YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 108 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 24 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) AND (7) QDP (8) MNOCProof [EQUIVALENT, 0 ms] (9) QDP (10) UsableRulesProof [EQUIVALENT, 0 ms] (11) QDP (12) QReductionProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPSizeChangeProof [EQUIVALENT, 0 ms] (15) YES (16) QDP (17) MNOCProof [EQUIVALENT, 0 ms] (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QReductionProof [EQUIVALENT, 0 ms] (22) QDP (23) QDPSizeChangeProof [EQUIVALENT, 0 ms] (24) YES (25) QDP (26) UsableRulesProof [EQUIVALENT, 0 ms] (27) QDP (28) QDPSizeChangeProof [EQUIVALENT, 0 ms] (29) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a12(a12(x1)) -> x1 a13(a13(x1)) -> x1 a14(a14(x1)) -> x1 a15(a15(x1)) -> x1 a16(a16(x1)) -> x1 a23(a23(x1)) -> x1 a24(a24(x1)) -> x1 a25(a25(x1)) -> x1 a26(a26(x1)) -> x1 a34(a34(x1)) -> x1 a35(a35(x1)) -> x1 a36(a36(x1)) -> x1 a45(a45(x1)) -> x1 a46(a46(x1)) -> x1 a56(a56(x1)) -> x1 a13(x1) -> a12(a23(a12(x1))) a14(x1) -> a12(a23(a34(a23(a12(x1))))) a15(x1) -> a12(a23(a34(a45(a34(a23(a12(x1))))))) a16(x1) -> a12(a23(a34(a45(a56(a45(a34(a23(a12(x1))))))))) a24(x1) -> a23(a34(a23(x1))) a25(x1) -> a23(a34(a45(a34(a23(x1))))) a26(x1) -> a23(a34(a45(a56(a45(a34(a23(x1))))))) a35(x1) -> a34(a45(a34(x1))) a36(x1) -> a34(a45(a56(a45(a34(x1))))) a46(x1) -> a45(a56(a45(x1))) a12(a23(a12(a23(a12(a23(x1)))))) -> x1 a23(a34(a23(a34(a23(a34(x1)))))) -> x1 a34(a45(a34(a45(a34(a45(x1)))))) -> x1 a45(a56(a45(a56(a45(a56(x1)))))) -> x1 a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a12(x_1)) = 1 + x_1 POL(a13(x_1)) = 4 + x_1 POL(a14(x_1)) = 6 + x_1 POL(a15(x_1)) = 8 + x_1 POL(a16(x_1)) = 10 + x_1 POL(a23(x_1)) = 1 + x_1 POL(a24(x_1)) = 4 + x_1 POL(a25(x_1)) = 6 + x_1 POL(a26(x_1)) = 8 + x_1 POL(a34(x_1)) = 1 + x_1 POL(a35(x_1)) = 4 + x_1 POL(a36(x_1)) = 6 + x_1 POL(a45(x_1)) = 1 + x_1 POL(a46(x_1)) = 4 + x_1 POL(a56(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: a12(a12(x1)) -> x1 a13(a13(x1)) -> x1 a14(a14(x1)) -> x1 a15(a15(x1)) -> x1 a16(a16(x1)) -> x1 a23(a23(x1)) -> x1 a24(a24(x1)) -> x1 a25(a25(x1)) -> x1 a26(a26(x1)) -> x1 a34(a34(x1)) -> x1 a35(a35(x1)) -> x1 a36(a36(x1)) -> x1 a45(a45(x1)) -> x1 a46(a46(x1)) -> x1 a56(a56(x1)) -> x1 a13(x1) -> a12(a23(a12(x1))) a14(x1) -> a12(a23(a34(a23(a12(x1))))) a15(x1) -> a12(a23(a34(a45(a34(a23(a12(x1))))))) a16(x1) -> a12(a23(a34(a45(a56(a45(a34(a23(a12(x1))))))))) a24(x1) -> a23(a34(a23(x1))) a25(x1) -> a23(a34(a45(a34(a23(x1))))) a26(x1) -> a23(a34(a45(a56(a45(a34(a23(x1))))))) a35(x1) -> a34(a45(a34(x1))) a36(x1) -> a34(a45(a56(a45(a34(x1))))) a46(x1) -> a45(a56(a45(x1))) a12(a23(a12(a23(a12(a23(x1)))))) -> x1 a23(a34(a23(a34(a23(a34(x1)))))) -> x1 a34(a45(a34(a45(a34(a45(x1)))))) -> x1 a45(a56(a45(a56(a45(a56(x1)))))) -> x1 ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(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: A12(a34(x1)) -> A34(a12(x1)) A12(a34(x1)) -> A12(x1) A12(a45(x1)) -> A12(x1) A12(a56(x1)) -> A12(x1) A23(a45(x1)) -> A23(x1) A23(a56(x1)) -> A23(x1) A34(a56(x1)) -> A34(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(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 3 SCCs with 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) Obligation: Q DP problem: The TRS P consists of the following rules: A34(a56(x1)) -> A34(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (8) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: A34(a56(x1)) -> A34(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) The set Q consists of the following terms: a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) 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. ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: A34(a56(x1)) -> A34(x1) R is empty. The set Q consists of the following terms: a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) 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]. a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: A34(a56(x1)) -> A34(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) 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: *A34(a56(x1)) -> A34(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (15) YES ---------------------------------------- (16) Obligation: Q DP problem: The TRS P consists of the following rules: A23(a56(x1)) -> A23(x1) A23(a45(x1)) -> A23(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (17) MNOCProof (EQUIVALENT) We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R. ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: A23(a56(x1)) -> A23(x1) A23(a45(x1)) -> A23(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) The set Q consists of the following terms: a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: A23(a56(x1)) -> A23(x1) A23(a45(x1)) -> A23(x1) R is empty. The set Q consists of the following terms: a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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]. a12(a34(x0)) a12(a45(x0)) a12(a56(x0)) a23(a45(x0)) a23(a56(x0)) a34(a56(x0)) ---------------------------------------- (22) Obligation: Q DP problem: The TRS P consists of the following rules: A23(a56(x1)) -> A23(x1) A23(a45(x1)) -> A23(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (23) 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: *A23(a56(x1)) -> A23(x1) The graph contains the following edges 1 > 1 *A23(a45(x1)) -> A23(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (24) YES ---------------------------------------- (25) Obligation: Q DP problem: The TRS P consists of the following rules: A12(a45(x1)) -> A12(x1) A12(a34(x1)) -> A12(x1) A12(a56(x1)) -> A12(x1) The TRS R consists of the following rules: a12(a34(x1)) -> a34(a12(x1)) a12(a45(x1)) -> a45(a12(x1)) a12(a56(x1)) -> a56(a12(x1)) a23(a45(x1)) -> a45(a23(x1)) a23(a56(x1)) -> a56(a23(x1)) a34(a56(x1)) -> a56(a34(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (26) 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. ---------------------------------------- (27) Obligation: Q DP problem: The TRS P consists of the following rules: A12(a45(x1)) -> A12(x1) A12(a34(x1)) -> A12(x1) A12(a56(x1)) -> A12(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (28) 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: *A12(a45(x1)) -> A12(x1) The graph contains the following edges 1 > 1 *A12(a34(x1)) -> A12(x1) The graph contains the following edges 1 > 1 *A12(a56(x1)) -> A12(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (29) YES