details

property	value
status	complete
benchmark	abaaaa-aaaaabaababab.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n151.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.66993 seconds
cpu usage	3.83969
user time	3.64455
system time	0.195135
max virtual memory	1.9011072E7
max residence set size	301312.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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(x1)))))) -> a(a(a(a(a(b(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(b(a(x1)))))) -> b(a(b(a(b(a(a(b(a(a(a(a(a(x1))))))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(a(a(a(b(a(x1)))))) -> b(a(b(a(b(a(a(b(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(b(x))))) -> b(a(b(a(b(a(a(b(a(a(a(a(x)))))))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(b(x))))) -> b(a(b(a(b(a(a(b(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(b(x))))) -> b(a(b(a(b(a(a(b(a(a(a(a(x)))))))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285 Node 262 is start node and node 263 is final node. Those nodes are connected through the following edges: * 262 to 264 labelled b_1(0)* 263 to 263 labelled #_1(0)* 264 to 265 labelled a_1(0)* 265 to 266 labelled b_1(0)* 266 to 267 labelled a_1(0)* 267 to 268 labelled b_1(0)* 268 to 269 labelled a_1(0)* 269 to 270 labelled a_1(0)* 270 to 271 labelled b_1(0)* 271 to 272 labelled a_1(0)* 271 to 275 labelled b_1(1)* 272 to 273 labelled a_1(0)* 272 to 275 labelled b_1(1)* 273 to 274 labelled a_1(0)* 273 to 275 labelled b_1(1)* 274 to 263 labelled a_1(0)* 274 to 275 labelled b_1(1)* 275 to 276 labelled a_1(1)* 276 to 277 labelled b_1(1)* 277 to 278 labelled a_1(1)* 278 to 279 labelled b_1(1)* 279 to 280 labelled a_1(1)* 280 to 281 labelled a_1(1)* 281 to 282 labelled b_1(1)* 282 to 283 labelled a_1(1)* 282 to 275 labelled b_1(1)* 283 to 284 labelled a_1(1)* 283 to 275 labelled b_1(1)* 284 to 285 labelled a_1(1)* 284 to 275 labelled b_1(1)* 285 to 263 labelled a_1(1)* 285 to 275 labelled b_1(1) ---------------------------------------- (6) YES popout

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