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TRS Standard pair #516967925
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
n079.star.cs.uiowa.edu
space
TCT_12
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.93862199783 seconds
cpu usage
6.98975997
max memory
5.01645312E8
stage attributes
key
value
output-size
3242
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 865 ms] (2) QTRS (3) RisEmptyProof [EQUIVALENT, 0 ms] (4) YES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
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