/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 { d ↦ 0, a ↦ 1, b ↦ 2, c ↦ 3 },
it remains to prove termination of the 4-rule system
{ 0 1 ⟶ 2 0 ,
  2 ⟶ 1 1 1 ,
  3 0 3 ⟶ 1 0 ,
  2 0 0 ⟶ 3 3 0 0 3 }

The system was reversed.

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (1,↑) ↦ 2, (2,↓) ↦ 3, (2,↑) ↦ 4, (0,↓) ↦ 5, (3,↑) ↦ 6, (3,↓) ↦ 7 },
it remains to prove termination of the 16-rule system
{ 0 1 ⟶ 2 3 ,
  0 1 ⟶ 4 ,
  4 ⟶ 0 5 5 ,
  4 ⟶ 0 5 ,
  4 ⟶ 0 ,
  6 1 7 ⟶ 2 5 ,
  6 1 7 ⟶ 0 ,
  2 1 3 ⟶ 6 1 1 7 7 ,
  2 1 3 ⟶ 2 1 7 7 ,
  2 1 3 ⟶ 2 7 7 ,
  2 1 3 ⟶ 6 7 ,
  2 1 3 ⟶ 6 ,
  5 1 →= 1 3 ,
  3 →= 5 5 5 ,
  7 1 7 →= 1 5 ,
  1 1 3 →= 7 1 1 7 7 }

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 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

4 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 6:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The system is trivially terminating.