details

property	value
status	complete
benchmark	abbaaaaa-aaaaaabbabb.srs.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n003.star.cs.uiowa.edu
space	Wenzel_16

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	3.6450510025 seconds
cpu usage	12.701736365
max memory	3.077275648E9

stage attributes

key	value
output-size	11054
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 }, it remains to prove termination of the 1-rule system { 0 1 1 0 0 0 0 0 ⟶ 0 0 0 0 0 0 1 1 0 1 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 0 1 1 0 ⟶ 1 1 0 1 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, (1,1) ↦ 2, (1,0) ↦ 3, (0,3) ↦ 4 }, it remains to prove termination of the 6-rule system { 0 0 0 0 0 1 2 3 0 ⟶ 1 2 3 1 2 3 0 0 0 0 0 0 , 0 0 0 0 0 1 2 3 1 ⟶ 1 2 3 1 2 3 0 0 0 0 0 1 , 0 0 0 0 0 1 2 3 4 ⟶ 1 2 3 1 2 3 0 0 0 0 0 4 , 3 0 0 0 0 1 2 3 0 ⟶ 2 2 3 1 2 3 0 0 0 0 0 0 , 3 0 0 0 0 1 2 3 1 ⟶ 2 2 3 1 2 3 0 0 0 0 0 1 , 3 0 0 0 0 1 2 3 4 ⟶ 2 2 3 1 2 3 0 0 0 0 0 4 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 10: 0 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 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 0 1 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 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 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 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 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 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 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 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 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 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 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 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 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 0 1 2 3 0 ⟶ 1 2 3 1 2 3 0 0 0 0 0 0 , popout

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