/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 },
it remains to prove termination of the 3-rule system
{ 0 0 0 0 ⟶ 1 0 0 0 ,
  0 1 1 1 ⟶ 1 0 1 0 ,
  1 1 0 0 ⟶ 0 1 1 0 }

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

After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (1,0) ↦ 2, (0,3) ↦ 3, (1,1) ↦ 4, (2,0) ↦ 5, (2,1) ↦ 6 },
it remains to prove termination of the 24-rule system
{ 0 0 0 0 0 ⟶ 1 2 0 0 0 ,
  0 0 0 0 1 ⟶ 1 2 0 0 1 ,
  0 0 0 0 3 ⟶ 1 2 0 0 3 ,
  2 0 0 0 0 ⟶ 4 2 0 0 0 ,
  2 0 0 0 1 ⟶ 4 2 0 0 1 ,
  2 0 0 0 3 ⟶ 4 2 0 0 3 ,
  5 0 0 0 0 ⟶ 6 2 0 0 0 ,
  5 0 0 0 1 ⟶ 6 2 0 0 1 ,
  5 0 0 0 3 ⟶ 6 2 0 0 3 ,
  0 1 4 4 2 ⟶ 1 2 1 2 0 ,
  0 1 4 4 4 ⟶ 1 2 1 2 1 ,
  2 1 4 4 2 ⟶ 4 2 1 2 0 ,
  2 1 4 4 4 ⟶ 4 2 1 2 1 ,
  5 1 4 4 2 ⟶ 6 2 1 2 0 ,
  5 1 4 4 4 ⟶ 6 2 1 2 1 ,
  1 4 2 0 0 ⟶ 0 1 4 2 0 ,
  1 4 2 0 1 ⟶ 0 1 4 2 1 ,
  1 4 2 0 3 ⟶ 0 1 4 2 3 ,
  4 4 2 0 0 ⟶ 2 1 4 2 0 ,
  4 4 2 0 1 ⟶ 2 1 4 2 1 ,
  4 4 2 0 3 ⟶ 2 1 4 2 3 ,
  6 4 2 0 0 ⟶ 5 1 4 2 0 ,
  6 4 2 0 1 ⟶ 5 1 4 2 1 ,
  6 4 2 0 3 ⟶ 5 1 4 2 3 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (3,↓) ↦ 2, (4,↓) ↦ 3, (2,↓) ↦ 4, (0,↓) ↦ 5, (3,↑) ↦ 6 },
it remains to prove termination of the 19-rule system
{ 0 1 1 1 1 ⟶ 0 1 1 1 ,
  0 1 1 1 2 ⟶ 0 1 1 2 ,
  0 1 1 1 3 ⟶ 0 1 1 3 ,
  0 2 4 4 5 ⟶ 0 2 5 1 ,
  0 2 4 4 5 ⟶ 6 5 1 ,
  0 2 4 4 5 ⟶ 0 1 ,
  6 4 5 1 1 ⟶ 6 4 5 1 ,
  6 4 5 1 1 ⟶ 0 1 ,
  6 4 5 1 2 ⟶ 6 4 5 2 ,
  6 4 5 1 2 ⟶ 0 2 ,
  6 4 5 1 3 ⟶ 6 4 5 3 ,
  6 4 5 1 3 ⟶ 0 3 ,
  5 1 1 1 1 →= 4 5 1 1 1 ,
  5 1 1 1 2 →= 4 5 1 1 2 ,
  5 1 1 1 3 →= 4 5 1 1 3 ,
  5 2 4 4 5 →= 4 5 2 5 1 ,
  2 4 5 1 1 →= 1 2 4 5 1 ,
  2 4 5 1 2 →= 1 2 4 5 2 ,
  2 4 5 1 3 →= 1 2 4 5 3 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

The system is trivially terminating.