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SRS Standard pair #516978059
details
property
value
status
complete
benchmark
aabaaaa-aaaaaabaab.srs.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
Wenzel_16
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
19.5969450474 seconds
cpu usage
68.922692198
max memory
5.589090304E9
stage attributes
key
value
output-size
9170
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 }, it remains to prove termination of the 1-rule system { 0 0 1 0 0 0 0 ⟶ 0 0 0 0 0 0 1 0 0 1 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 1-rule system { 0 0 0 0 1 0 0 ⟶ 1 0 0 1 0 0 0 0 0 0 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,0) ↦ 2, (0,3) ↦ 3, (1,1) ↦ 4 }, it remains to prove termination of the 6-rule system { 0 0 0 0 1 2 0 0 ⟶ 1 2 0 1 2 0 0 0 0 0 0 , 0 0 0 0 1 2 0 1 ⟶ 1 2 0 1 2 0 0 0 0 0 1 , 0 0 0 0 1 2 0 3 ⟶ 1 2 0 1 2 0 0 0 0 0 3 , 2 0 0 0 1 2 0 0 ⟶ 4 2 0 1 2 0 0 0 0 0 0 , 2 0 0 0 1 2 0 1 ⟶ 4 2 0 1 2 0 0 0 0 0 1 , 2 0 0 0 1 2 0 3 ⟶ 4 2 0 1 2 0 0 0 0 0 3 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 9: 0 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 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 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 1 ⎟ ⎜ 0 1 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 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 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 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 1 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 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 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 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 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 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 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 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, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 4 ↦ 4 }, it remains to prove termination of the 5-rule system { 0 0 0 0 1 2 0 0 ⟶ 1 2 0 1 2 0 0 0 0 0 0 , 0 0 0 0 1 2 0 1 ⟶ 1 2 0 1 2 0 0 0 0 0 1 , 0 0 0 0 1 2 0 3 ⟶ 1 2 0 1 2 0 0 0 0 0 3 , 2 0 0 0 1 2 0 1 ⟶ 4 2 0 1 2 0 0 0 0 0 1 , 2 0 0 0 1 2 0 3 ⟶ 4 2 0 1 2 0 0 0 0 0 3 }
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