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SRS Standard pair #516974987
details
property
value
status
complete
benchmark
size-12-alpha-3-num-526.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n168.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
2.95451903343 seconds
cpu usage
9.998606561
max memory
1.848324096E9
stage attributes
key
value
output-size
43467
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 2-rule system { 0 0 1 ⟶ 1 1 2 0 0 0 , 1 2 0 ⟶ } The system was reversed. After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 2-rule system { 0 1 1 ⟶ 1 1 1 2 0 0 , 1 2 0 ⟶ } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (1,↑) ↦ 2, (2,↓) ↦ 3, (0,↓) ↦ 4 }, it remains to prove termination of the 7-rule system { 0 1 1 ⟶ 2 1 1 3 4 4 , 0 1 1 ⟶ 2 1 3 4 4 , 0 1 1 ⟶ 2 3 4 4 , 0 1 1 ⟶ 0 4 , 0 1 1 ⟶ 0 , 4 1 1 →= 1 1 1 3 4 4 , 1 3 4 →= } 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 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 4 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 4-rule system { 0 1 1 ⟶ 0 2 , 0 1 1 ⟶ 0 , 2 1 1 →= 1 1 1 3 2 2 , 1 3 2 →= } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (4,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (0,2) ↦ 3, (2,1) ↦ 4, (1,2) ↦ 5, (2,2) ↦ 6, (1,3) ↦ 7, (2,3) ↦ 8, (1,5) ↦ 9, (2,5) ↦ 10, (0,3) ↦ 11, (0,5) ↦ 12, (3,2) ↦ 13, (3,1) ↦ 14, (4,2) ↦ 15, (4,1) ↦ 16, (3,3) ↦ 17, (3,5) ↦ 18, (4,3) ↦ 19, (4,5) ↦ 20 }, it remains to prove termination of the 48-rule system { 0 1 2 2 ⟶ 0 3 4 , 0 1 2 5 ⟶ 0 3 6 , 0 1 2 7 ⟶ 0 3 8 , 0 1 2 9 ⟶ 0 3 10 , 0 1 2 2 ⟶ 0 1 , 0 1 2 5 ⟶ 0 3 , 0 1 2 7 ⟶ 0 11 , 0 1 2 9 ⟶ 0 12 , 3 4 2 2 →= 1 2 2 7 13 6 4 , 3 4 2 5 →= 1 2 2 7 13 6 6 , 3 4 2 7 →= 1 2 2 7 13 6 8 , 3 4 2 9 →= 1 2 2 7 13 6 10 , 5 4 2 2 →= 2 2 2 7 13 6 4 , 5 4 2 5 →= 2 2 2 7 13 6 6 , 5 4 2 7 →= 2 2 2 7 13 6 8 , 5 4 2 9 →= 2 2 2 7 13 6 10 , 6 4 2 2 →= 4 2 2 7 13 6 4 , 6 4 2 5 →= 4 2 2 7 13 6 6 , 6 4 2 7 →= 4 2 2 7 13 6 8 , 6 4 2 9 →= 4 2 2 7 13 6 10 , 13 4 2 2 →= 14 2 2 7 13 6 4 , 13 4 2 5 →= 14 2 2 7 13 6 6 , 13 4 2 7 →= 14 2 2 7 13 6 8 , 13 4 2 9 →= 14 2 2 7 13 6 10 , 15 4 2 2 →= 16 2 2 7 13 6 4 , 15 4 2 5 →= 16 2 2 7 13 6 6 , 15 4 2 7 →= 16 2 2 7 13 6 8 , 15 4 2 9 →= 16 2 2 7 13 6 10 , 1 7 13 4 →= 1 ,
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