/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 ⟶ ,
  2 1 ⟶ 1 2 0 }

The system was reversed.

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

Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019].

After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (0,2) ↦ 2, (1,0) ↦ 3, (1,1) ↦ 4, (1,2) ↦ 5, (2,0) ↦ 6, (2,1) ↦ 7, (2,2) ↦ 8, (3,0) ↦ 9, (3,1) ↦ 10, (3,2) ↦ 11 },
it remains to prove termination of the 48-rule system
{ 0 0 ⟶ 0 ,
  0 1 ⟶ 1 ,
  0 2 ⟶ 2 ,
  3 0 ⟶ 3 ,
  3 1 ⟶ 4 ,
  3 2 ⟶ 5 ,
  6 0 ⟶ 6 ,
  6 1 ⟶ 7 ,
  6 2 ⟶ 8 ,
  9 0 ⟶ 9 ,
  9 1 ⟶ 10 ,
  9 2 ⟶ 11 ,
  0 0 0 ⟶ 1 5 8 6 ,
  0 0 1 ⟶ 1 5 8 7 ,
  0 0 2 ⟶ 1 5 8 8 ,
  3 0 0 ⟶ 4 5 8 6 ,
  3 0 1 ⟶ 4 5 8 7 ,
  3 0 2 ⟶ 4 5 8 8 ,
  6 0 0 ⟶ 7 5 8 6 ,
  6 0 1 ⟶ 7 5 8 7 ,
  6 0 2 ⟶ 7 5 8 8 ,
  9 0 0 ⟶ 10 5 8 6 ,
  9 0 1 ⟶ 10 5 8 7 ,
  9 0 2 ⟶ 10 5 8 8 ,
  1 3 ⟶ 0 ,
  1 4 ⟶ 1 ,
  1 5 ⟶ 2 ,
  4 3 ⟶ 3 ,
  4 4 ⟶ 4 ,
  4 5 ⟶ 5 ,
  7 3 ⟶ 6 ,
  7 4 ⟶ 7 ,
  7 5 ⟶ 8 ,
  10 3 ⟶ 9 ,
  10 4 ⟶ 10 ,
  10 5 ⟶ 11 ,
  2 7 3 ⟶ 0 1 5 6 ,
  2 7 4 ⟶ 0 1 5 7 ,
  2 7 5 ⟶ 0 1 5 8 ,
  5 7 3 ⟶ 3 1 5 6 ,
  5 7 4 ⟶ 3 1 5 7 ,
  5 7 5 ⟶ 3 1 5 8 ,
  8 7 3 ⟶ 6 1 5 6 ,
  8 7 4 ⟶ 6 1 5 7 ,
  8 7 5 ⟶ 6 1 5 8 ,
  11 7 3 ⟶ 9 1 5 6 ,
  11 7 4 ⟶ 9 1 5 7 ,
  11 7 5 ⟶ 9 1 5 8 }

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

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

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

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

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

4 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

5 ↦ ⎛     ⎞
    ⎜ 1 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

11 ↦ ⎛     ⎞
     ⎜ 1 2 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (5,↓) ↦ 1, (1,↓) ↦ 2, (6,↑) ↦ 3, (7,↓) ↦ 4, (4,↓) ↦ 5, (7,↑) ↦ 6, (3,↑) ↦ 7, (1,↑) ↦ 8, (6,↓) ↦ 9, (8,↑) ↦ 10, (0,↓) ↦ 11, (3,↓) ↦ 12, (2,↓) ↦ 13, (8,↓) ↦ 14 },
it remains to prove termination of the 50-rule system
{ 0 1 2 ⟶ 3 4 5 ,
  0 1 2 ⟶ 6 5 ,
  7 1 2 ⟶ 3 4 5 ,
  7 1 2 ⟶ 6 5 ,
  8 9 ⟶ 10 ,
  10 5 11 ⟶ 8 9 12 ,
  10 5 11 ⟶ 3 12 ,
  10 5 11 ⟶ 7 ,
  10 5 13 ⟶ 8 9 5 ,
  10 5 13 ⟶ 3 5 ,
  10 5 9 ⟶ 8 9 4 ,
  10 5 9 ⟶ 3 4 ,
  10 5 9 ⟶ 6 ,
  3 5 11 ⟶ 0 2 9 12 ,
  3 5 11 ⟶ 8 9 12 ,
  3 5 11 ⟶ 3 12 ,
  3 5 11 ⟶ 7 ,
  3 5 13 ⟶ 0 2 9 5 ,
  3 5 13 ⟶ 8 9 5 ,
  3 5 13 ⟶ 3 5 ,
  3 5 9 ⟶ 0 2 9 4 ,
  3 5 9 ⟶ 8 9 4 ,
  3 5 9 ⟶ 3 4 ,
  3 5 9 ⟶ 6 ,
  6 5 11 ⟶ 7 2 9 12 ,
  6 5 11 ⟶ 8 9 12 ,
  6 5 11 ⟶ 3 12 ,
  6 5 11 ⟶ 7 ,
  6 5 13 ⟶ 7 2 9 5 ,
  6 5 13 ⟶ 8 9 5 ,
  6 5 13 ⟶ 3 5 ,
  6 5 9 ⟶ 7 2 9 4 ,
  6 5 9 ⟶ 8 9 4 ,
  6 5 9 ⟶ 3 4 ,
  6 5 9 ⟶ 6 ,
  11 2 →= 13 ,
  12 2 →= 5 ,
  11 1 2 →= 13 9 4 5 ,
  12 1 2 →= 5 9 4 5 ,
  2 11 →= 1 ,
  2 9 →= 14 ,
  14 5 11 →= 1 2 9 12 ,
  14 5 13 →= 1 2 9 5 ,
  14 5 9 →= 1 2 9 4 ,
  9 5 11 →= 11 2 9 12 ,
  9 5 13 →= 11 2 9 5 ,
  9 5 9 →= 11 2 9 4 ,
  4 5 11 →= 12 2 9 12 ,
  4 5 13 →= 12 2 9 5 ,
  4 5 9 →= 12 2 9 4 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

5 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

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

13 ↦ ⎛     ⎞
     ⎜ 1 2 ⎟
     ⎜ 0 1 ⎟
     ⎝     ⎠

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

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

The system is trivially terminating.