/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, c ↦ 1, b ↦ 2 },
it remains to prove termination of the 3-rule system
{ 0 1 ⟶ 1 2 ,
  0 ⟶ 2 2 2 ,
  2 1 2 ⟶ 0 1 }

The system was reversed.

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

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

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

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

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

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

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

3 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 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, 5 ↦ 4 },
it remains to prove termination of the 6-rule system
{ 0 1 ⟶ 2 3 ,
  0 1 ⟶ 0 ,
  2 3 4 ⟶ 0 1 ,
  3 1 →= 4 3 ,
  1 →= 4 4 4 ,
  4 3 4 →= 3 1 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 3:

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

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

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

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

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

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

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 4:

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

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

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

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

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

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

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 0 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

The system is trivially terminating.