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SRS Relative pair #487521496
details
property
value
status
complete
benchmark
random-141.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n110.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
128.70627594 seconds
cpu usage
510.544290498
max memory
1.0919940096E10
stage attributes
key
value
output-size
2425
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be disproven: (0) RelTRS (1) RelTRSLoopFinderProof [COMPLETE, 30.5 s] (2) NO ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: a(c(c(x1))) -> c(c(a(x1))) a(b(b(x1))) -> b(b(a(x1))) a(a(c(x1))) -> b(b(a(x1))) b(c(a(x1))) -> a(c(c(x1))) The relative TRS consists of the following S rules: a(c(a(x1))) -> a(c(b(x1))) b(a(a(x1))) -> a(a(a(x1))) ---------------------------------------- (1) RelTRSLoopFinderProof (COMPLETE) The following loop was found: ---------- Loop: ---------- b(b(a(a(c(a(c(a(x1)))))))) -> b(b(a(a(c(b(c(a(x1)))))))) with rule a(c(a(x1'))) -> a(c(b(x1'))) at position [0,0,0] and matcher [x1' / c(a(x1))] b(b(a(a(c(b(c(a(x1)))))))) -> b(b(a(a(c(a(c(c(x1)))))))) with rule b(c(a(x1'))) -> a(c(c(x1'))) at position [0,0,0,0,0] and matcher [x1' / x1] b(b(a(a(c(a(c(c(x1)))))))) -> b(b(a(a(c(c(c(a(x1)))))))) with rule a(c(c(x1'))) -> c(c(a(x1'))) at position [0,0,0,0,0] and matcher [x1' / x1] b(b(a(a(c(c(c(a(x1)))))))) -> b(a(a(a(c(c(c(a(x1)))))))) with rule b(a(a(x1'))) -> a(a(a(x1'))) at position [0] and matcher [x1' / c(c(c(a(x1))))] b(a(a(a(c(c(c(a(x1)))))))) -> b(a(a(c(c(a(c(a(x1)))))))) with rule a(c(c(x1'))) -> c(c(a(x1'))) at position [0,0,0] and matcher [x1' / c(a(x1))] b(a(a(c(c(a(c(a(x1)))))))) -> a(a(a(c(c(a(c(a(x1)))))))) with rule b(a(a(x1'))) -> a(a(a(x1'))) at position [] and matcher [x1' / c(c(a(c(a(x1)))))] a(a(a(c(c(a(c(a(x1)))))))) -> a(b(b(a(c(a(c(a(x1)))))))) with rule a(a(c(x1'))) -> b(b(a(x1'))) at position [0] and matcher [x1' / c(a(c(a(x1))))] a(b(b(a(c(a(c(a(x1)))))))) -> b(b(a(a(c(a(c(a(x1)))))))) with rule a(b(b(x1'))) -> b(b(a(x1'))) at position [] and matcher [x1' / a(c(a(c(a(x1)))))] Now an instance of the first term with Matcher [ ] occurs in the last term at position []. Context: [] Therefore, the relative TRS problem does not terminate. ---------------------------------------- (2) NO
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