/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 4-rule system
{ 0 ⟶ ,
  0 1 ⟶ 0 2 1 1 2 ,
  1 ⟶ ,
  2 2 ⟶ 0 }

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 1 1 2 0 ,
  1 ⟶ ,
  2 2 ⟶ 0 }

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

After renaming modulo the bijection { (1,↑) ↦ 0, (0,↓) ↦ 1, (2,↑) ↦ 2, (1,↓) ↦ 3, (2,↓) ↦ 4, (0,↑) ↦ 5 },
it remains to prove termination of the 9-rule system
{ 0 1 ⟶ 2 3 3 4 1 ,
  0 1 ⟶ 0 3 4 1 ,
  0 1 ⟶ 0 4 1 ,
  0 1 ⟶ 2 1 ,
  2 4 ⟶ 5 ,
  1 →= ,
  3 1 →= 4 3 3 4 1 ,
  3 →= ,
  4 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 ⎟
    ⎝     ⎠

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

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

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

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

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, (2,3) ↦ 4, (3,1) ↦ 5, (1,2) ↦ 6, (1,3) ↦ 7, (1,5) ↦ 8, (0,3) ↦ 9, (0,5) ↦ 10, (2,1) ↦ 11, (2,2) ↦ 12, (2,5) ↦ 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 88-rule system
{ 0 1 2 ⟶ 0 3 4 5 2 ,
  0 1 6 ⟶ 0 3 4 5 6 ,
  0 1 7 ⟶ 0 3 4 5 7 ,
  0 1 8 ⟶ 0 3 4 5 8 ,
  0 1 2 ⟶ 0 9 5 2 ,
  0 1 6 ⟶ 0 9 5 6 ,
  0 1 7 ⟶ 0 9 5 7 ,
  0 1 8 ⟶ 0 9 5 8 ,
  1 2 →= 1 ,
  1 6 →= 3 ,
  1 7 →= 9 ,
  1 8 →= 10 ,
  2 2 →= 2 ,
  2 6 →= 6 ,
  2 7 →= 7 ,
  2 8 →= 8 ,
  11 2 →= 11 ,
  11 6 →= 12 ,
  11 7 →= 4 ,
  11 8 →= 13 ,
  5 2 →= 5 ,
  5 6 →= 14 ,
  5 7 →= 15 ,
  5 8 →= 16 ,
  17 2 →= 17 ,
  17 6 →= 18 ,
  17 7 →= 19 ,
  17 8 →= 20 ,
  3 11 2 →= 9 14 12 4 5 2 ,
  3 11 6 →= 9 14 12 4 5 6 ,
  3 11 7 →= 9 14 12 4 5 7 ,
  3 11 8 →= 9 14 12 4 5 8 ,
  6 11 2 →= 7 14 12 4 5 2 ,
  6 11 6 →= 7 14 12 4 5 6 ,
  6 11 7 →= 7 14 12 4 5 7 ,
  6 11 8 →= 7 14 12 4 5 8 ,
  12 11 2 →= 4 14 12 4 5 2 ,
  12 11 6 →= 4 14 12 4 5 6 ,
  12 11 7 →= 4 14 12 4 5 7 ,
  12 11 8 →= 4 14 12 4 5 8 ,
  14 11 2 →= 15 14 12 4 5 2 ,
  14 11 6 →= 15 14 12 4 5 6 ,
  14 11 7 →= 15 14 12 4 5 7 ,
  14 11 8 →= 15 14 12 4 5 8 ,
  18 11 2 →= 19 14 12 4 5 2 ,
  18 11 6 →= 19 14 12 4 5 6 ,
  18 11 7 →= 19 14 12 4 5 7 ,
  18 11 8 →= 19 14 12 4 5 8 ,
  3 11 →= 1 ,
  3 12 →= 3 ,
  3 4 →= 9 ,
  3 13 →= 10 ,
  6 11 →= 2 ,
  6 12 →= 6 ,
  6 4 →= 7 ,
  6 13 →= 8 ,
  12 11 →= 11 ,
  12 12 →= 12 ,
  12 4 →= 4 ,
  12 13 →= 13 ,
  14 11 →= 5 ,
  14 12 →= 14 ,
  14 4 →= 15 ,
  14 13 →= 16 ,
  18 11 →= 17 ,
  18 12 →= 18 ,
  18 4 →= 19 ,
  18 13 →= 20 ,
  9 15 5 →= 1 2 ,
  9 15 14 →= 1 6 ,
  9 15 15 →= 1 7 ,
  9 15 16 →= 1 8 ,
  7 15 5 →= 2 2 ,
  7 15 14 →= 2 6 ,
  7 15 15 →= 2 7 ,
  7 15 16 →= 2 8 ,
  4 15 5 →= 11 2 ,
  4 15 14 →= 11 6 ,
  4 15 15 →= 11 7 ,
  4 15 16 →= 11 8 ,
  15 15 5 →= 5 2 ,
  15 15 14 →= 5 6 ,
  15 15 15 →= 5 7 ,
  15 15 16 →= 5 8 ,
  19 15 5 →= 17 2 ,
  19 15 14 →= 17 6 ,
  19 15 15 →= 17 7 ,
  19 15 16 →= 17 8 }

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 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

6 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

7 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

11 ↦ ⎛     ⎞
     ⎜ 1 3 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

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

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

14 ↦ ⎛     ⎞
     ⎜ 1 3 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

15 ↦ ⎛     ⎞
     ⎜ 1 3 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

16 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

17 ↦ ⎛     ⎞
     ⎜ 1 1 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

18 ↦ ⎛     ⎞
     ⎜ 1 1 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

19 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

20 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1, 6 ↦ 2, 7 ↦ 3, 8 ↦ 4, 11 ↦ 5, 5 ↦ 6, 14 ↦ 7, 15 ↦ 8, 16 ↦ 9, 17 ↦ 10, 12 ↦ 11, 4 ↦ 12, 3 ↦ 13, 13 ↦ 14, 18 ↦ 15 },
it remains to prove termination of the 37-rule system
{ 0 1 →= 0 ,
  1 1 →= 1 ,
  1 2 →= 2 ,
  1 3 →= 3 ,
  1 4 →= 4 ,
  5 1 →= 5 ,
  6 1 →= 6 ,
  6 2 →= 7 ,
  6 3 →= 8 ,
  6 4 →= 9 ,
  10 1 →= 10 ,
  2 5 1 →= 3 7 11 12 6 1 ,
  2 5 2 →= 3 7 11 12 6 2 ,
  2 5 3 →= 3 7 11 12 6 3 ,
  2 5 4 →= 3 7 11 12 6 4 ,
  11 5 1 →= 12 7 11 12 6 1 ,
  11 5 2 →= 12 7 11 12 6 2 ,
  11 5 3 →= 12 7 11 12 6 3 ,
  11 5 4 →= 12 7 11 12 6 4 ,
  7 5 1 →= 8 7 11 12 6 1 ,
  7 5 2 →= 8 7 11 12 6 2 ,
  7 5 3 →= 8 7 11 12 6 3 ,
  7 5 4 →= 8 7 11 12 6 4 ,
  13 11 →= 13 ,
  2 11 →= 2 ,
  2 12 →= 3 ,
  11 5 →= 5 ,
  11 11 →= 11 ,
  11 12 →= 12 ,
  11 14 →= 14 ,
  7 11 →= 7 ,
  7 12 →= 8 ,
  15 11 →= 15 ,
  12 8 6 →= 5 1 ,
  12 8 7 →= 5 2 ,
  12 8 8 →= 5 3 ,
  12 8 9 →= 5 4 }

The system is trivially terminating.