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SRS Standard pair #516969100
details
property
value
status
complete
benchmark
uni-4.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
Waldmann_06_SRS
run statistics
property
value
solver
AProVE21
configuration
standard
runtime (wallclock)
2.07540798187 seconds
cpu usage
5.494232089
max memory
4.6936064E8
stage attributes
key
value
output-size
1800
starexec-result
NO
output
/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) NonTerminationProof [COMPLETE, 34 ms] (2) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(a(a(x1)))) -> b(a(a(a(x1)))) b(b(x1)) -> a(a(b(x1))) Q is empty. ---------------------------------------- (1) NonTerminationProof (COMPLETE) We used the non-termination processor [OPPELT08] to show that the SRS problem is infinite. Found the self-embedding DerivationStructure: "b b b b b -> b b b b b a a a b" b b b b b -> b b b b b a a a b by OverlapClosure OC 3"b b b b b -> b b b b a a a a b by OverlapClosure OC 3"b b b b b -> b b b a a a a a b by OverlapClosure OC 3"b b b b b -> b b a a a a a a b by OverlapClosure OC 3"b b b b b -> b a a a a a a a b by OverlapClosure OC 2"b b b -> b a a a b by OverlapClosure OC 3"b b b -> a a a a b by OverlapClosure OC 2"b b -> a a b by original rule (OC 1)""b b -> a a b by original rule (OC 1)"""a a a a -> b a a a by original rule (OC 1)"""b b b -> a a a a b by OverlapClosure OC 2"b b -> a a b by original rule (OC 1)""b b -> a a b by original rule (OC 1)""""a a a a -> b a a a by original rule (OC 1)"""a a a a -> b a a a by original rule (OC 1)"""a a a a -> b a a a by original rule (OC 1)"""a a a a -> b a a a by original rule (OC 1)" ---------------------------------------- (2) NO
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