details

property	value
status	complete
benchmark	z083.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n134.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.29762411118 seconds
cpu usage	17.544075214
max memory	1.874628608E9

stage attributes

key	value
output-size	6608
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 60 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 2 ms] (6) QDP (7) UsableRulesProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 11 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 90 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(d(b(x1))) -> c(d(b(x1))) b(a(c(x1))) -> b(c(x1)) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> b(b(x1)) Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(a(x_1)) = 1 + x_1 POL(b(x_1)) = 1 + x_1 POL(c(x_1)) = 1 + x_1 POL(d(x_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(c(x1))) -> b(c(x1)) ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(d(b(x1))) -> c(d(b(x1))) a(d(x1)) -> d(c(x1)) b(b(b(x1))) -> a(b(c(x1))) d(c(x1)) -> b(d(x1)) d(c(x1)) -> d(b(d(x1))) d(a(c(x1))) -> 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: A(d(x1)) -> D(c(x1)) B(b(b(x1))) -> A(b(c(x1))) B(b(b(x1))) -> B(c(x1)) D(c(x1)) -> B(d(x1)) D(c(x1)) -> D(x1) D(c(x1)) -> D(b(d(x1))) D(a(c(x1))) -> B(b(x1)) D(a(c(x1))) -> B(x1) popout

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job log

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actions

all output return to SRS Standard