------------- (36) Obligation: Q DP problem: The TRS P consists of the following rules: U3_AA(==_out_aa) -> P_IN_AA P_IN_AA -> U3_AA(==_out_aa) R is empty. Q is empty. We have to consider all (P,Q,R)-chains. ---------------------------------------- (37) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: 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 P_IN_AA -> U3_AA(==_out_aa) with rule P_IN_AA -> U3_AA(==_out_aa) at position [] and matcher [ ] U3_AA(==_out_aa) -> P_IN_AA with rule U3_AA(==_out_aa) -> P_IN_AA Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (38) NO