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SRS Standard pair #516975071
details
property
value
status
complete
benchmark
size-12-alpha-3-num-61.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n084.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
2.04658293724 seconds
cpu usage
6.446041326
max memory
1.092947968E9
stage attributes
key
value
output-size
8792
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, c ↦ 2 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 0 0 ⟶ 0 1 2 , 1 ⟶ , 2 1 ⟶ 1 0 2 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (2,↓) ↦ 3, (1,↑) ↦ 4, (2,↑) ↦ 5 }, it remains to prove termination of the 10-rule system { 0 1 ⟶ 0 2 3 , 0 1 ⟶ 4 3 , 0 1 ⟶ 5 , 5 2 ⟶ 4 1 3 , 5 2 ⟶ 0 3 , 5 2 ⟶ 5 , 1 →= , 1 1 →= 1 2 3 , 2 →= , 3 2 →= 2 1 3 } 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 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 3 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 8-rule system { 0 1 ⟶ 0 2 3 , 0 1 ⟶ 4 , 4 2 ⟶ 0 3 , 4 2 ⟶ 4 , 1 →= , 1 1 →= 1 2 3 , 2 →= , 3 2 →= 2 1 3 } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (5,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (0,2) ↦ 3, (2,3) ↦ 4, (3,1) ↦ 5, (1,2) ↦ 6, (3,2) ↦ 7, (1,3) ↦ 8, (3,3) ↦ 9, (5,4) ↦ 10, (4,1) ↦ 11, (4,2) ↦ 12, (4,3) ↦ 13, (2,1) ↦ 14, (0,3) ↦ 15, (2,2) ↦ 16, (5,1) ↦ 17, (5,2) ↦ 18, (5,3) ↦ 19 }, it remains to prove termination of the 84-rule system { 0 1 2 ⟶ 0 3 4 5 , 0 1 6 ⟶ 0 3 4 7 , 0 1 8 ⟶ 0 3 4 9 , 0 1 2 ⟶ 10 11 , 0 1 6 ⟶ 10 12 , 0 1 8 ⟶ 10 13 , 10 12 14 ⟶ 0 15 5 , 10 12 16 ⟶ 0 15 7 , 10 12 4 ⟶ 0 15 9 , 10 12 14 ⟶ 10 11 , 10 12 16 ⟶ 10 12 , 10 12 4 ⟶ 10 13 , 1 2 →= 1 , 1 6 →= 3 , 1 8 →= 15 , 2 2 →= 2 , 2 6 →= 6 , 2 8 →= 8 , 14 2 →= 14 , 14 6 →= 16 , 14 8 →= 4 , 5 2 →= 5 ,
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