/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 0 ⟶ 1 ,
  1 2 ⟶ ,
  2 1 ⟶ 0 1 2 2 }

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

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

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 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

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

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

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

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 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

4 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

7 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

9 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

14 ↦ ⎛     ⎞
     ⎜ 1 2 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

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

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

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

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

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

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

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

The system is trivially terminating.