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SRS Standard pair #516975851
details
property
value
status
complete
benchmark
abc.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n067.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.70341300964 seconds
cpu usage
5.170401259
max memory
1.10743552E9
stage attributes
key
value
output-size
5433
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, c ↦ 2 }, it remains to prove termination of the 4-rule system { 0 1 2 ⟶ 2 2 1 1 0 0 , 0 ⟶ , 1 ⟶ , 2 ⟶ } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (2,↓) ↦ 2, (2,↑) ↦ 3, (0,↓) ↦ 4, (1,↑) ↦ 5 }, it remains to prove termination of the 10-rule system { 0 1 2 ⟶ 3 2 1 1 4 4 , 0 1 2 ⟶ 3 1 1 4 4 , 0 1 2 ⟶ 5 1 4 4 , 0 1 2 ⟶ 5 4 4 , 0 1 2 ⟶ 0 4 , 0 1 2 ⟶ 0 , 4 1 2 →= 2 2 1 1 4 4 , 4 →= , 1 →= , 2 →= } 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 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2, 4 ↦ 3 }, it remains to prove termination of the 6-rule system { 0 1 2 ⟶ 0 3 , 0 1 2 ⟶ 0 , 3 1 2 →= 2 2 1 1 3 3 , 3 →= , 1 →= , 2 →= } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (4,0) ↦ 0, (0,1) ↦ 1, (1,2) ↦ 2, (2,1) ↦ 3, (0,3) ↦ 4, (3,1) ↦ 5, (2,2) ↦ 6, (3,2) ↦ 7, (2,3) ↦ 8, (3,3) ↦ 9, (2,5) ↦ 10, (3,5) ↦ 11, (0,2) ↦ 12, (0,5) ↦ 13, (1,1) ↦ 14, (1,3) ↦ 15, (4,3) ↦ 16, (4,2) ↦ 17, (1,5) ↦ 18, (4,1) ↦ 19, (4,5) ↦ 20 }, it remains to prove termination of the 88-rule system { 0 1 2 3 ⟶ 0 4 5 , 0 1 2 6 ⟶ 0 4 7 , 0 1 2 8 ⟶ 0 4 9 , 0 1 2 10 ⟶ 0 4 11 , 0 1 2 3 ⟶ 0 1 , 0 1 2 6 ⟶ 0 12 , 0 1 2 8 ⟶ 0 4 , 0 1 2 10 ⟶ 0 13 , 4 5 2 3 →= 12 6 3 14 15 9 5 , 4 5 2 6 →= 12 6 3 14 15 9 7 , 4 5 2 8 →= 12 6 3 14 15 9 9 , 4 5 2 10 →= 12 6 3 14 15 9 11 , 15 5 2 3 →= 2 6 3 14 15 9 5 , 15 5 2 6 →= 2 6 3 14 15 9 7 , 15 5 2 8 →= 2 6 3 14 15 9 9 , 15 5 2 10 →= 2 6 3 14 15 9 11 , 8 5 2 3 →= 6 6 3 14 15 9 5 , 8 5 2 6 →= 6 6 3 14 15 9 7 , 8 5 2 8 →= 6 6 3 14 15 9 9 , 8 5 2 10 →= 6 6 3 14 15 9 11 , 9 5 2 3 →= 7 6 3 14 15 9 5 , 9 5 2 6 →= 7 6 3 14 15 9 7 , 9 5 2 8 →= 7 6 3 14 15 9 9 , 9 5 2 10 →= 7 6 3 14 15 9 11 ,
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