details

property	value
status	complete
benchmark	abaaaaa-aaaaaababab.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n142.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.65252 seconds
cpu usage	3.77329
user time	3.60559
system time	0.167697
max virtual memory	1.947702E7
max residence set size	293560.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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(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(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(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(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(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(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: 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293 Node 274 is start node and node 275 is final node. Those nodes are connected through the following edges: * 274 to 276 labelled b_1(0)* 275 to 275 labelled #_1(0)* 276 to 277 labelled a_1(0)* 277 to 278 labelled b_1(0)* 278 to 279 labelled a_1(0)* 279 to 280 labelled b_1(0)* 280 to 281 labelled a_1(0)* 280 to 285 labelled b_1(1)* 281 to 282 labelled a_1(0)* 281 to 285 labelled b_1(1)* 282 to 283 labelled a_1(0)* 282 to 285 labelled b_1(1)* 283 to 284 labelled a_1(0)* 283 to 285 labelled b_1(1)* 284 to 275 labelled a_1(0)* 284 to 285 labelled b_1(1)* 285 to 286 labelled a_1(1)* 286 to 287 labelled b_1(1)* 287 to 288 labelled a_1(1)* 288 to 289 labelled b_1(1)* 289 to 290 labelled a_1(1)* 289 to 285 labelled b_1(1)* 290 to 291 labelled a_1(1)* 290 to 285 labelled b_1(1)* 291 to 292 labelled a_1(1)* 291 to 285 labelled b_1(1)* 292 to 293 labelled a_1(1)* 292 to 285 labelled b_1(1)* 293 to 275 labelled a_1(1)* 293 to 285 labelled b_1(1) ---------------------------------------- (6) YES popout

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