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SRS Standard pair #516975557
details
property
value
status
complete
benchmark
secr10.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.84429883957 seconds
cpu usage
5.596380878
max memory
1.356111872E9
stage attributes
key
value
output-size
7347
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, c ↦ 1, b ↦ 2 }, it remains to prove termination of the 6-rule system { 0 0 ⟶ 1 2 0 2 0 , 2 0 2 ⟶ 2 , 0 0 0 ⟶ 1 1 0 , 1 1 ⟶ 0 2 1 2 0 , 0 1 0 ⟶ 1 1 0 , 1 0 1 ⟶ 0 0 1 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 2 ↦ 1, 1 ↦ 2 }, it remains to prove termination of the 6-rule system { 0 0 ⟶ 0 1 0 1 2 , 1 0 1 ⟶ 1 , 0 0 0 ⟶ 0 2 2 , 2 2 ⟶ 0 1 2 1 0 , 0 2 0 ⟶ 0 2 2 , 2 0 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, (2,↓) ↦ 3, (1,↑) ↦ 4, (2,↑) ↦ 5 }, it remains to prove termination of the 25-rule system { 0 1 ⟶ 0 2 1 2 3 , 0 1 ⟶ 4 1 2 3 , 0 1 ⟶ 0 2 3 , 0 1 ⟶ 4 3 , 0 1 ⟶ 5 , 0 1 1 ⟶ 0 3 3 , 0 1 1 ⟶ 5 3 , 0 1 1 ⟶ 5 , 5 3 ⟶ 0 2 3 2 1 , 5 3 ⟶ 4 3 2 1 , 5 3 ⟶ 5 2 1 , 5 3 ⟶ 4 1 , 5 3 ⟶ 0 , 0 3 1 ⟶ 0 3 3 , 0 3 1 ⟶ 5 3 , 0 3 1 ⟶ 5 , 5 1 3 ⟶ 5 1 1 , 5 1 3 ⟶ 0 1 , 5 1 3 ⟶ 0 , 1 1 →= 1 2 1 2 3 , 2 1 2 →= 2 , 1 1 1 →= 1 3 3 , 3 3 →= 1 2 3 2 1 , 1 3 1 →= 1 3 3 , 3 1 3 →= 3 1 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 21-rule system { 0 1 ⟶ 0 2 1 2 3 , 0 1 ⟶ 0 2 3 , 0 1 ⟶ 4 , 0 1 1 ⟶ 0 3 3 , 0 1 1 ⟶ 4 3 , 0 1 1 ⟶ 4 , 4 3 ⟶ 0 2 3 2 1 ,
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