23.20/7.00 YES 25.39/7.65 proof of /export/starexec/sandbox/benchmark/theBenchmark.xml 25.39/7.65 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 25.39/7.65 25.39/7.65 25.39/7.65 Termination w.r.t. Q of the given QTRS could be proven: 25.39/7.65 25.39/7.65 (0) QTRS 25.39/7.65 (1) QTRSRRRProof [EQUIVALENT, 33 ms] 25.39/7.65 (2) QTRS 25.39/7.65 (3) QTRSRRRProof [EQUIVALENT, 2 ms] 25.39/7.65 (4) QTRS 25.39/7.65 (5) DependencyPairsProof [EQUIVALENT, 32 ms] 25.39/7.65 (6) QDP 25.39/7.65 (7) DependencyGraphProof [EQUIVALENT, 0 ms] 25.39/7.65 (8) QDP 25.39/7.65 (9) UsableRulesProof [EQUIVALENT, 1 ms] 25.39/7.65 (10) QDP 25.39/7.65 (11) MRRProof [EQUIVALENT, 32 ms] 25.39/7.65 (12) QDP 25.39/7.65 (13) PisEmptyProof [EQUIVALENT, 0 ms] 25.39/7.65 (14) YES 25.39/7.65 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (0) 25.39/7.65 Obligation: 25.39/7.65 Q restricted rewrite system: 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 R(x1) -> r(x1) 25.39/7.65 r(p(x1)) -> p(p(r(P(x1)))) 25.39/7.65 r(r(x1)) -> x1 25.39/7.65 r(P(P(x1))) -> P(P(r(x1))) 25.39/7.65 p(P(x1)) -> x1 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 r(R(x1)) -> x1 25.39/7.65 R(r(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (1) QTRSRRRProof (EQUIVALENT) 25.39/7.65 Used ordering: 25.39/7.65 Polynomial interpretation [POLO]: 25.39/7.65 25.39/7.65 POL(P(x_1)) = x_1 25.39/7.65 POL(R(x_1)) = 1 + x_1 25.39/7.65 POL(p(x_1)) = x_1 25.39/7.65 POL(r(x_1)) = x_1 25.39/7.65 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 25.39/7.65 25.39/7.65 R(x1) -> r(x1) 25.39/7.65 r(R(x1)) -> x1 25.39/7.65 R(r(x1)) -> x1 25.39/7.65 25.39/7.65 25.39/7.65 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (2) 25.39/7.65 Obligation: 25.39/7.65 Q restricted rewrite system: 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 r(p(x1)) -> p(p(r(P(x1)))) 25.39/7.65 r(r(x1)) -> x1 25.39/7.65 r(P(P(x1))) -> P(P(r(x1))) 25.39/7.65 p(P(x1)) -> x1 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (3) QTRSRRRProof (EQUIVALENT) 25.39/7.65 Used ordering: 25.39/7.65 Polynomial interpretation [POLO]: 25.39/7.65 25.39/7.65 POL(P(x_1)) = x_1 25.39/7.65 POL(p(x_1)) = x_1 25.39/7.65 POL(r(x_1)) = 1 + x_1 25.39/7.65 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: 25.39/7.65 25.39/7.65 r(r(x1)) -> x1 25.39/7.65 25.39/7.65 25.39/7.65 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (4) 25.39/7.65 Obligation: 25.39/7.65 Q restricted rewrite system: 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 r(p(x1)) -> p(p(r(P(x1)))) 25.39/7.65 r(P(P(x1))) -> P(P(r(x1))) 25.39/7.65 p(P(x1)) -> x1 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (5) DependencyPairsProof (EQUIVALENT) 25.39/7.65 Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (6) 25.39/7.65 Obligation: 25.39/7.65 Q DP problem: 25.39/7.65 The TRS P consists of the following rules: 25.39/7.65 25.39/7.65 R(p(x1)) -> P^1(p(r(P(x1)))) 25.39/7.65 R(p(x1)) -> P^1(r(P(x1))) 25.39/7.65 R(p(x1)) -> R(P(x1)) 25.39/7.65 R(p(x1)) -> P^2(x1) 25.39/7.65 R(P(P(x1))) -> P^2(P(r(x1))) 25.39/7.65 R(P(P(x1))) -> P^2(r(x1)) 25.39/7.65 R(P(P(x1))) -> R(x1) 25.39/7.65 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 r(p(x1)) -> p(p(r(P(x1)))) 25.39/7.65 r(P(P(x1))) -> P(P(r(x1))) 25.39/7.65 p(P(x1)) -> x1 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 We have to consider all minimal (P,Q,R)-chains. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (7) DependencyGraphProof (EQUIVALENT) 25.39/7.65 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (8) 25.39/7.65 Obligation: 25.39/7.65 Q DP problem: 25.39/7.65 The TRS P consists of the following rules: 25.39/7.65 25.39/7.65 R(P(P(x1))) -> R(x1) 25.39/7.65 R(p(x1)) -> R(P(x1)) 25.39/7.65 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 r(p(x1)) -> p(p(r(P(x1)))) 25.39/7.65 r(P(P(x1))) -> P(P(r(x1))) 25.39/7.65 p(P(x1)) -> x1 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 We have to consider all minimal (P,Q,R)-chains. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (9) UsableRulesProof (EQUIVALENT) 25.39/7.65 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. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (10) 25.39/7.65 Obligation: 25.39/7.65 Q DP problem: 25.39/7.65 The TRS P consists of the following rules: 25.39/7.65 25.39/7.65 R(P(P(x1))) -> R(x1) 25.39/7.65 R(p(x1)) -> R(P(x1)) 25.39/7.65 25.39/7.65 The TRS R consists of the following rules: 25.39/7.65 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Q is empty. 25.39/7.65 We have to consider all minimal (P,Q,R)-chains. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (11) MRRProof (EQUIVALENT) 25.39/7.65 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. 25.39/7.65 25.39/7.65 Strictly oriented dependency pairs: 25.39/7.65 25.39/7.65 R(P(P(x1))) -> R(x1) 25.39/7.65 R(p(x1)) -> R(P(x1)) 25.39/7.65 25.39/7.65 Strictly oriented rules of the TRS R: 25.39/7.65 25.39/7.65 P(p(x1)) -> x1 25.39/7.65 25.39/7.65 Used ordering: Polynomial interpretation [POLO]: 25.39/7.65 25.39/7.65 POL(P(x_1)) = 1 + 3*x_1 25.39/7.65 POL(R(x_1)) = 2*x_1 25.39/7.65 POL(p(x_1)) = 2 + 3*x_1 25.39/7.65 25.39/7.65 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (12) 25.39/7.65 Obligation: 25.39/7.65 Q DP problem: 25.39/7.65 P is empty. 25.39/7.65 R is empty. 25.39/7.65 Q is empty. 25.39/7.65 We have to consider all minimal (P,Q,R)-chains. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (13) PisEmptyProof (EQUIVALENT) 25.39/7.65 The TRS P is empty. Hence, there is no (P,Q,R) chain. 25.39/7.65 ---------------------------------------- 25.39/7.65 25.39/7.65 (14) 25.39/7.65 YES 28.35/10.76 EOF