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SRS Standard pair #516973127
details
property
value
status
complete
benchmark
size-11-alpha-3-num-20.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n059.star.cs.uiowa.edu
space
Waldmann_07_size11
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.4630510807 seconds
cpu usage
4.018851162
max memory
7.36116736E8
stage attributes
key
value
output-size
6130
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 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 0 1 ⟶ 1 1 0 2 , 1 ⟶ , 2 2 ⟶ 0 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 1 0 ⟶ 2 0 1 1 , 1 ⟶ , 2 2 ⟶ 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (1,↑) ↦ 0, (0,↓) ↦ 1, (2,↑) ↦ 2, (1,↓) ↦ 3, (0,↑) ↦ 4, (2,↓) ↦ 5 }, it remains to prove termination of the 9-rule system { 0 1 ⟶ 2 1 3 3 , 0 1 ⟶ 4 3 3 , 0 1 ⟶ 0 3 , 0 1 ⟶ 0 , 2 5 ⟶ 4 , 1 →= , 3 1 →= 5 1 3 3 , 3 →= , 5 5 →= 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 3 ↦ 2, 2 ↦ 3, 5 ↦ 4, 4 ↦ 5 }, it remains to prove termination of the 7-rule system { 0 1 ⟶ 0 2 , 0 1 ⟶ 0 , 3 4 ⟶ 5 , 1 →= , 2 1 →= 4 1 2 2 , 2 →= , 4 4 →= 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞
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