SRS Standard pair #516975203

details

property	value
status	complete
benchmark	size-12-alpha-3-num-543.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n098.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	MnM 3.18b
configuration	default
runtime (wallclock)	2.50828504562 seconds
cpu usage	8.26490007
max memory	1.8394112E9

stage attributes

key	value
output-size	68540
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 2-rule system
{ 0 1 ⟶ ,
  0 1 2 ⟶ 1 2 1 2 0 0 1 }

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 ⟶ ,
  2 0 1 ⟶ 0 1 1 2 0 2 0 }

Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols.

After renaming modulo the bijection { (2,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (0,↑) ↦ 3, (2,↓) ↦ 4 },
it remains to prove termination of the 7-rule system
{ 0 1 2 ⟶ 3 2 2 4 1 4 1 ,
  0 1 2 ⟶ 0 1 4 1 ,
  0 1 2 ⟶ 3 4 1 ,
  0 1 2 ⟶ 0 1 ,
  0 1 2 ⟶ 3 ,
  1 2 →= ,
  4 1 2 →= 1 2 2 4 1 4 1 }

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, 2 ↦ 2, 4 ↦ 3 },
it remains to prove termination of the 4-rule system
{ 0 1 2 ⟶ 0 1 3 1 ,
  0 1 2 ⟶ 0 1 ,
  1 2 →= ,
  3 1 2 →= 1 2 2 3 1 3 1 }

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, (1,3) ↦ 4, (3,1) ↦ 5, (1,1) ↦ 6, (2,2) ↦ 7, (2,3) ↦ 8, (2,5) ↦ 9, (1,5) ↦ 10, (0,2) ↦ 11, (0,3) ↦ 12, (0,5) ↦ 13, (3,2) ↦ 14, (3,3) ↦ 15, (3,5) ↦ 16, (4,1) ↦ 17, (4,2) ↦ 18, (4,3) ↦ 19, (4,5) ↦ 20 },
it remains to prove termination of the 48-rule system
{ 0 1 2 3 ⟶ 0 1 4 5 6 ,
  0 1 2 7 ⟶ 0 1 4 5 2 ,
  0 1 2 8 ⟶ 0 1 4 5 4 ,
  0 1 2 9 ⟶ 0 1 4 5 10 ,
  0 1 2 3 ⟶ 0 1 6 ,
  0 1 2 7 ⟶ 0 1 2 ,
  0 1 2 8 ⟶ 0 1 4 ,
  0 1 2 9 ⟶ 0 1 10 ,
  1 2 3 →= 1 ,
  1 2 7 →= 11 ,
  1 2 8 →= 12 ,
  1 2 9 →= 13 ,
  6 2 3 →= 6 ,
  6 2 7 →= 2 ,
  6 2 8 →= 4 ,
  6 2 9 →= 10 ,
  3 2 3 →= 3 ,
  3 2 7 →= 7 ,
  3 2 8 →= 8 ,
  3 2 9 →= 9 ,
  5 2 3 →= 5 ,
  5 2 7 →= 14 ,
  5 2 8 →= 15 ,
  5 2 9 →= 16 ,
  17 2 3 →= 17 ,
  17 2 7 →= 18 ,
  17 2 8 →= 19 ,
  17 2 9 →= 20 ,
  12 5 2 3 →= 1 2 7 8 5 4 5 6 ,

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