details

property	value
status	complete
benchmark	z028.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.75025 seconds
cpu usage	4.16121
user time	3.98251
system time	0.178703
max virtual memory	1.9142144E7
max residence set size	347480.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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, 4 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(b(a(a(a(x1))))))) -> 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(b(a(b(a(x1))))))) -> b(a(b(a(b(a(a(a(a(x1))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(a(a(b(a(b(a(x1))))))) -> b(a(b(a(b(a(a(a(a(x1))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(a(a(b(a(b(x)))))) -> b(a(b(a(b(a(a(a(x)))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(b(a(b(x)))))) -> b(a(b(a(b(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(b(a(b(x)))))) -> b(a(b(a(b(a(a(a(x)))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 561, 562, 563, 564, 565, 566, 567, 568, 569, 570, 571, 572, 573, 574, 575, 576 Node 561 is start node and node 562 is final node. Those nodes are connected through the following edges: * 561 to 563 labelled b_1(0)* 562 to 562 labelled #_1(0)* 563 to 564 labelled a_1(0)* 564 to 565 labelled b_1(0)* 565 to 566 labelled a_1(0)* 566 to 567 labelled b_1(0)* 567 to 568 labelled a_1(0)* 567 to 570 labelled b_1(1)* 568 to 569 labelled a_1(0)* 568 to 570 labelled b_1(1)* 569 to 562 labelled a_1(0)* 569 to 570 labelled b_1(1)* 570 to 571 labelled a_1(1)* 571 to 572 labelled b_1(1)* 572 to 573 labelled a_1(1)* 573 to 574 labelled b_1(1)* 574 to 575 labelled a_1(1)* 574 to 570 labelled b_1(1)* 575 to 576 labelled a_1(1)* 575 to 570 labelled b_1(1)* 576 to 562 labelled a_1(1)* 576 to 570 labelled b_1(1) ---------------------------------------- (6) YES popout

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