/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) Strip Symbols Proof [SOUND, 0 ms] (2) QTRS (3) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(x))) -> b(a(x)) Q is empty. ---------------------------------------- (1) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(a(x))) -> b(a(x)) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(b(x)) -> b(x) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(x)) -> b(x) Q is empty. ---------------------------------------- (3) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 0. This implies Q-termination of R. The following rules were used to construct the certificate: a(b(x)) -> b(x) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 5, 6 Node 5 is start node and node 6 is final node. Those nodes are connected through the following edges: * 5 to 6 labelled b_1(0)* 6 to 6 labelled #_1(0) ---------------------------------------- (4) YES