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SRS Stand 10685 pair #381721539
details
property
value
status
complete
benchmark
size-12-alpha-3-num-179.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.57072997093 seconds
cpu usage
3.458566129
max memory
2.24587776E8
stage attributes
key
value
output-size
1739
starexec-result
NO
output
/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination w.r.t. Q of the given QTRS could be disproven: (0) QTRS (1) NonTerminationProof [COMPLETE, 0 ms] (2) NO ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> x1 a(b(x1)) -> b(a(c(b(a(a(x1)))))) c(b(c(x1))) -> 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: "a b b c b -> b c b a b b c b a" a b b c b -> b c b a b b c b a by OverlapClosure OC 3"a b b c b -> b c b a b b a c b a by OverlapClosure OC 2"a b b c -> b c b a b a by OverlapClosure OC 2"a b b -> b c b a b a c b by OverlapClosure OC 2"a b b -> b c b a b a c b a by OverlapClosure OC 2"a b -> b c b a a by OverlapClosure OC 3"a b -> b a c b a a by original rule (OC 1)""a -> by original rule (OC 1)"""a b -> b a c b a by OverlapClosure OC 2"a b -> b a c b a a by original rule (OC 1)""a -> by original rule (OC 1)""""a -> by original rule (OC 1)"""c b c -> by original rule (OC 1)"""a b -> b a c b a by OverlapClosure OC 2"a b -> b a c b a a by original rule (OC 1)""a -> by original rule (OC 1)""""a -> by original rule (OC 1)" ---------------------------------------- (2) NO
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