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SRS Standard pair #516977759
details
property
value
status
complete
benchmark
abaaaa-aaaaababaabab.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n076.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
3.14011287689 seconds
cpu usage
10.735715726
max memory
2.38157824E9
stage attributes
key
value
output-size
1980
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 1-rule system { 0 1 0 0 0 0 ⟶ 0 0 0 0 0 1 0 1 0 0 1 0 1 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 1-rule system { 0 0 0 0 1 0 ⟶ 1 0 1 0 0 1 0 1 0 0 0 0 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2 }, it remains to prove termination of the 9-rule system { 0 1 1 1 2 1 ⟶ 0 2 1 1 2 1 2 1 1 1 1 1 , 0 1 1 1 2 1 ⟶ 0 1 2 1 2 1 1 1 1 1 , 0 1 1 1 2 1 ⟶ 0 2 1 2 1 1 1 1 1 , 0 1 1 1 2 1 ⟶ 0 2 1 1 1 1 1 , 0 1 1 1 2 1 ⟶ 0 1 1 1 1 , 0 1 1 1 2 1 ⟶ 0 1 1 1 , 0 1 1 1 2 1 ⟶ 0 1 1 , 0 1 1 1 2 1 ⟶ 0 1 , 1 1 1 1 2 1 →= 2 1 2 1 1 2 1 2 1 1 1 1 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 7: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 2 0 0 0 ⎟ ⎜ 0 0 0 0 1 0 0 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 0 2 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 1 1 1 1 1 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1 }, it remains to prove termination of the 1-rule system { 0 0 0 0 1 0 →= 1 0 1 0 0 1 0 1 0 0 0 0 0 } The system is trivially terminating.
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