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SRS Standard pair #487518935
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
n132.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.73667407036 seconds
cpu usage
4.084965917
max memory
3.0302208E8
stage attributes
key
value
output-size
3484
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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(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: 310, 311, 320, 321, 322, 323, 324, 325, 326, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355 Node 310 is start node and node 311 is final node. Those nodes are connected through the following edges: * 310 to 320 labelled b_1(0)* 311 to 311 labelled #_1(0)* 320 to 321 labelled a_1(0)* 321 to 322 labelled a_1(0)* 322 to 323 labelled b_1(0)* 323 to 324 labelled b_1(0)* 324 to 325 labelled a_1(0)* 324 to 342 labelled b_1(1)* 325 to 326 labelled a_1(0)* 325 to 335 labelled b_1(1)* 326 to 311 labelled b_1(0)* 335 to 336 labelled a_1(1)* 336 to 337 labelled a_1(1)* 337 to 338 labelled b_1(1)* 338 to 339 labelled b_1(1)* 339 to 340 labelled a_1(1)* 339 to 349 labelled b_1(2)* 340 to 341 labelled a_1(1)* 340 to 335 labelled b_1(1)* 341 to 311 labelled b_1(1)* 342 to 343 labelled a_1(1)* 343 to 344 labelled a_1(1)* 344 to 345 labelled b_1(1)* 345 to 346 labelled b_1(1)* 346 to 347 labelled a_1(1)* 346 to 349 labelled b_1(2)* 347 to 348 labelled a_1(1)* 347 to 335 labelled b_1(1)* 348 to 339 labelled b_1(1)* 349 to 350 labelled a_1(2)* 350 to 351 labelled a_1(2)* 351 to 352 labelled b_1(2)* 352 to 353 labelled b_1(2)* 353 to 354 labelled a_1(2)* 353 to 349 labelled b_1(2)* 354 to 355 labelled a_1(2)* 354 to 335 labelled b_1(1)* 355 to 339 labelled b_1(2) ---------------------------------------- (6) YES
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