19.79/5.95 YES 20.30/6.10 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 20.30/6.10 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 20.30/6.10 20.30/6.10 20.30/6.10 Termination w.r.t. Q of the given QTRS could be proven: 20.30/6.10 20.30/6.10 (0) QTRS 20.30/6.10 (1) QTRS Reverse [EQUIVALENT, 0 ms] 20.30/6.10 (2) QTRS 20.30/6.10 (3) QTRSRRRProof [EQUIVALENT, 3 ms] 20.30/6.10 (4) QTRS 20.30/6.10 (5) DependencyPairsProof [EQUIVALENT, 1 ms] 20.30/6.10 (6) QDP 20.30/6.10 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 20.30/6.10 (8) AND 20.30/6.10 (9) QDP 20.30/6.10 (10) MRRProof [EQUIVALENT, 42 ms] 20.30/6.10 (11) QDP 20.30/6.10 (12) QDPOrderProof [EQUIVALENT, 0 ms] 20.30/6.10 (13) QDP 20.30/6.10 (14) PisEmptyProof [EQUIVALENT, 0 ms] 20.30/6.10 (15) YES 20.30/6.10 (16) QDP 20.30/6.10 (17) UsableRulesProof [EQUIVALENT, 0 ms] 20.30/6.10 (18) QDP 20.30/6.10 (19) QDPSizeChangeProof [EQUIVALENT, 1 ms] 20.30/6.10 (20) YES 20.30/6.10 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (0) 20.30/6.10 Obligation: 20.30/6.10 Q restricted rewrite system: 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> c(d(b(x1))) 20.30/6.10 b(a(c(x1))) -> b(c(x1)) 20.30/6.10 a(d(x1)) -> d(c(x1)) 20.30/6.10 b(b(b(x1))) -> a(b(c(x1))) 20.30/6.10 d(c(x1)) -> b(d(x1)) 20.30/6.10 d(c(x1)) -> d(b(d(x1))) 20.30/6.10 d(a(c(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (1) QTRS Reverse (EQUIVALENT) 20.30/6.10 We applied the QTRS Reverse Processor [REVERSE]. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (2) 20.30/6.10 Obligation: 20.30/6.10 Q restricted rewrite system: 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 c(a(b(x1))) -> c(b(x1)) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (3) QTRSRRRProof (EQUIVALENT) 20.30/6.10 Used ordering: 20.30/6.10 Polynomial interpretation [POLO]: 20.30/6.10 20.30/6.10 POL(a(x_1)) = 1 + x_1 20.30/6.10 POL(b(x_1)) = 1 + x_1 20.30/6.10 POL(c(x_1)) = 1 + x_1 20.30/6.10 POL(d(x_1)) = x_1 20.30/6.10 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 20.30/6.10 20.30/6.10 c(a(b(x1))) -> c(b(x1)) 20.30/6.10 20.30/6.10 20.30/6.10 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (4) 20.30/6.10 Obligation: 20.30/6.10 Q restricted rewrite system: 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (5) DependencyPairsProof (EQUIVALENT) 20.30/6.10 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (6) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 The TRS P consists of the following rules: 20.30/6.10 20.30/6.10 B(d(b(x1))) -> B(d(c(x1))) 20.30/6.10 B(d(b(x1))) -> D(c(x1)) 20.30/6.10 B(d(b(x1))) -> C(x1) 20.30/6.10 D(a(x1)) -> C(d(x1)) 20.30/6.10 D(a(x1)) -> D(x1) 20.30/6.10 B(b(b(x1))) -> C(b(a(x1))) 20.30/6.10 B(b(b(x1))) -> B(a(x1)) 20.30/6.10 C(d(x1)) -> D(b(x1)) 20.30/6.10 C(d(x1)) -> B(x1) 20.30/6.10 C(d(x1)) -> D(b(d(x1))) 20.30/6.10 C(d(x1)) -> B(d(x1)) 20.30/6.10 C(a(d(x1))) -> B(b(x1)) 20.30/6.10 C(a(d(x1))) -> B(x1) 20.30/6.10 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (7) DependencyGraphProof (EQUIVALENT) 20.30/6.10 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (8) 20.30/6.10 Complex Obligation (AND) 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (9) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 The TRS P consists of the following rules: 20.30/6.10 20.30/6.10 B(d(b(x1))) -> C(x1) 20.30/6.10 C(d(x1)) -> B(x1) 20.30/6.10 B(d(b(x1))) -> B(d(c(x1))) 20.30/6.10 C(d(x1)) -> B(d(x1)) 20.30/6.10 C(a(d(x1))) -> B(b(x1)) 20.30/6.10 C(a(d(x1))) -> B(x1) 20.30/6.10 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (10) MRRProof (EQUIVALENT) 20.30/6.10 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. 