/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 { a12 ↦ 0, a13 ↦ 1, a14 ↦ 2, a15 ↦ 3, a16 ↦ 4, a23 ↦ 5, a24 ↦ 6, a25 ↦ 7, a26 ↦ 8, a34 ↦ 9, a35 ↦ 10, a36 ↦ 11, a45 ↦ 12, a46 ↦ 13, a56 ↦ 14 },
it remains to prove termination of the 35-rule system
{ 0 0 0 0 ⟶ ,
  1 1 1 1 ⟶ ,
  2 2 2 2 ⟶ ,
  3 3 3 3 ⟶ ,
  4 4 4 4 ⟶ ,
  5 5 5 5 ⟶ ,
  6 6 6 6 ⟶ ,
  7 7 7 7 ⟶ ,
  8 8 8 8 ⟶ ,
  9 9 9 9 ⟶ ,
  10 10 10 10 ⟶ ,
  11 11 11 11 ⟶ ,
  12 12 12 12 ⟶ ,
  13 13 13 13 ⟶ ,
  14 14 14 14 ⟶ ,
  1 1 ⟶ 0 0 5 5 0 0 ,
  2 2 ⟶ 0 0 5 5 9 9 5 5 0 0 ,
  3 3 ⟶ 0 0 5 5 9 9 12 12 9 9 5 5 0 0 ,
  4 4 ⟶ 0 0 5 5 9 9 12 12 14 14 12 12 9 9 5 5 0 0 ,
  6 6 ⟶ 5 5 9 9 5 5 ,
  7 7 ⟶ 5 5 9 9 12 12 9 9 5 5 ,
  8 8 ⟶ 5 5 9 9 12 12 14 14 12 12 9 9 5 5 ,
  10 10 ⟶ 9 9 12 12 9 9 ,
  11 11 ⟶ 9 9 12 12 14 14 12 12 9 9 ,
  13 13 ⟶ 12 12 14 14 12 12 ,
  0 0 5 5 0 0 5 5 0 0 5 5 ⟶ ,
  5 5 9 9 5 5 9 9 5 5 9 9 ⟶ ,
  9 9 12 12 9 9 12 12 9 9 12 12 ⟶ ,
  12 12 14 14 12 12 14 14 12 12 14 14 ⟶ ,
  0 0 9 9 ⟶ 9 9 0 0 ,
  0 0 12 12 ⟶ 12 12 0 0 ,
  0 0 14 14 ⟶ 14 14 0 0 ,
  5 5 12 12 ⟶ 12 12 5 5 ,
  5 5 14 14 ⟶ 14 14 5 5 ,
  9 9 14 14 ⟶ 14 14 9 9 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 2:

0 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

1 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

3 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

4 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

5 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

6 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

7 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

8 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

9 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

10 ↦ ⎛     ⎞
     ⎜ 1 1 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

11 ↦ ⎛     ⎞
     ⎜ 1 1 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

12 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

13 ↦ ⎛     ⎞
     ⎜ 1 1 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

14 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

After renaming modulo the bijection { 0 ↦ 0, 5 ↦ 1, 9 ↦ 2, 12 ↦ 3, 14 ↦ 4 },
it remains to prove termination of the 15-rule system
{ 0 0 0 0 ⟶ ,
  1 1 1 1 ⟶ ,
  2 2 2 2 ⟶ ,
  3 3 3 3 ⟶ ,
  4 4 4 4 ⟶ ,
  0 0 1 1 0 0 1 1 0 0 1 1 ⟶ ,
  1 1 2 2 1 1 2 2 1 1 2 2 ⟶ ,
  2 2 3 3 2 2 3 3 2 2 3 3 ⟶ ,
  3 3 4 4 3 3 4 4 3 3 4 4 ⟶ ,
  0 0 2 2 ⟶ 2 2 0 0 ,
  0 0 3 3 ⟶ 3 3 0 0 ,
  0 0 4 4 ⟶ 4 4 0 0 ,
  1 1 3 3 ⟶ 3 3 1 1 ,
  1 1 4 4 ⟶ 4 4 1 1 ,
  2 2 4 4 ⟶ 4 4 2 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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

3 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

4 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

After renaming modulo the bijection { 0 ↦ 0, 2 ↦ 1, 3 ↦ 2, 4 ↦ 3, 1 ↦ 4 },
it remains to prove termination of the 6-rule system
{ 0 0 1 1 ⟶ 1 1 0 0 ,
  0 0 2 2 ⟶ 2 2 0 0 ,
  0 0 3 3 ⟶ 3 3 0 0 ,
  4 4 2 2 ⟶ 2 2 4 4 ,
  4 4 3 3 ⟶ 3 3 4 4 ,
  1 1 3 3 ⟶ 3 3 1 1 }

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

After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (1,↑) ↦ 3, (2,↓) ↦ 4, (3,↓) ↦ 5, (4,↑) ↦ 6, (4,↓) ↦ 7 },
it remains to prove termination of the 20-rule system
{ 0 1 2 2 ⟶ 3 2 1 1 ,
  0 1 2 2 ⟶ 3 1 1 ,
  0 1 2 2 ⟶ 0 1 ,
  0 1 2 2 ⟶ 0 ,
  0 1 4 4 ⟶ 0 1 ,
  0 1 4 4 ⟶ 0 ,
  0 1 5 5 ⟶ 0 1 ,
  0 1 5 5 ⟶ 0 ,
  6 7 4 4 ⟶ 6 7 ,
  6 7 4 4 ⟶ 6 ,
  6 7 5 5 ⟶ 6 7 ,
  6 7 5 5 ⟶ 6 ,
  3 2 5 5 ⟶ 3 2 ,
  3 2 5 5 ⟶ 3 ,
  1 1 2 2 →= 2 2 1 1 ,
  1 1 4 4 →= 4 4 1 1 ,
  1 1 5 5 →= 5 5 1 1 ,
  7 7 4 4 →= 4 4 7 7 ,
  7 7 5 5 →= 5 5 7 7 ,
  2 2 5 5 →= 5 5 2 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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

3 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

4 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

5 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

6 ↦ ⎛     ⎞
    ⎜ 1 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

7 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1, 4 ↦ 2, 5 ↦ 3, 7 ↦ 4 },
it remains to prove termination of the 6-rule system
{ 0 0 1 1 →= 1 1 0 0 ,
  0 0 2 2 →= 2 2 0 0 ,
  0 0 3 3 →= 3 3 0 0 ,
  4 4 2 2 →= 2 2 4 4 ,
  4 4 3 3 →= 3 3 4 4 ,
  1 1 3 3 →= 3 3 1 1 }

The system is trivially terminating.