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SRS Standard pair #487090672
details
property
value
status
complete
benchmark
abbaaba-abaabbaab.srs.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
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.79478 seconds
cpu usage
4.40579
user time
4.19391
system time
0.211881
max virtual memory
1.9009016E7
max residence set size
405100.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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, 5 ms] (6) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(b(a(a(b(a(x1))))))) -> a(b(a(a(b(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(a(b(b(a(x1))))))) -> b(a(a(b(b(a(a(b(a(x1))))))))) Q is empty. ---------------------------------------- (3) Strip Symbols Proof (SOUND) We were given the following TRS: a(b(a(a(b(b(a(x1))))))) -> b(a(a(b(b(a(a(b(a(x1))))))))) By stripping symbols from the only rule of the system, we obtained the following TRS [ENDRULLIS]: a(b(a(a(b(b(x)))))) -> b(a(a(b(b(a(a(b(x)))))))) ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(b(a(a(b(b(x)))))) -> b(a(a(b(b(a(a(b(x)))))))) Q is empty. ---------------------------------------- (5) RFCMatchBoundsTRSProof (EQUIVALENT) Termination of the TRS R could be shown with a Match Bound [MATCHBOUNDS1,MATCHBOUNDS2] of 2. This implies Q-termination of R. The following rules were used to construct the certificate: a(b(a(a(b(b(x)))))) -> b(a(a(b(b(a(a(b(x)))))))) The certificate found is represented by the following graph. The certificate consists of the following enumerated nodes: 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422 Node 393 is start node and node 394 is final node. Those nodes are connected through the following edges: * 393 to 395 labelled b_1(0)* 394 to 394 labelled #_1(0)* 395 to 396 labelled a_1(0)* 396 to 397 labelled a_1(0)* 397 to 398 labelled b_1(0)* 398 to 399 labelled b_1(0)* 399 to 400 labelled a_1(0)* 399 to 409 labelled b_1(1)* 400 to 401 labelled a_1(0)* 400 to 402 labelled b_1(1)* 401 to 394 labelled b_1(0)* 402 to 403 labelled a_1(1)* 403 to 404 labelled a_1(1)* 404 to 405 labelled b_1(1)* 405 to 406 labelled b_1(1)* 406 to 407 labelled a_1(1)* 406 to 416 labelled b_1(2)* 407 to 408 labelled a_1(1)* 407 to 402 labelled b_1(1)* 408 to 394 labelled b_1(1)* 409 to 410 labelled a_1(1)* 410 to 411 labelled a_1(1)* 411 to 412 labelled b_1(1)* 412 to 413 labelled b_1(1)* 413 to 414 labelled a_1(1)* 413 to 416 labelled b_1(2)* 414 to 415 labelled a_1(1)* 414 to 402 labelled b_1(1)* 415 to 406 labelled b_1(1)* 416 to 417 labelled a_1(2)* 417 to 418 labelled a_1(2)* 418 to 419 labelled b_1(2)* 419 to 420 labelled b_1(2)* 420 to 421 labelled a_1(2)* 420 to 416 labelled b_1(2)* 421 to 422 labelled a_1(2)* 421 to 402 labelled b_1(1)* 422 to 406 labelled b_1(2) ---------------------------------------- (6) YES
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