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SRS Standard pair #516976007
details
property
value
status
complete
benchmark
dup11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n177.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
0.922112941742 seconds
cpu usage
2.097738481
max memory
6.0166144E8
stage attributes
key
value
output-size
4117
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES After renaming modulo the bijection { a ↦ 0, b ↦ 1, C ↦ 2, c ↦ 3, A ↦ 4, B ↦ 5 }, it remains to prove termination of the 24-rule system { 0 0 1 1 ⟶ 2 2 , 1 1 3 3 ⟶ 4 4 , 3 3 0 0 ⟶ 5 5 , 4 4 2 2 ⟶ 1 1 , 2 2 5 5 ⟶ 0 0 , 5 5 4 4 ⟶ 3 3 , 0 0 0 0 0 0 0 0 0 0 ⟶ 4 4 4 4 4 4 , 4 4 4 4 4 4 4 4 ⟶ 0 0 0 0 0 0 0 0 , 1 1 1 1 1 1 1 1 1 1 ⟶ 5 5 5 5 5 5 , 5 5 5 5 5 5 5 5 ⟶ 1 1 1 1 1 1 1 1 , 3 3 3 3 3 3 3 3 3 3 ⟶ 2 2 2 2 2 2 , 2 2 2 2 2 2 2 2 ⟶ 3 3 3 3 3 3 3 3 , 5 5 0 0 0 0 0 0 0 0 ⟶ 3 3 4 4 4 4 4 4 , 4 4 4 4 4 4 1 1 ⟶ 0 0 0 0 0 0 0 0 2 2 , 2 2 1 1 1 1 1 1 1 1 ⟶ 0 0 5 5 5 5 5 5 , 5 5 5 5 5 5 3 3 ⟶ 1 1 1 1 1 1 1 1 4 4 , 4 4 3 3 3 3 3 3 3 3 ⟶ 1 1 2 2 2 2 2 2 , 2 2 2 2 2 2 0 0 ⟶ 3 3 3 3 3 3 3 3 5 5 , 0 0 4 4 ⟶ , 4 4 0 0 ⟶ , 1 1 5 5 ⟶ , 5 5 1 1 ⟶ , 3 3 2 2 ⟶ , 2 2 3 3 ⟶ } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 5 ↦ 0, 0 ↦ 1, 3 ↦ 2, 4 ↦ 3, 1 ↦ 4, 2 ↦ 5 }, it remains to prove termination of the 6-rule system { 0 0 1 1 1 1 1 1 1 1 ⟶ 2 2 3 3 3 3 3 3 , 3 3 3 3 3 3 4 4 ⟶ 1 1 1 1 1 1 1 1 5 5 , 5 5 4 4 4 4 4 4 4 4 ⟶ 1 1 0 0 0 0 0 0 , 0 0 0 0 0 0 2 2 ⟶ 4 4 4 4 4 4 4 4 3 3 , 3 3 2 2 2 2 2 2 2 2 ⟶ 4 4 5 5 5 5 5 5 , 5 5 5 5 5 5 1 1 ⟶ 2 2 2 2 2 2 2 2 0 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (3,↑) ↦ 3, (3,↓) ↦ 4, (4,↓) ↦ 5, (5,↑) ↦ 6, (5,↓) ↦ 7, (2,↓) ↦ 8 }, it remains to prove termination of the 30-rule system { 0 1 2 2 2 2 2 2 2 2 ⟶ 3 4 4 4 4 4 , 0 1 2 2 2 2 2 2 2 2 ⟶ 3 4 4 4 4 , 0 1 2 2 2 2 2 2 2 2 ⟶ 3 4 4 4 , 0 1 2 2 2 2 2 2 2 2 ⟶ 3 4 4 , 0 1 2 2 2 2 2 2 2 2 ⟶ 3 4 , 0 1 2 2 2 2 2 2 2 2 ⟶ 3 , 3 4 4 4 4 4 5 5 ⟶ 6 7 , 3 4 4 4 4 4 5 5 ⟶ 6 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 1 1 1 1 1 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 1 1 1 1 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 1 1 1 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 1 1 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 1 , 6 7 5 5 5 5 5 5 5 5 ⟶ 0 , 0 1 1 1 1 1 8 8 ⟶ 3 4 , 0 1 1 1 1 1 8 8 ⟶ 3 , 3 4 8 8 8 8 8 8 8 8 ⟶ 6 7 7 7 7 7 , 3 4 8 8 8 8 8 8 8 8 ⟶ 6 7 7 7 7 , 3 4 8 8 8 8 8 8 8 8 ⟶ 6 7 7 7 ,
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