/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 { b ↦ 0, a ↦ 1, c ↦ 2 },
it remains to prove termination of the 5-rule system
{ 0 1 1 ⟶ 1 0 2 ,
  2 1 ⟶ 1 2 ,
  0 2 1 ⟶ 1 0 2 ,
  2 0 ⟶ 0 1 ,
  1 2 0 ⟶ 2 0 1 }

The length-preserving system was inverted.

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

The system was reversed.

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (2,↓) ↦ 2, (2,↑) ↦ 3, (0,↓) ↦ 4 },
it remains to prove termination of the 14-rule system
{ 0 1 2 ⟶ 3 2 1 ,
  0 1 2 ⟶ 3 1 ,
  0 2 ⟶ 3 4 ,
  0 2 ⟶ 0 ,
  0 1 2 ⟶ 3 4 1 ,
  0 1 2 ⟶ 0 1 ,
  3 1 ⟶ 0 ,
  3 1 4 ⟶ 0 2 ,
  3 1 4 ⟶ 3 ,
  4 1 2 →= 2 2 1 ,
  4 2 →= 2 4 ,
  4 1 2 →= 2 4 1 ,
  2 1 →= 1 4 ,
  2 1 4 →= 1 4 2 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

The system is trivially terminating.