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SRS Standard pair #516974099
details
property
value
status
complete
benchmark
size-12-alpha-3-num-131.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n143.star.cs.uiowa.edu
space
Waldmann_07_size12
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
3.03323507309 seconds
cpu usage
10.16898021
max memory
1.842823168E9
stage attributes
key
value
output-size
40425
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 ⟶ , 0 1 ⟶ , 0 2 ⟶ 2 2 0 0 1 , 1 1 ⟶ } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 1 0 ⟶ , 2 0 ⟶ 1 0 0 2 2 , 1 1 ⟶ } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (2,↑) ↦ 0, (0,↓) ↦ 1, (1,↑) ↦ 2, (2,↓) ↦ 3, (0,↑) ↦ 4, (1,↓) ↦ 5 }, it remains to prove termination of the 9-rule system { 0 1 ⟶ 2 1 1 3 3 , 0 1 ⟶ 4 1 3 3 , 0 1 ⟶ 4 3 3 , 0 1 ⟶ 0 3 , 0 1 ⟶ 0 , 1 →= , 5 1 →= , 3 1 →= 5 1 1 3 3 , 5 5 →= } 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, 3 ↦ 2, 5 ↦ 3 }, it remains to prove termination of the 6-rule system { 0 1 ⟶ 0 2 , 0 1 ⟶ 0 , 1 →= , 3 1 →= , 2 1 →= 3 1 1 2 2 , 3 3 →= } 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, (0,3) ↦ 9, (3,1) ↦ 10, (3,2) ↦ 11, (3,3) ↦ 12, (4,1) ↦ 13, (4,2) ↦ 14, (4,3) ↦ 15 }, it remains to prove termination of the 66-rule system { 0 1 2 ⟶ 0 3 4 , 0 1 5 ⟶ 0 3 6 , 0 1 7 ⟶ 0 3 8 , 0 1 2 ⟶ 0 1 , 0 1 5 ⟶ 0 3 , 0 1 7 ⟶ 0 9 , 1 2 →= 1 , 1 5 →= 3 , 1 7 →= 9 , 2 2 →= 2 , 2 5 →= 5 , 2 7 →= 7 , 4 2 →= 4 , 4 5 →= 6 , 4 7 →= 8 , 10 2 →= 10 ,
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