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SRS Relat 19260 pair #381725301
details
property
value
status
complete
benchmark
zr13.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n112.star.cs.uiowa.edu
space
Mixed_relative_SRS
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.68654799461 seconds
cpu usage
3.851485485
max memory
2.01060352E8
stage attributes
key
value
output-size
1997
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given RelTRS could be disproven: (0) RelTRS (1) RelTRS Reverse [EQUIVALENT, 0 ms] (2) RelTRS (3) RelTRSLoopFinderProof [COMPLETE, 3 ms] (4) NO ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: a(b(c(x1))) -> c(x1) The relative TRS consists of the following S rules: c(x1) -> c(b(a(x1))) ---------------------------------------- (1) RelTRS Reverse (EQUIVALENT) We have reversed the following relative TRS [REVERSE]: The set of rules R is a(b(c(x1))) -> c(x1) The set of rules S is c(x1) -> c(b(a(x1))) We have obtained the following relative TRS: The set of rules R is c(b(a(x1))) -> c(x1) The set of rules S is c(x1) -> a(b(c(x1))) ---------------------------------------- (2) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: c(b(a(x1))) -> c(x1) The relative TRS consists of the following S rules: c(x1) -> a(b(c(x1))) ---------------------------------------- (3) RelTRSLoopFinderProof (COMPLETE) The following loop was found: ---------- Loop: ---------- c(b(c(x1'))) -> c(b(a(b(c(x1'))))) with rule c(x1'') -> a(b(c(x1''))) at position [0,0] and matcher [x1'' / x1'] c(b(a(b(c(x1'))))) -> c(b(c(x1'))) with rule c(b(a(x1))) -> c(x1) at position [] and matcher [x1 / b(c(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. ---------------------------------------- (4) NO
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