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SRS Stand 10685 pair #381720953
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
n055.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.69680190086 seconds
cpu usage
4.170162487
max memory
3.26955008E8
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Strip Symbols Proof [SOUND, 0 ms] (4)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: 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370 Node 341 is start node and node 342 is final node. Those nodes are connected through the following edges: * 341 to 343 labelled b_1(0)* 342 to 342 labelled #_1(0)* 343 to 344 labelled a_1(0)* 344 to 345 labelled a_1(0)* 345 to 346 labelled b_1(0)* 346 to 347 labelled b_1(0)* 347 to 348 labelled a_1(0)* 347 to 357 labelled b_1(1)* 348 to 349 labelled a_1(0)* 348 to 350 labelled b_1(1)* 349 to 342 labelled b_1(0)* 350 to 351 labelled a_1(1)* 351 to 352 labelled a_1(1)* 352 to 353 labelled b_1(1)* 353 to 354 labelled b_1(1)* 354 to 355 labelled a_1(1)* 354 to 364 labelled b_1(2)* 355 to 356 labelled a_1(1)* 355 to 350 labelled b_1(1)* 356 to 342 labelled b_1(1)* 357 to 358 labelled a_1(1)* 358 to 359 labelled a_1(1)* 359 to 360 labelled b_1(1)* 360 to 361 labelled b_1(1)* 361 to 362 labelled a_1(1)* 361 to 364 labelled b_1(2)* 362 to 363 labelled a_1(1)* 362 to 350 labelled b_1(1)* 363 to 354 labelled b_1(1)* 364 to 365 labelled a_1(2)* 365 to 366 labelled a_1(2)* 366 to 367 labelled b_1(2)* 367 to 368 labelled b_1(2)* 368 to 369 labelled a_1(2)* 368 to 364 labelled b_1(2)* 369 to 370 labelled a_1(2)* 369 to 350 labelled b_1(1)* 370 to 354 labelled b_1(2) ---------------------------------------- (6) YES
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