/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- MAYBE proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Quasi decreasingness of the given CTRS could not be shown: (0) CTRS (1) CTRSToQTRSProof [SOUND, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) NonTerminationLoopProof [COMPLETE, 0 ms] (10) NO ---------------------------------------- (0) Obligation: Conditional term rewrite system: The TRS R consists of the following rules: a -> b f(a) -> b The conditional TRS C consists of the following conditional rules: g(x) -> g(a) <= f(x) -> x ---------------------------------------- (1) CTRSToQTRSProof (SOUND) The conditional rules have been transormed into unconditional rules according to [CTRS,AAECCNOC]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: g(x) -> U1(f(x)) U1(x) -> g(a) a -> b f(a) -> b 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: G(x) -> U1^1(f(x)) G(x) -> F(x) U1^1(x) -> G(a) U1^1(x) -> A The TRS R consists of the following rules: g(x) -> U1(f(x)) U1(x) -> g(a) a -> b f(a) -> b Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: U1^1(x) -> G(a) G(x) -> U1^1(f(x)) The TRS R consists of the following rules: g(x) -> U1(f(x)) U1(x) -> g(a) a -> b f(a) -> b Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (7) UsableRulesProof (EQUIVALENT) 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. ---------------------------------------- (8) Obligation: Q DP problem: The TRS P consists of the following rules: U1^1(x) -> G(a) G(x) -> U1^1(f(x)) The TRS R consists of the following rules: f(a) -> b a -> b Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (9) NonTerminationLoopProof (COMPLETE) We used the non-termination processor [FROCOS05] to show that the DP problem is infinite. Found a loop by narrowing to the left: s = G(x') evaluates to t =G(a) Thus s starts an infinite chain as s semiunifies with t with the following substitutions: * Matcher: [x' / a] * Semiunifier: [ ] -------------------------------------------------------------------------------- Rewriting sequence G(x') -> U1^1(f(x')) with rule G(x'') -> U1^1(f(x'')) at position [] and matcher [x'' / x'] U1^1(f(x')) -> G(a) with rule U1^1(x) -> G(a) Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence All these steps are and every following step will be a correct step w.r.t to Q. ---------------------------------------- (10) NO