/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4) QTRS (5) RFCMatchBoundsTRSProof [EQUIVALENT, 0 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(a(a(a(x1))))))))) -> a(a(a(a(a(a(b(a(b(a(b(x1))))))))))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(a(b(a(b(a(x1))))))))) -> b(a(b(a(b(a(a(a(a(a(a(x1))))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(a(a(a(a(b(a(b(a(x1))))))))) -> b(a(b(a(b(a(a(a(a(a(a(x1))))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(a(a(a(a(b(a(b(x)))))))) -> b(a(b(a(b(a(a(a(a(a(x)))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(a(b(a(b(x)))))))) -> b(a(b(a(b(a(a(a(a(a(x)))))))))) Q is empty. ---------------------------------------- (5) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 1. This implies Q-termination of R. The following rules were used to construct the certificate: a(a(a(a(a(b(a(b(x)))))))) -> b(a(b(a(b(a(a(a(a(a(x)))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 7, 8, 28, 29, 30, 31, 32, 33, 34, 35, 36, 77, 78, 79, 80, 81, 82, 83, 84, 85 Node 7 is start node and node 8 is final node. Those nodes are connected through the following edges: * 7 to 28 labelled b_1(0)* 8 to 8 labelled #_1(0)* 28 to 29 labelled a_1(0)* 29 to 30 labelled b_1(0)* 30 to 31 labelled a_1(0)* 31 to 32 labelled b_1(0)* 32 to 33 labelled a_1(0)* 32 to 77 labelled b_1(1)* 33 to 34 labelled a_1(0)* 33 to 77 labelled b_1(1)* 34 to 35 labelled a_1(0)* 34 to 77 labelled b_1(1)* 35 to 36 labelled a_1(0)* 35 to 77 labelled b_1(1)* 36 to 8 labelled a_1(0)* 36 to 77 labelled b_1(1)* 77 to 78 labelled a_1(1)* 78 to 79 labelled b_1(1)* 79 to 80 labelled a_1(1)* 80 to 81 labelled b_1(1)* 81 to 82 labelled a_1(1)* 81 to 77 labelled b_1(1)* 82 to 83 labelled a_1(1)* 82 to 77 labelled b_1(1)* 83 to 84 labelled a_1(1)* 83 to 77 labelled b_1(1)* 84 to 85 labelled a_1(1)* 84 to 77 labelled b_1(1)* 85 to 8 labelled a_1(1)* 85 to 77 labelled b_1(1) ---------------------------------------- (6) YES