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	n190.star.cs.uiowa.edu
space	TCT_12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.50777 seconds
cpu usage	5.10751
user time	4.90857
system time	0.198942
max virtual memory	1.8368932E7
max residence set size	267224.0

stage attributes

key	value
starexec-result	YES

output

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 846 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4)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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