/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) QDP (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] (11) YES (12) QDP (13) QDPOrderProof [EQUIVALENT, 986 ms] (14) QDP (15) SemLabProof [SOUND, 58 ms] (16) QDP (17) DependencyGraphProof [EQUIVALENT, 0 ms] (18) QDP (19) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(b(x1)))) -> b(b(b(a(x1)))) b(a(x1)) -> a(a(a(a(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(a(a(x1)))) -> A(b(b(b(x1)))) B(b(a(a(x1)))) -> B(b(b(x1))) B(b(a(a(x1)))) -> B(b(x1)) B(b(a(a(x1)))) -> B(x1) A(b(x1)) -> A(a(a(a(x1)))) A(b(x1)) -> A(a(a(x1))) A(b(x1)) -> A(a(x1)) A(b(x1)) -> A(x1) The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(x1)) -> A(x1) The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) UsableRulesProof (EQUIVALENT) We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: A(b(x1)) -> A(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPSizeChangeProof (EQUIVALENT) By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. From the DPs we obtained the following set of size-change graphs: *A(b(x1)) -> A(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (11) YES ---------------------------------------- (12) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(a(a(x1)))) -> B(b(x1)) B(b(a(a(x1)))) -> B(b(b(x1))) B(b(a(a(x1)))) -> B(x1) The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (13) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(b(a(a(x1)))) -> B(b(b(x1))) B(b(a(a(x1)))) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]: <<< POL(B(x_1)) = [[0A]] + [[-I, -1A, -1A]] * x_1 >>> <<< POL(b(x_1)) = [[0A], [-1A], [-I]] + [[-I, -I, 0A], [-I, -I, -1A], [0A, 1A, -1A]] * x_1 >>> <<< POL(a(x_1)) = [[1A], [1A], [0A]] + [[-1A, 0A, 0A], [0A, -1A, -1A], [-1A, -1A, -1A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) ---------------------------------------- (14) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(a(a(x1)))) -> B(b(x1)) The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(b(b(x1)))) a(b(x1)) -> a(a(a(a(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (15) SemLabProof (SOUND) We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1. a: 0 b: 1 + x0 B: 0 By semantic labelling [SEMLAB] we obtain the following labelled QDP problem. ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: B.1(b.0(a.0(a.0(x1)))) -> B.1(b.0(x1)) B.1(b.0(a.0(a.1(x1)))) -> B.0(b.1(x1)) The TRS R consists of the following rules: b.1(b.0(a.0(a.0(x1)))) -> a.1(b.0(b.1(b.0(x1)))) b.1(b.0(a.0(a.1(x1)))) -> a.0(b.1(b.0(b.1(x1)))) a.1(b.0(x1)) -> a.0(a.0(a.0(a.0(x1)))) a.0(b.1(x1)) -> a.0(a.0(a.0(a.1(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: B.1(b.0(a.0(a.0(x1)))) -> B.1(b.0(x1)) The TRS R consists of the following rules: b.1(b.0(a.0(a.0(x1)))) -> a.1(b.0(b.1(b.0(x1)))) b.1(b.0(a.0(a.1(x1)))) -> a.0(b.1(b.0(b.1(x1)))) a.1(b.0(x1)) -> a.0(a.0(a.0(a.0(x1)))) a.0(b.1(x1)) -> a.0(a.0(a.0(a.1(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) UsableRulesReductionPairsProof (EQUIVALENT) By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well. The following dependency pairs can be deleted: B.1(b.0(a.0(a.0(x1)))) -> B.1(b.0(x1)) The following rules are removed from R: b.1(b.0(a.0(a.0(x1)))) -> a.1(b.0(b.1(b.0(x1)))) b.1(b.0(a.0(a.1(x1)))) -> a.0(b.1(b.0(b.1(x1)))) a.1(b.0(x1)) -> a.0(a.0(a.0(a.0(x1)))) a.0(b.1(x1)) -> a.0(a.0(a.0(a.1(x1)))) Used ordering: POLO with Polynomial interpretation [POLO]: POL(B.1(x_1)) = x_1 POL(a.0(x_1)) = x_1 POL(b.0(x_1)) = x_1 ---------------------------------------- (20) Obligation: Q DP problem: P is empty. R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (22) YES