27.58/8.00 YES 28.01/8.03 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 28.01/8.03 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 28.01/8.03 28.01/8.03 28.01/8.03 Termination w.r.t. Q of the given QTRS could be proven: 28.01/8.03 28.01/8.03 (0) QTRS 28.01/8.03 (1) QTRSRRRProof [EQUIVALENT, 58 ms] 28.01/8.03 (2) QTRS 28.01/8.03 (3) DependencyPairsProof [EQUIVALENT, 8 ms] 28.01/8.03 (4) QDP 28.01/8.03 (5) MRRProof [EQUIVALENT, 43 ms] 28.01/8.03 (6) QDP 28.01/8.03 (7) QDPOrderProof [EQUIVALENT, 31 ms] 28.01/8.03 (8) QDP 28.01/8.03 (9) DependencyGraphProof [EQUIVALENT, 0 ms] 28.01/8.03 (10) TRUE 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (0) 28.01/8.03 Obligation: 28.01/8.03 Q restricted rewrite system: 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 a(b(c(x1))) -> b(x1) 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (1) QTRSRRRProof (EQUIVALENT) 28.01/8.03 Used ordering: 28.01/8.03 Polynomial interpretation [POLO]: 28.01/8.03 28.01/8.03 POL(a(x_1)) = 3 + x_1 28.01/8.03 POL(b(x_1)) = 1 + x_1 28.01/8.03 POL(c(x_1)) = 1 + x_1 28.01/8.03 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 28.01/8.03 28.01/8.03 a(b(c(x1))) -> b(x1) 28.01/8.03 28.01/8.03 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (2) 28.01/8.03 Obligation: 28.01/8.03 Q restricted rewrite system: 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (3) DependencyPairsProof (EQUIVALENT) 28.01/8.03 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (4) 28.01/8.03 Obligation: 28.01/8.03 Q DP problem: 28.01/8.03 The TRS P consists of the following rules: 28.01/8.03 28.01/8.03 C(b(b(x1))) -> A(x1) 28.01/8.03 A(a(x1)) -> C(b(a(c(x1)))) 28.01/8.03 A(a(x1)) -> A(c(x1)) 28.01/8.03 A(a(x1)) -> C(x1) 28.01/8.03 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 We have to consider all minimal (P,Q,R)-chains. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (5) MRRProof (EQUIVALENT) 28.01/8.03 By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented. 28.01/8.03 28.01/8.03 Strictly oriented dependency pairs: 28.01/8.03 28.01/8.03 A(a(x1)) -> A(c(x1)) 28.01/8.03 A(a(x1)) -> C(x1) 28.01/8.03 28.01/8.03 28.01/8.03 Used ordering: Polynomial interpretation [POLO]: 28.01/8.03 28.01/8.03 POL(A(x_1)) = 2 + x_1 28.01/8.03 POL(C(x_1)) = x_1 28.01/8.03 POL(a(x_1)) = 3 + x_1 28.01/8.03 POL(b(x_1)) = 1 + x_1 28.01/8.03 POL(c(x_1)) = 1 + x_1 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (6) 28.01/8.03 Obligation: 28.01/8.03 Q DP problem: 28.01/8.03 The TRS P consists of the following rules: 28.01/8.03 28.01/8.03 C(b(b(x1))) -> A(x1) 28.01/8.03 A(a(x1)) -> C(b(a(c(x1)))) 28.01/8.03 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 We have to consider all minimal (P,Q,R)-chains. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (7) QDPOrderProof (EQUIVALENT) 28.01/8.03 We use the reduction pair processor [LPAR04,JAR06]. 28.01/8.03 28.01/8.03 28.01/8.03 The following pairs can be oriented strictly and are deleted. 28.01/8.03 28.01/8.03 A(a(x1)) -> C(b(a(c(x1)))) 28.01/8.03 The remaining pairs can at least be oriented weakly. 28.01/8.03 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 28.01/8.03 28.01/8.03 <<< 28.01/8.03 POL(C(x_1)) = [[-I]] + [[0A, -I, -I]] * x_1 28.01/8.03 >>> 28.01/8.03 28.01/8.03 <<< 28.01/8.03 POL(b(x_1)) = [[0A], [-I], [0A]] + [[-I, 0A, -I], [0A, -I, -I], [0A, 0A, 0A]] * x_1 28.01/8.03 >>> 28.01/8.03 28.01/8.03 <<< 28.01/8.03 POL(A(x_1)) = [[0A]] + [[0A, -I, -I]] * x_1 28.01/8.03 >>> 28.01/8.03 28.01/8.03 <<< 28.01/8.03 POL(a(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 0A], [-I, -I, -I], [0A, 0A, 0A]] * x_1 28.01/8.03 >>> 28.01/8.03 28.01/8.03 <<< 28.01/8.03 POL(c(x_1)) = [[1A], [0A], [0A]] + [[0A, 0A, 0A], [0A, -I, -I], [0A, 0A, 0A]] * x_1 28.01/8.03 >>> 28.01/8.03 28.01/8.03 28.01/8.03 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 28.01/8.03 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 28.01/8.03 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (8) 28.01/8.03 Obligation: 28.01/8.03 Q DP problem: 28.01/8.03 The TRS P consists of the following rules: 28.01/8.03 28.01/8.03 C(b(b(x1))) -> A(x1) 28.01/8.03 28.01/8.03 The TRS R consists of the following rules: 28.01/8.03 28.01/8.03 c(b(b(x1))) -> a(x1) 28.01/8.03 c(x1) -> b(x1) 28.01/8.03 a(a(x1)) -> c(b(a(c(x1)))) 28.01/8.03 28.01/8.03 Q is empty. 28.01/8.03 We have to consider all minimal (P,Q,R)-chains. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (9) DependencyGraphProof (EQUIVALENT) 28.01/8.03 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 28.01/8.03 ---------------------------------------- 28.01/8.03 28.01/8.03 (10) 28.01/8.03 TRUE 28.33/8.19 EOF