details

property	value
status	complete
benchmark	un10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n088.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	181.475432873 seconds
cpu usage	612.634967236
max memory	1.5891398656E10

stage attributes

key	value
output-size	6639
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 27 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 8782 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 8414 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 7607 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(b(b(a(b(b(b(b(x1)))))))))) -> b(b(b(b(b(a(b(b(a(b(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: b(b(b(b(a(b(b(a(b(b(x1)))))))))) -> b(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(a(b(b(a(b(b(b(b(b(x1))))))))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(a(b(b(a(b(b(b(b(b(x1)))))))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(a(b(b(b(b(b(x1)))))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(a(b(b(b(b(b(x1))))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(b(b(x1))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(b(x1)))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(x1))) The TRS R consists of the following rules: b(b(b(b(a(b(b(a(b(b(x1)))))))))) -> b(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(b(x1)))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(b(b(x1))))) B(b(b(b(a(b(b(a(b(b(x1)))))))))) -> B(b(b(x1))) The TRS R consists of the following rules: b(b(b(b(a(b(b(a(b(b(x1)))))))))) -> b(a(b(b(a(b(b(a(b(b(b(b(b(x1))))))))))))) popout

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job log

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actions

all output return to SRS Standard