details

property	value
status	complete
benchmark	z081.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n052.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.35956192017 seconds
cpu usage	14.108835981
max memory	2.287763456E9

stage attributes

key	value
output-size	3996
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 29 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 11 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 9 ms] (6) QDP (7) MRRProof [EQUIVALENT, 20 ms] (8) QDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(c(a(x1))) -> a(b(x1)) b(b(b(x1))) -> c(a(c(x1))) c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) d(c(x1)) -> b(b(b(x1))) c(b(x1)) -> d(a(x1)) d(b(c(x1))) -> a(a(x1)) d(a(x1)) -> b(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 POL(c(x_1)) = 1 + x_1 POL(d(x_1)) = 2 + x_1 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: b(c(a(x1))) -> a(b(x1)) b(b(b(x1))) -> c(a(c(x1))) d(b(c(x1))) -> a(a(x1)) d(a(x1)) -> b(x1) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) d(c(x1)) -> b(b(b(x1))) c(b(x1)) -> d(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: C(d(x1)) -> D(c(x1)) C(d(x1)) -> C(x1) C(d(b(x1))) -> D(c(c(x1))) C(d(b(x1))) -> C(c(x1)) C(d(b(x1))) -> C(x1) C(b(x1)) -> D(a(x1)) The TRS R consists of the following rules: c(d(x1)) -> d(c(x1)) c(d(b(x1))) -> d(c(c(x1))) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to SRS Stand 10685