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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 17 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 52 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 499 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(0(1(1(x1)))) 0(1(0(1(x1)))) -> 0(0(1(0(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: 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(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: 0^1(0(0(0(x1)))) -> 1^1(1(0(1(x1)))) 0^1(0(0(0(x1)))) -> 1^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(1(0(x1)))) -> 0^1(1(0(0(x1)))) 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 1^1(0(1(0(x1)))) -> 0^1(0(x1)) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 0^1(0(0(0(x1)))) -> 1^1(0(1(x1))) 0^1(0(0(0(x1)))) -> 0^1(1(x1)) 0^1(0(0(0(x1)))) -> 1^1(x1) 1^1(0(1(0(x1)))) -> 1^1(0(0(x1))) 1^1(0(1(0(x1)))) -> 0^1(0(x1)) The remaining pairs can at least be oriented weakly. Used ordering: Polynomial interpretation [POLO]: POL(0(x_1)) = 1 + x_1 POL(0^1(x_1)) = 1 + x_1 POL(1(x_1)) = 1 + x_1 POL(1^1(x_1)) = 1 + x_1 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(0(0(0(x1)))) -> 1^1(1(0(1(x1)))) 1^1(0(1(0(x1)))) -> 0^1(1(0(0(x1)))) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) QDPOrderProof (EQUIVALENT) We use the reduction pair processor [LPAR04,JAR06]. The following pairs can be oriented strictly and are deleted. 1^1(0(1(0(x1)))) -> 0^1(1(0(0(x1)))) The remaining pairs can at least be oriented weakly. Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: <<< POL(0^1(x_1)) = [[0A]] + [[-I, -I, 0A]] * x_1 >>> <<< POL(0(x_1)) = [[0A], [-I], [-I]] + [[0A, 1A, 1A], [0A, -I, 0A], [0A, 0A, 1A]] * x_1 >>> <<< POL(1^1(x_1)) = [[1A]] + [[0A, 0A, 0A]] * x_1 >>> <<< POL(1(x_1)) = [[0A], [-I], [-I]] + [[1A, 0A, 0A], [0A, 0A, 1A], [0A, 0A, 0A]] * x_1 >>> The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: 0^1(0(0(0(x1)))) -> 1^1(1(0(1(x1)))) The TRS R consists of the following rules: 0(0(0(0(x1)))) -> 1(1(0(1(x1)))) 1(0(1(0(x1)))) -> 0(1(0(0(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (10) TRUE