20.30/6.10 20.30/6.10 Strictly oriented dependency pairs: 20.30/6.10 20.30/6.10 B(d(b(x1))) -> C(x1) 20.30/6.10 C(d(x1)) -> B(x1) 20.30/6.10 C(d(x1)) -> B(d(x1)) 20.30/6.10 C(a(d(x1))) -> B(b(x1)) 20.30/6.10 C(a(d(x1))) -> B(x1) 20.30/6.10 20.30/6.10 20.30/6.10 Used ordering: Polynomial interpretation [POLO]: 20.30/6.10 20.30/6.10 POL(B(x_1)) = 2*x_1 20.30/6.10 POL(C(x_1)) = 1 + 2*x_1 20.30/6.10 POL(a(x_1)) = 2 + x_1 20.30/6.10 POL(b(x_1)) = 2 + x_1 20.30/6.10 POL(c(x_1)) = 2 + x_1 20.30/6.10 POL(d(x_1)) = x_1 20.30/6.10 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (11) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 The TRS P consists of the following rules: 20.30/6.10 20.30/6.10 B(d(b(x1))) -> B(d(c(x1))) 20.30/6.10 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (12) QDPOrderProof (EQUIVALENT) 20.30/6.10 We use the reduction pair processor [LPAR04,JAR06]. 20.30/6.10 20.30/6.10 20.30/6.10 The following pairs can be oriented strictly and are deleted. 20.30/6.10 20.30/6.10 B(d(b(x1))) -> B(d(c(x1))) 20.30/6.10 The remaining pairs can at least be oriented weakly. 20.30/6.10 Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]: 20.30/6.10 20.30/6.10 <<< 20.30/6.10 POL(B(x_1)) = [[-I]] + [[0A, 0A, -I]] * x_1 20.30/6.10 >>> 20.30/6.10 20.30/6.10 <<< 20.30/6.10 POL(d(x_1)) = [[0A], [0A], [0A]] + [[-I, -I, 0A], [0A, -I, 0A], [-I, 0A, 0A]] * x_1 20.30/6.10 >>> 20.30/6.10 20.30/6.10 <<< 20.30/6.10 POL(b(x_1)) = [[1A], [0A], [0A]] + [[0A, 1A, 1A], [0A, 0A, -I], [0A, 0A, -I]] * x_1 20.30/6.10 >>> 20.30/6.10 20.30/6.10 <<< 20.30/6.10 POL(c(x_1)) = [[0A], [0A], [0A]] + [[-I, 0A, 0A], [-I, 1A, 1A], [-I, 0A, 0A]] * x_1 20.30/6.10 >>> 20.30/6.10 20.30/6.10 <<< 20.30/6.10 POL(a(x_1)) = [[0A], [0A], [1A]] + [[0A, 0A, 0A], [-I, 0A, 0A], [1A, 1A, 1A]] * x_1 20.30/6.10 >>> 20.30/6.10 20.30/6.10 20.30/6.10 The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented: 20.30/6.10 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (13) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 P is empty. 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (14) PisEmptyProof (EQUIVALENT) 20.30/6.10 The TRS P is empty. Hence, there is no (P,Q,R) chain. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (15) 20.30/6.10 YES 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (16) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 The TRS P consists of the following rules: 20.30/6.10 20.30/6.10 D(a(x1)) -> D(x1) 20.30/6.10 20.30/6.10 The TRS R consists of the following rules: 20.30/6.10 20.30/6.10 b(d(b(x1))) -> b(d(c(x1))) 20.30/6.10 d(a(x1)) -> c(d(x1)) 20.30/6.10 b(b(b(x1))) -> c(b(a(x1))) 20.30/6.10 c(d(x1)) -> d(b(x1)) 20.30/6.10 c(d(x1)) -> d(b(d(x1))) 20.30/6.10 c(a(d(x1))) -> b(b(x1)) 20.30/6.10 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (17) UsableRulesProof (EQUIVALENT) 20.30/6.10 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. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (18) 20.30/6.10 Obligation: 20.30/6.10 Q DP problem: 20.30/6.10 The TRS P consists of the following rules: 20.30/6.10 20.30/6.10 D(a(x1)) -> D(x1) 20.30/6.10 20.30/6.10 R is empty. 20.30/6.10 Q is empty. 20.30/6.10 We have to consider all minimal (P,Q,R)-chains. 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (19) QDPSizeChangeProof (EQUIVALENT) 20.30/6.10 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. 20.30/6.10 20.30/6.10 From the DPs we obtained the following set of size-change graphs: 20.30/6.10 *D(a(x1)) -> D(x1) 20.30/6.10 The graph contains the following edges 1 > 1 20.30/6.10 20.30/6.10 20.30/6.10 ---------------------------------------- 20.30/6.10 20.30/6.10 (20) 20.30/6.10 YES 20.68/6.23 EOF