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


--------------------------------------------------------------------------------


YES

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

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

After renaming modulo the bijection { (1,↑) ↦ 0, (0,↓) ↦ 1, (0,↑) ↦ 2, (2,↓) ↦ 3, (1,↓) ↦ 4, (2,↑) ↦ 5 },
it remains to prove termination of the 10-rule system
{ 0 1 ⟶ 2 3 4 3 4 3 ,
  0 1 ⟶ 5 4 3 4 3 ,
  0 1 ⟶ 0 3 4 3 ,
  0 1 ⟶ 5 4 3 ,
  0 1 ⟶ 0 3 ,
  0 1 ⟶ 5 ,
  5 3 ⟶ 2 ,
  1 →= ,
  4 1 →= 1 3 4 3 4 3 ,
  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 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 3 ↦ 2, 4 ↦ 3, 5 ↦ 4, 2 ↦ 5 },
it remains to prove termination of the 6-rule system
{ 0 1 ⟶ 0 2 3 2 ,
  0 1 ⟶ 0 2 ,
  4 2 ⟶ 5 ,
  1 →= ,
  3 1 →= 1 2 3 2 3 2 ,
  2 2 →= 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 5-rule system
{ 0 1 ⟶ 0 2 3 2 ,
  0 1 ⟶ 0 2 ,
  1 →= ,
  3 1 →= 1 2 3 2 3 2 ,
  2 2 →= 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, (3,2) ↦ 5, (2,1) ↦ 6, (1,2) ↦ 7, (2,2) ↦ 8, (1,3) ↦ 9, (1,5) ↦ 10, (2,5) ↦ 11, (0,3) ↦ 12, (0,5) ↦ 13, (3,1) ↦ 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 68-rule system
{ 0 1 2 ⟶ 0 3 4 5 6 ,
  0 1 7 ⟶ 0 3 4 5 8 ,
  0 1 9 ⟶ 0 3 4 5 4 ,
  0 1 10 ⟶ 0 3 4 5 11 ,
  0 1 2 ⟶ 0 3 6 ,
  0 1 7 ⟶ 0 3 8 ,
  0 1 9 ⟶ 0 3 4 ,
  0 1 10 ⟶ 0 3 11 ,
  1 2 →= 1 ,
  1 7 →= 3 ,
  1 9 →= 12 ,
  1 10 →= 13 ,
  2 2 →= 2 ,
  2 7 →= 7 ,
  2 9 →= 9 ,
  2 10 →= 10 ,
  6 2 →= 6 ,
  6 7 →= 8 ,
  6 9 →= 4 ,
  6 10 →= 11 ,
  14 2 →= 14 ,
  14 7 →= 5 ,
  14 9 →= 15 ,
  14 10 →= 16 ,
  17 2 →= 17 ,
  17 7 →= 18 ,
  17 9 →= 19 ,
  17 10 →= 20 ,
  12 14 2 →= 1 7 4 5 4 5 6 ,
  12 14 7 →= 1 7 4 5 4 5 8 ,
  12 14 9 →= 1 7 4 5 4 5 4 ,
  12 14 10 →= 1 7 4 5 4 5 11 ,
  9 14 2 →= 2 7 4 5 4 5 6 ,
  9 14 7 →= 2 7 4 5 4 5 8 ,
  9 14 9 →= 2 7 4 5 4 5 4 ,
  9 14 10 →= 2 7 4 5 4 5 11 ,
  4 14 2 →= 6 7 4 5 4 5 6 ,
  4 14 7 →= 6 7 4 5 4 5 8 ,
  4 14 9 →= 6 7 4 5 4 5 4 ,
  4 14 10 →= 6 7 4 5 4 5 11 ,
  15 14 2 →= 14 7 4 5 4 5 6 ,
  15 14 7 →= 14 7 4 5 4 5 8 ,
  15 14 9 →= 14 7 4 5 4 5 4 ,
  15 14 10 →= 14 7 4 5 4 5 11 ,
  19 14 2 →= 17 7 4 5 4 5 6 ,
  19 14 7 →= 17 7 4 5 4 5 8 ,
  19 14 9 →= 17 7 4 5 4 5 4 ,
  19 14 10 →= 17 7 4 5 4 5 11 ,
  3 8 6 →= 1 2 ,
  3 8 8 →= 1 7 ,
  3 8 4 →= 1 9 ,
  3 8 11 →= 1 10 ,
  7 8 6 →= 2 2 ,
  7 8 8 →= 2 7 ,
  7 8 4 →= 2 9 ,
  7 8 11 →= 2 10 ,
  8 8 6 →= 6 2 ,
  8 8 8 →= 6 7 ,
  8 8 4 →= 6 9 ,
  8 8 11 →= 6 10 ,
  5 8 6 →= 14 2 ,
  5 8 8 →= 14 7 ,
  5 8 4 →= 14 9 ,
  5 8 11 →= 14 10 ,
  18 8 6 →= 17 2 ,
  18 8 8 →= 17 7 ,
  18 8 4 →= 17 9 ,
  18 8 11 →= 17 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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

8 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

15 ↦ ⎛     ⎞
     ⎜ 1 2 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

16 ↦ ⎛     ⎞
     ⎜ 1 0 ⎟
     ⎜ 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 { 1 ↦ 0, 2 ↦ 1, 9 ↦ 2, 12 ↦ 3, 7 ↦ 4, 10 ↦ 5, 6 ↦ 6, 8 ↦ 7, 4 ↦ 8, 11 ↦ 9, 14 ↦ 10, 15 ↦ 11, 17 ↦ 12, 19 ↦ 13, 5 ↦ 14 },
it remains to prove termination of the 38-rule system
{ 0 1 →= 0 ,
  0 2 →= 3 ,
  1 1 →= 1 ,
  1 4 →= 4 ,
  1 2 →= 2 ,
  1 5 →= 5 ,
  6 1 →= 6 ,
  6 4 →= 7 ,
  6 2 →= 8 ,
  6 5 →= 9 ,
  10 1 →= 10 ,
  10 2 →= 11 ,
  12 1 →= 12 ,
  12 2 →= 13 ,
  3 10 1 →= 0 4 8 14 8 14 6 ,
  3 10 4 →= 0 4 8 14 8 14 7 ,
  3 10 2 →= 0 4 8 14 8 14 8 ,
  3 10 5 →= 0 4 8 14 8 14 9 ,
  2 10 1 →= 1 4 8 14 8 14 6 ,
  2 10 4 →= 1 4 8 14 8 14 7 ,
  2 10 2 →= 1 4 8 14 8 14 8 ,
  2 10 5 →= 1 4 8 14 8 14 9 ,
  8 10 1 →= 6 4 8 14 8 14 6 ,
  8 10 4 →= 6 4 8 14 8 14 7 ,
  8 10 2 →= 6 4 8 14 8 14 8 ,
  8 10 5 →= 6 4 8 14 8 14 9 ,
  11 10 1 →= 10 4 8 14 8 14 6 ,
  11 10 4 →= 10 4 8 14 8 14 7 ,
  11 10 2 →= 10 4 8 14 8 14 8 ,
  11 10 5 →= 10 4 8 14 8 14 9 ,
  13 10 1 →= 12 4 8 14 8 14 6 ,
  13 10 4 →= 12 4 8 14 8 14 7 ,
  13 10 2 →= 12 4 8 14 8 14 8 ,
  13 10 5 →= 12 4 8 14 8 14 9 ,
  14 7 6 →= 10 1 ,
  14 7 7 →= 10 4 ,
  14 7 8 →= 10 2 ,
  14 7 9 →= 10 5 }

The system is trivially terminating.