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SRS Standard pair #516976199
details
property
value
status
complete
benchmark
random-529.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
Waldmann_19
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.11793804169 seconds
cpu usage
2.697835516
max memory
7.0682624E8
stage attributes
key
value
output-size
1754
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { a ↦ 0, b ↦ 1 }, it remains to prove termination of the 3-rule system { 0 1 1 0 ⟶ 0 0 0 0 , 1 0 0 0 ⟶ 0 0 1 1 , 1 1 0 0 ⟶ 1 0 1 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (0,↓) ↦ 2, (1,↑) ↦ 3 }, it remains to prove termination of the 13-rule system { 0 1 1 2 ⟶ 0 2 2 2 , 0 1 1 2 ⟶ 0 2 2 , 0 1 1 2 ⟶ 0 2 , 3 2 2 2 ⟶ 0 2 1 1 , 3 2 2 2 ⟶ 0 1 1 , 3 2 2 2 ⟶ 3 1 , 3 2 2 2 ⟶ 3 , 3 1 2 2 ⟶ 3 2 1 2 , 3 1 2 2 ⟶ 0 1 2 , 3 1 2 2 ⟶ 3 2 , 2 1 1 2 →= 2 2 2 2 , 1 2 2 2 →= 2 2 1 1 , 1 1 2 2 →= 1 2 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 5-rule system { 0 1 1 2 ⟶ 0 2 2 2 , 3 1 2 2 ⟶ 3 2 1 2 , 2 1 1 2 →= 2 2 2 2 , 1 2 2 2 →= 2 2 1 1 , 1 1 2 2 →= 1 2 1 2 } Applying sparse untiling TROCU(2) after reversal [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 3-rule system { 0 1 1 0 →= 0 0 0 0 , 1 0 0 0 →= 0 0 1 1 , 1 1 0 0 →= 1 0 1 0 } The system is trivially terminating.
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