details

property	value
status	complete
benchmark	random-284.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	1.07599592209 seconds
cpu usage	2.524255375
max memory	7.60082432E8

stage attributes

key	value
output-size	3233
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 3-rule system { 0 1 0 1 ⟶ 1 1 0 0 , 1 0 1 0 ⟶ 0 1 0 1 , 1 0 1 1 ⟶ 0 1 0 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 1 0 1 ⟶ 1 1 0 0 , 1 0 1 0 ⟶ 0 1 0 1 , 0 0 1 0 ⟶ 1 1 0 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 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1 }, it remains to prove termination of the 2-rule system { 0 1 0 1 ⟶ 1 1 0 0 , 1 0 1 0 ⟶ 0 1 0 1 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (0,↓) ↦ 2, (1,↑) ↦ 3 }, it remains to prove termination of the 10-rule system { 0 1 2 1 ⟶ 3 1 2 2 , 0 1 2 1 ⟶ 3 2 2 , 0 1 2 1 ⟶ 0 2 , 0 1 2 1 ⟶ 0 , 3 2 1 2 ⟶ 0 1 2 1 , 3 2 1 2 ⟶ 3 2 1 , 3 2 1 2 ⟶ 0 1 , 3 2 1 2 ⟶ 3 , 2 1 2 1 →= 1 1 2 2 , 1 2 1 2 →= 2 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 4-rule system { 0 1 2 1 ⟶ 3 1 2 2 , 3 2 1 2 ⟶ 0 1 2 1 , 2 1 2 1 →= 1 1 2 2 , 1 2 1 2 →= 2 1 2 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 5: 0 ↦ ⎛ ⎞ ⎜ 1 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 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 ⎟ ⎜ 0 0 1 0 0 ⎟ popout

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