details

property	value
status	complete
benchmark	random-540.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.84105300903 seconds
cpu usage	11.186964849
max memory	1.767841792E9

stage attributes

key	value
output-size	6840
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 30 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 9 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 5 ms] (6) AND (7) QDP (8) UsableRulesProof [EQUIVALENT, 1 ms] (9) QDP (10) DependencyGraphProof [EQUIVALENT, 0 ms] (11) QDP (12) UsableRulesProof [EQUIVALENT, 1 ms] (13) QDP (14) MRRProof [EQUIVALENT, 0 ms] (15) QDP (16) PisEmptyProof [EQUIVALENT, 0 ms] (17) YES (18) QDP (19) UsableRulesProof [EQUIVALENT, 1 ms] (20) QDP (21) QDPSizeChangeProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(b(b(x1)))) -> a(b(a(b(x1)))) b(b(a(a(x1)))) -> a(b(a(b(x1)))) b(a(a(b(x1)))) -> b(a(b(a(x1)))) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = x_1 POL(b(x_1)) = 1 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: b(a(b(b(x1)))) -> a(b(a(b(x1)))) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(a(b(x1)))) b(a(a(b(x1)))) -> b(a(b(a(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(a(a(x1)))) -> B(a(b(x1))) B(b(a(a(x1)))) -> B(x1) B(a(a(b(x1)))) -> B(a(b(a(x1)))) B(a(a(b(x1)))) -> B(a(x1)) The TRS R consists of the following rules: b(b(a(a(x1)))) -> a(b(a(b(x1)))) b(a(a(b(x1)))) -> b(a(b(a(x1)))) Q is empty. popout

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