/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 { c ↦ 0, a ↦ 1, d ↦ 2, b ↦ 3 },
it remains to prove termination of the 5-rule system
{ 0 0 0 1 ⟶ 2 2 ,
  2 3 ⟶ 0 0 ,
  3 0 ⟶ 3 1 0 ,
  0 ⟶ 1 1 ,
  2 ⟶ 3 0 }

The system was reversed.

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

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 3 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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 4 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

3 ↦ ⎛     ⎞
    ⎜ 1 6 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

6 ↦ ⎛     ⎞
    ⎜ 1 2 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

The system is trivially terminating.