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SRS Standard pair #516974735
details
property
value
status
complete
benchmark
size-12-alpha-3-num-265.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.38485097885 seconds
cpu usage
3.923178349
max memory
9.21214976E8
stage attributes
key
value
output-size
5656
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 1 ⟶ 2 , 1 ⟶ , 2 2 ⟶ 1 1 0 0 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, (1,↑) ↦ 4, (0,↓) ↦ 5 }, it remains to prove termination of the 9-rule system { 0 1 ⟶ 2 , 2 3 ⟶ 4 1 5 5 3 , 2 3 ⟶ 4 5 5 3 , 2 3 ⟶ 0 5 3 , 2 3 ⟶ 0 3 , 5 →= , 5 1 →= 3 , 1 →= , 3 3 →= 1 1 5 5 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 1 ⎟ ⎜ 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, 3 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 7-rule system { 0 1 ⟶ 2 , 2 3 ⟶ 0 4 3 , 2 3 ⟶ 0 3 , 4 →= , 4 1 →= 3 , 1 →= , 3 3 →= 1 1 4 4 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, (5,2) ↦ 3, (2,1) ↦ 4, (1,3) ↦ 5, (2,3) ↦ 6, (1,4) ↦ 7, (2,4) ↦ 8, (3,1) ↦ 9, (0,4) ↦ 10, (4,3) ↦ 11, (3,3) ↦ 12, (3,4) ↦ 13, (3,6) ↦ 14, (0,3) ↦ 15, (4,1) ↦ 16, (4,4) ↦ 17, (5,4) ↦ 18, (5,1) ↦ 19, (5,3) ↦ 20 }, it remains to prove termination of the 89-rule system { 0 1 2 ⟶ 3 4 , 0 1 5 ⟶ 3 6 , 0 1 7 ⟶ 3 8 , 3 6 9 ⟶ 0 10 11 9 , 3 6 12 ⟶ 0 10 11 12 , 3 6 13 ⟶ 0 10 11 13 , 3 6 14 ⟶ 0 10 11 14 , 3 6 9 ⟶ 0 15 9 , 3 6 12 ⟶ 0 15 12 , 3 6 13 ⟶ 0 15 13 , 3 6 14 ⟶ 0 15 14 , 10 16 →= 1 , 10 11 →= 15 , 10 17 →= 10 , 7 16 →= 2 , 7 11 →= 5 , 7 17 →= 7 , 8 16 →= 4 , 8 11 →= 6 , 8 17 →= 8 , 13 16 →= 9 , 13 11 →= 12 , 13 17 →= 13 , 17 16 →= 16 ,
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