details

property	value
status	complete
benchmark	beans6.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_06

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	5.78951096535 seconds
cpu usage	21.061526678
max memory	3.809456128E9

stage attributes

key	value
output-size	240446
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 { b ↦ 0, a ↦ 1, c ↦ 2, d ↦ 3 }, it remains to prove termination of the 7-rule system { 0 1 1 ⟶ 1 0 2 , 2 1 ⟶ 1 2 , 2 0 ⟶ 0 1 , 1 1 0 ⟶ 3 0 1 , 1 3 ⟶ 3 1 , 0 3 ⟶ 1 0 , 1 1 ⟶ 1 0 1 } The system was reversed. After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 7-rule system { 0 0 1 ⟶ 2 1 0 , 0 2 ⟶ 2 0 , 1 2 ⟶ 0 1 , 1 0 0 ⟶ 0 1 3 , 3 0 ⟶ 0 3 , 3 1 ⟶ 1 0 , 0 0 ⟶ 0 1 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (1,↑) ↦ 3, (2,↓) ↦ 4, (3,↓) ↦ 5, (3,↑) ↦ 6 }, it remains to prove termination of the 21-rule system { 0 1 2 ⟶ 3 1 , 0 1 2 ⟶ 0 , 0 4 ⟶ 0 , 3 4 ⟶ 0 2 , 3 4 ⟶ 3 , 3 1 1 ⟶ 0 2 5 , 3 1 1 ⟶ 3 5 , 3 1 1 ⟶ 6 , 6 1 ⟶ 0 5 , 6 1 ⟶ 6 , 6 2 ⟶ 3 1 , 6 2 ⟶ 0 , 0 1 ⟶ 0 2 1 , 0 1 ⟶ 3 1 , 1 1 2 →= 4 2 1 , 1 4 →= 4 1 , 2 4 →= 1 2 , 2 1 1 →= 1 2 5 , 5 1 →= 1 5 , 5 2 →= 2 1 , 1 1 →= 1 2 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 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 4 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 8-rule system { 0 1 ⟶ 0 2 1 , 1 1 2 →= 3 2 1 , 1 3 →= 3 1 , 2 3 →= 1 2 , popout

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