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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) QTRSRRRProof [EQUIVALENT, 11 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 1 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) AND (9) QDP (10) QDPOrderProof [EQUIVALENT, 33 ms] (11) QDP (12) DependencyGraphProof [EQUIVALENT, 0 ms] (13) QDP (14) QDPOrderProof [EQUIVALENT, 175 ms] (15) QDP (16) PisEmptyProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) UsableRulesProof [EQUIVALENT, 0 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(d(b(x1))) -> c(d(b(x1))) b(a(c(x1))) -> b(c(x1)) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(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(d(b(x1))) -> b(d(c(x1))) c(a(b(x1))) -> c(b(x1)) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. ---------------------------------------- (3) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 POL(d(x_1)) = x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: c(a(b(x1))) -> c(b(x1)) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(d(b(x1))) -> B(d(c(x1))) B(d(b(x1))) -> D(c(x1)) B(d(b(x1))) -> C(x1) D(a(x1)) -> C(d(x1)) D(a(x1)) -> D(x1) B(b(b(x1))) -> C(b(a(x1))) B(b(b(x1))) -> B(a(x1)) C(d(x1)) -> D(b(x1)) C(d(x1)) -> B(x1) C(d(x1)) -> D(b(d(x1))) C(d(x1)) -> B(d(x1)) C(a(d(x1))) -> B(b(x1)) C(a(d(x1))) -> B(x1) The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes. ---------------------------------------- (8) Complex Obligation (AND) ---------------------------------------- (9) Obligation: Q DP problem: The TRS P consists of the following rules: B(d(b(x1))) -> C(x1) C(d(x1)) -> B(x1) B(d(b(x1))) -> B(d(c(x1))) C(d(x1)) -> B(d(x1)) C(a(d(x1))) -> B(b(x1)) C(a(d(x1))) -> B(x1) The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (10) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(d(b(x1))) -> C(x1) C(a(d(x1))) -> B(x1) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(B(x_1)) = x_1 POL(C(x_1)) = x_1 POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 POL(d(x_1)) = x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(d(x1)) -> d(b(x1)) d(a(x1)) -> c(d(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) b(d(b(x1))) -> b(d(c(x1))) b(b(b(x1))) -> c(b(a(x1))) ---------------------------------------- (11) Obligation: Q DP problem: The TRS P consists of the following rules: C(d(x1)) -> B(x1) B(d(b(x1))) -> B(d(c(x1))) C(d(x1)) -> B(d(x1)) C(a(d(x1))) -> B(b(x1)) The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (12) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes. ---------------------------------------- (13) Obligation: Q DP problem: The TRS P consists of the following rules: B(d(b(x1))) -> B(d(c(x1))) The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (14) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. B(d(b(x1))) -> B(d(c(x1))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(B(x_1)) = [[-I]] + [[0A, 0A, -I]] * x_1 >>> <<< POL(d(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, 0A], [0A, -I, 0A], [-I, 0A, 0A]] * x_1 >>> <<< POL(b(x_1)) = [[1A], [0A], [0A]] + [[0A, 1A, 1A], [0A, 0A, -I], [0A, 0A, -I]] * x_1 >>> <<< POL(c(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, 0A], [-I, 1A, 1A], [-I, 0A, 0A]] * x_1 >>> <<< POL(a(x_1)) = [[0A], [0A], [1A]] + [[0A, 0A, 0A], [-I, 0A, 0A], [1A, 1A, 1A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: c(d(x1)) -> d(b(x1)) d(a(x1)) -> c(d(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) b(d(b(x1))) -> b(d(c(x1))) b(b(b(x1))) -> c(b(a(x1))) ---------------------------------------- (15) Obligation: Q DP problem: P is empty. The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (16) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (17) YES ---------------------------------------- (18) Obligation: Q DP problem: The TRS P consists of the following rules: D(a(x1)) -> D(x1) The TRS R consists of the following rules: b(d(b(x1))) -> b(d(c(x1))) d(a(x1)) -> c(d(x1)) b(b(b(x1))) -> c(b(a(x1))) c(d(x1)) -> d(b(x1)) c(d(x1)) -> d(b(d(x1))) c(a(d(x1))) -> b(b(x1)) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (19) 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. ---------------------------------------- (20) Obligation: Q DP problem: The TRS P consists of the following rules: D(a(x1)) -> D(x1) R is empty. Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (21) 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: *D(a(x1)) -> D(x1) The graph contains the following edges 1 > 1 ---------------------------------------- (22) YES