28.17/8.07 YES 28.35/8.10 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 28.35/8.10 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 28.35/8.10 28.35/8.10 28.35/8.10 Termination w.r.t. Q of the given QTRS could be proven: 28.35/8.10 28.35/8.10 (0) QTRS 28.35/8.10 (1) QTRS Reverse [EQUIVALENT, 0 ms] 28.35/8.10 (2) QTRS 28.35/8.10 (3) DependencyPairsProof [EQUIVALENT, 23 ms] 28.35/8.10 (4) QDP 28.35/8.10 (5) DependencyGraphProof [EQUIVALENT, 7 ms] 28.35/8.10 (6) AND 28.35/8.10 (7) QDP 28.35/8.10 (8) UsableRulesProof [EQUIVALENT, 0 ms] 28.35/8.10 (9) QDP 28.35/8.10 (10) QDPSizeChangeProof [EQUIVALENT, 0 ms] 28.35/8.10 (11) YES 28.35/8.10 (12) QDP 28.35/8.10 (13) MRRProof [EQUIVALENT, 25 ms] 28.35/8.10 (14) QDP 28.35/8.10 (15) MRRProof [EQUIVALENT, 5 ms] 28.35/8.10 (16) QDP 28.35/8.10 (17) QDPOrderProof [EQUIVALENT, 181 ms] 28.35/8.10 (18) QDP 28.35/8.10 (19) PisEmptyProof [EQUIVALENT, 0 ms] 28.35/8.10 (20) YES 28.35/8.10 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (0) 28.35/8.10 Obligation: 28.35/8.10 Q restricted rewrite system: 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 a(b(a(b(x1)))) -> b(b(a(a(x1)))) 28.35/8.10 a(a(a(b(x1)))) -> a(a(b(a(x1)))) 28.35/8.10 b(b(a(b(x1)))) -> b(a(b(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (1) QTRS Reverse (EQUIVALENT) 28.35/8.10 We applied the QTRS Reverse Processor [REVERSE]. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (2) 28.35/8.10 Obligation: 28.35/8.10 Q restricted rewrite system: 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (3) DependencyPairsProof (EQUIVALENT) 28.35/8.10 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (4) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(b(a(x1)))) -> B(b(x1)) 28.35/8.10 B(a(b(a(x1)))) -> B(x1) 28.35/8.10 B(a(a(a(x1)))) -> B(a(a(x1))) 28.35/8.10 B(a(b(b(x1)))) -> B(b(a(b(x1)))) 28.35/8.10 B(a(b(b(x1)))) -> B(a(b(x1))) 28.35/8.10 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (5) DependencyGraphProof (EQUIVALENT) 28.35/8.10 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (6) 28.35/8.10 Complex Obligation (AND) 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (7) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(a(a(x1)))) -> B(a(a(x1))) 28.35/8.10 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (8) UsableRulesProof (EQUIVALENT) 28.35/8.10 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. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (9) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(a(a(x1)))) -> B(a(a(x1))) 28.35/8.10 28.35/8.10 R is empty. 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (10) QDPSizeChangeProof (EQUIVALENT) 28.35/8.10 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. 28.35/8.10 28.35/8.10 From the DPs we obtained the following set of size-change graphs: 28.35/8.10 *B(a(a(a(x1)))) -> B(a(a(x1))) 28.35/8.10 The graph contains the following edges 1 > 1 28.35/8.10 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (11) 28.35/8.10 YES 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (12) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(b(a(x1)))) -> B(x1) 28.35/8.10 B(a(b(a(x1)))) -> B(b(x1)) 28.35/8.10 B(a(b(b(x1)))) -> B(b(a(b(x1)))) 28.35/8.10 B(a(b(b(x1)))) -> B(a(b(x1))) 28.35/8.10 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (13) MRRProof (EQUIVALENT) 28.35/8.10 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. 28.35/8.10 28.35/8.10 Strictly oriented dependency pairs: 28.35/8.10 28.35/8.10 B(a(b(a(x1)))) -> B(x1) 28.35/8.10 B(a(b(a(x1)))) -> B(b(x1)) 28.35/8.10 28.35/8.10 28.35/8.10 Used ordering: Polynomial interpretation [POLO]: 28.35/8.10 28.35/8.10 POL(B(x_1)) = 2*x_1 28.35/8.10 POL(a(x_1)) = 2 + 2*x_1 28.35/8.10 POL(b(x_1)) = x_1 28.35/8.10 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (14) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(b(b(x1)))) -> B(b(a(b(x1)))) 28.35/8.10 B(a(b(b(x1)))) -> B(a(b(x1))) 28.35/8.10 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (15) MRRProof (EQUIVALENT) 28.35/8.10 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. 28.35/8.10 28.35/8.10 Strictly oriented dependency pairs: 28.35/8.10 28.35/8.10 B(a(b(b(x1)))) -> B(a(b(x1))) 28.35/8.10 28.35/8.10 28.35/8.10 Used ordering: Polynomial interpretation [POLO]: 28.35/8.10 28.35/8.10 POL(B(x_1)) = x_1 28.35/8.10 POL(a(x_1)) = x_1 28.35/8.10 POL(b(x_1)) = 2 + x_1 28.35/8.10 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (16) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 The TRS P consists of the following rules: 28.35/8.10 28.35/8.10 B(a(b(b(x1)))) -> B(b(a(b(x1)))) 28.35/8.10 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (17) QDPOrderProof (EQUIVALENT) 28.35/8.10 We use the reduction pair processor [LPAR04,JAR06]. 28.35/8.10 28.35/8.10 28.35/8.10 The following pairs can be oriented strictly and are deleted. 28.35/8.10 28.35/8.10 B(a(b(b(x1)))) -> B(b(a(b(x1)))) 28.35/8.10 The remaining pairs can at least be oriented weakly. 28.35/8.10 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 28.35/8.10 28.35/8.10 <<< 28.35/8.10 POL(B(x_1)) = [[-I]] + [[0A, 0A, 0A]] * x_1 28.35/8.10 >>> 28.35/8.10 28.35/8.10 <<< 28.35/8.10 POL(a(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, 0A], [-I, -I, 1A], [-I, -I, -I]] * x_1 28.35/8.10 >>> 28.35/8.10 28.35/8.10 <<< 28.35/8.10 POL(b(x_1)) = [[1A], [1A], [0A]] + [[0A, -I, 0A], [0A, -I, 0A], [0A, -I, 0A]] * x_1 28.35/8.10 >>> 28.35/8.10 28.35/8.10 28.35/8.10 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (18) 28.35/8.10 Obligation: 28.35/8.10 Q DP problem: 28.35/8.10 P is empty. 28.35/8.10 The TRS R consists of the following rules: 28.35/8.10 28.35/8.10 b(a(b(a(x1)))) -> a(a(b(b(x1)))) 28.35/8.10 b(a(a(a(x1)))) -> a(b(a(a(x1)))) 28.35/8.10 b(a(b(b(x1)))) -> b(b(a(b(x1)))) 28.35/8.10 28.35/8.10 Q is empty. 28.35/8.10 We have to consider all minimal (P,Q,R)-chains. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (19) PisEmptyProof (EQUIVALENT) 28.35/8.10 The TRS P is empty. Hence, there is no (P,Q,R) chain. 28.35/8.10 ---------------------------------------- 28.35/8.10 28.35/8.10 (20) 28.35/8.10 YES 28.53/8.19 EOF