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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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(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(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(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(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(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(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: 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 513, 514, 515, 516, 517, 518, 519, 520, 521, 522, 523, 524, 525 Node 500 is start node and node 501 is final node. Those nodes are connected through the following edges: * 500 to 502 labelled b_1(0)* 501 to 501 labelled #_1(0)* 502 to 503 labelled a_1(0)* 503 to 504 labelled b_1(0)* 504 to 505 labelled a_1(0)* 505 to 506 labelled a_1(0)* 506 to 507 labelled b_1(0)* 507 to 508 labelled a_1(0)* 508 to 509 labelled b_1(0)* 509 to 510 labelled a_1(0)* 509 to 514 labelled b_1(1)* 510 to 511 labelled a_1(0)* 510 to 514 labelled b_1(1)* 511 to 512 labelled a_1(0)* 511 to 514 labelled b_1(1)* 512 to 513 labelled a_1(0)* 512 to 514 labelled b_1(1)* 513 to 501 labelled a_1(0)* 513 to 514 labelled b_1(1)* 514 to 515 labelled a_1(1)* 515 to 516 labelled b_1(1)* 516 to 517 labelled a_1(1)* 517 to 518 labelled a_1(1)* 518 to 519 labelled b_1(1)* 519 to 520 labelled a_1(1)* 520 to 521 labelled b_1(1)* 521 to 522 labelled a_1(1)* 521 to 514 labelled b_1(1)* 522 to 523 labelled a_1(1)* 522 to 514 labelled b_1(1)* 523 to 524 labelled a_1(1)* 523 to 514 labelled b_1(1)* 524 to 525 labelled a_1(1)* 524 to 514 labelled b_1(1)* 525 to 501 labelled a_1(1)* 525 to 514 labelled b_1(1) ---------------------------------------- (6) YES