details

property	value
status	complete
benchmark	random-169.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n136.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	1.19767379761 seconds
cpu usage	3.069233541
max memory	7.10049792E8

stage attributes

key	value
output-size	2213
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 { b ↦ 0, a ↦ 1 }, it remains to prove termination of the 3-rule system { 0 1 1 1 ⟶ 0 0 1 1 , 1 0 1 1 ⟶ 0 0 0 1 , 0 0 0 1 ⟶ 1 1 1 0 } The system was reversed. After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1 }, it remains to prove termination of the 3-rule system { 0 0 0 1 ⟶ 0 0 1 1 , 0 0 1 0 ⟶ 0 1 1 1 , 0 1 1 1 ⟶ 1 0 0 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2 }, it remains to prove termination of the 9-rule system { 0 1 1 2 ⟶ 0 1 2 2 , 0 1 1 2 ⟶ 0 2 2 , 0 1 2 1 ⟶ 0 2 2 2 , 0 2 2 2 ⟶ 0 1 1 , 0 2 2 2 ⟶ 0 1 , 0 2 2 2 ⟶ 0 , 1 1 1 2 →= 1 1 2 2 , 1 1 2 1 →= 1 2 2 2 , 1 2 2 2 →= 2 1 1 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 5-rule system { 0 1 1 2 ⟶ 0 1 2 2 , 0 1 2 1 ⟶ 0 2 2 2 , 1 1 1 2 →= 1 1 2 2 , 1 1 2 1 →= 1 2 2 2 , 1 2 2 2 →= 2 1 1 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 1 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1 }, it remains to prove termination of the 3-rule system { 0 0 0 1 →= 0 0 1 1 , 0 0 1 0 →= 0 1 1 1 , 0 1 1 1 →= 1 0 0 0 } The system is trivially terminating. popout

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