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SRS Standard pair #516976061
details
property
value
status
complete
benchmark
hom01.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n072.star.cs.uiowa.edu
space
Trafo_06
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.18046212196 seconds
cpu usage
2.998778601
max memory
6.857728E8
stage attributes
key
value
output-size
2755
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, d ↦ 2, c ↦ 3 }, it remains to prove termination of the 3-rule system { 0 0 1 2 1 2 0 ⟶ 0 0 3 0 0 1 2 , 0 0 3 ⟶ 3 3 0 0 , 3 3 3 ⟶ 1 2 3 1 2 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 8: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 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 ⎟ ⎜ 0 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 1 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 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 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 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 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 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 1 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 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 ⎟ ⎜ 0 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 3 ↦ 1, 1 ↦ 2, 2 ↦ 3 }, it remains to prove termination of the 2-rule system { 0 0 1 ⟶ 1 1 0 0 , 1 1 1 ⟶ 2 3 1 2 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 4: 0 ↦ ⎛ ⎞ ⎜ 2 0 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 1 1 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 1 ⎟ ⎜ 0 1 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 ⎟ ⎜ 0 1 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎜ 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 1-rule system { 0 0 1 ⟶ 1 1 0 0 }
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