details

property	value
status	complete
benchmark	polycounter-10.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n016.star.cs.uiowa.edu
space	TCT_12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.50851893425 seconds
cpu usage	5.05968071
max memory	2.81870336E8

stage attributes

key	value
output-size	3245
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 842 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: f(s(x1), x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f(0, s(x2), x3, x4, x5, x6, x7, x8, x9, x10) -> f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10) f(0, 0, s(x3), x4, x5, x6, x7, x8, x9, x10) -> f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10) f(0, 0, 0, s(x4), x5, x6, x7, x8, x9, x10) -> f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10) f(0, 0, 0, 0, s(x5), x6, x7, x8, x9, x10) -> f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10) f(0, 0, 0, 0, 0, s(x6), x7, x8, x9, x10) -> f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10) f(0, 0, 0, 0, 0, 0, s(x7), x8, x9, x10) -> f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10) f(0, 0, 0, 0, 0, 0, 0, s(x8), x9, x10) -> f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10) f(0, 0, 0, 0, 0, 0, 0, 0, s(x9), x10) -> f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10) f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(x10)) -> f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10) f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) -> 0 Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Quasi precedence: s_1 > [f_10, 0] Status: f_10: [9,10,8,7,6,5,4,3,2,1] s_1: multiset status 0: multiset status With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: f(s(x1), x2, x3, x4, x5, x6, x7, x8, x9, x10) -> f(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10) f(0, s(x2), x3, x4, x5, x6, x7, x8, x9, x10) -> f(x2, x2, x3, x4, x5, x6, x7, x8, x9, x10) f(0, 0, s(x3), x4, x5, x6, x7, x8, x9, x10) -> f(x3, x3, x3, x4, x5, x6, x7, x8, x9, x10) f(0, 0, 0, s(x4), x5, x6, x7, x8, x9, x10) -> f(x4, x4, x4, x4, x5, x6, x7, x8, x9, x10) f(0, 0, 0, 0, s(x5), x6, x7, x8, x9, x10) -> f(x5, x5, x5, x5, x5, x6, x7, x8, x9, x10) f(0, 0, 0, 0, 0, s(x6), x7, x8, x9, x10) -> f(x6, x6, x6, x6, x6, x6, x7, x8, x9, x10) f(0, 0, 0, 0, 0, 0, s(x7), x8, x9, x10) -> f(x7, x7, x7, x7, x7, x7, x7, x8, x9, x10) f(0, 0, 0, 0, 0, 0, 0, s(x8), x9, x10) -> f(x8, x8, x8, x8, x8, x8, x8, x8, x9, x10) f(0, 0, 0, 0, 0, 0, 0, 0, s(x9), x10) -> f(x9, x9, x9, x9, x9, x9, x9, x9, x9, x10) f(0, 0, 0, 0, 0, 0, 0, 0, 0, s(x10)) -> f(x10, x10, x10, x10, x10, x10, x10, x10, x10, x10) f(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) -> 0 ---------------------------------------- (2) Obligation: Q restricted rewrite system: R is empty. Q is empty. ---------------------------------------- (3) RisEmptyProof (EQUIVALENT) The TRS R is empty. Hence, termination is trivially proven. ---------------------------------------- (4) YES popout

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