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SRS Standard pair #516975647
details
property
value
status
complete
benchmark
secr7.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n075.star.cs.uiowa.edu
space
Secret_06_SRS
run statistics
property
value
solver
MnM 3.18b
configuration
default
runtime (wallclock)
1.54661107063 seconds
cpu usage
4.464785058
max memory
8.77449216E8
stage attributes
key
value
output-size
5638
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 1 0 0 2 1 , 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 1 4 4 2 1 , 0 1 2 ⟶ 5 4 4 2 1 , 0 1 2 ⟶ 0 4 2 1 , 0 1 2 ⟶ 0 2 1 , 0 1 2 ⟶ 3 1 , 0 1 2 ⟶ 5 , 4 1 2 →= 2 1 4 4 2 1 , 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 2 1 , 0 1 2 ⟶ 0 2 1 , 3 1 2 →= 2 1 3 3 2 1 , 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, (3,2) ↦ 5, (1,1) ↦ 6, (2,2) ↦ 7, (2,3) ↦ 8, (1,3) ↦ 9, (2,5) ↦ 10, (1,5) ↦ 11, (0,2) ↦ 12, (3,1) ↦ 13, (3,3) ↦ 14, (4,3) ↦ 15, (4,2) ↦ 16, (3,5) ↦ 17, (0,5) ↦ 18, (4,1) ↦ 19, (4,5) ↦ 20 }, it remains to prove termination of the 88-rule system { 0 1 2 3 ⟶ 0 4 5 3 6 , 0 1 2 7 ⟶ 0 4 5 3 2 , 0 1 2 8 ⟶ 0 4 5 3 9 , 0 1 2 10 ⟶ 0 4 5 3 11 , 0 1 2 3 ⟶ 0 12 3 6 , 0 1 2 7 ⟶ 0 12 3 2 , 0 1 2 8 ⟶ 0 12 3 9 , 0 1 2 10 ⟶ 0 12 3 11 , 4 13 2 3 →= 12 3 9 14 5 3 6 , 4 13 2 7 →= 12 3 9 14 5 3 2 , 4 13 2 8 →= 12 3 9 14 5 3 9 , 4 13 2 10 →= 12 3 9 14 5 3 11 , 9 13 2 3 →= 2 3 9 14 5 3 6 , 9 13 2 7 →= 2 3 9 14 5 3 2 , 9 13 2 8 →= 2 3 9 14 5 3 9 , 9 13 2 10 →= 2 3 9 14 5 3 11 , 8 13 2 3 →= 7 3 9 14 5 3 6 , 8 13 2 7 →= 7 3 9 14 5 3 2 , 8 13 2 8 →= 7 3 9 14 5 3 9 , 8 13 2 10 →= 7 3 9 14 5 3 11 , 14 13 2 3 →= 5 3 9 14 5 3 6 , 14 13 2 7 →= 5 3 9 14 5 3 2 , 14 13 2 8 →= 5 3 9 14 5 3 9 , 14 13 2 10 →= 5 3 9 14 5 3 11 ,
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