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SRS Standard pair #516968459
details
property
value
status
complete
benchmark
z066.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n008.star.cs.uiowa.edu
space
Zantema_04
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
0.734188079834 seconds
cpu usage
1.458478101
max memory
4.84278272E8
stage attributes
key
value
output-size
8228
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, c ↦ 2, C ↦ 3, B ↦ 4, A ↦ 5 }, it remains to prove termination of the 12-rule system { 0 1 2 ⟶ 2 1 0 , 3 4 5 ⟶ 5 4 3 , 1 0 3 ⟶ 3 0 1 , 2 5 4 ⟶ 4 5 2 , 5 2 1 ⟶ 1 2 5 , 4 3 0 ⟶ 0 3 4 , 0 5 ⟶ , 5 0 ⟶ , 1 4 ⟶ , 4 1 ⟶ , 2 3 ⟶ , 3 2 ⟶ } The system was reversed. After renaming modulo the bijection { 2 ↦ 0, 1 ↦ 1, 0 ↦ 2, 5 ↦ 3, 4 ↦ 4, 3 ↦ 5 }, it remains to prove termination of the 12-rule system { 0 1 2 ⟶ 2 1 0 , 3 4 5 ⟶ 5 4 3 , 5 2 1 ⟶ 1 2 5 , 4 3 0 ⟶ 0 3 4 , 1 0 3 ⟶ 3 0 1 , 2 5 4 ⟶ 4 5 2 , 3 2 ⟶ , 2 3 ⟶ , 4 1 ⟶ , 1 4 ⟶ , 5 0 ⟶ , 0 5 ⟶ } 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4, 5 ↦ 5 }, it remains to prove termination of the 6-rule system { 0 1 2 ⟶ 2 1 0 , 3 4 5 ⟶ 5 4 3 , 5 2 1 ⟶ 1 2 5 , 4 3 0 ⟶ 0 3 4 , 1 0 3 ⟶ 3 0 1 , 2 5 4 ⟶ 4 5 2 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (2,↓) ↦ 2, (2,↑) ↦ 3, (0,↓) ↦ 4, (1,↑) ↦ 5, (3,↑) ↦ 6, (4,↓) ↦ 7, (5,↓) ↦ 8, (5,↑) ↦ 9, (3,↓) ↦ 10, (4,↑) ↦ 11 }, it remains to prove termination of the 24-rule system { 0 1 2 ⟶ 3 1 4 , 0 1 2 ⟶ 5 4 , 0 1 2 ⟶ 0 , 6 7 8 ⟶ 9 7 10 , 6 7 8 ⟶ 11 10 , 6 7 8 ⟶ 6 , 9 2 1 ⟶ 5 2 8 , 9 2 1 ⟶ 3 8 , 9 2 1 ⟶ 9 , 11 10 4 ⟶ 0 10 7 , 11 10 4 ⟶ 6 7 , 11 10 4 ⟶ 11 , 5 4 10 ⟶ 6 4 1 , 5 4 10 ⟶ 0 1 ,
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