details

property	value
status	complete
benchmark	abaababa-aabababaab.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.6862 seconds
cpu usage	3.94457
user time	3.75684
system time	0.187731
max virtual memory	1.894496E7
max residence set size	301364.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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(a(b(a(b(a(x1)))))))) -> a(a(b(a(b(a(b(a(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(b(a(b(a(a(b(a(x1)))))))) -> b(a(a(b(a(b(a(b(a(a(x1)))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(a(b(a(a(b(a(x1)))))))) -> b(a(a(b(a(b(a(b(a(a(x1)))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(a(x))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(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(b(a(b(a(a(b(x))))))) -> b(a(a(b(a(b(a(b(a(x))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 591, 592, 593, 594, 595, 596, 597, 598, 599, 600, 601, 602, 603, 604, 605, 606, 607, 608 Node 591 is start node and node 592 is final node. Those nodes are connected through the following edges: * 591 to 593 labelled b_1(0)* 592 to 592 labelled #_1(0)* 593 to 594 labelled a_1(0)* 594 to 595 labelled a_1(0)* 595 to 596 labelled b_1(0)* 596 to 597 labelled a_1(0)* 596 to 601 labelled b_1(1)* 597 to 598 labelled b_1(0)* 598 to 599 labelled a_1(0)* 598 to 601 labelled b_1(1)* 599 to 600 labelled b_1(0)* 600 to 592 labelled a_1(0)* 600 to 601 labelled b_1(1)* 601 to 602 labelled a_1(1)* 602 to 603 labelled a_1(1)* 603 to 604 labelled b_1(1)* 604 to 605 labelled a_1(1)* 604 to 601 labelled b_1(1)* 605 to 606 labelled b_1(1)* 606 to 607 labelled a_1(1)* 606 to 601 labelled b_1(1)* 607 to 608 labelled b_1(1)* 608 to 592 labelled a_1(1)* 608 to 601 labelled b_1(1) ---------------------------------------- (6) YES popout

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