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

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

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

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 5:

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

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

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

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 2 1 ⟶ 0 1 2 3 1 2 ,
  0 1 2 1 ⟶ 0 1 2 3 1 2 3 1 2 ,
  3 1 2 1 →= 1 2 1 1 3 1 2 3 1 2 ,
  3 1 2 1 →= 1 2 1 1 3 1 2 3 1 2 3 1 2 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 8:

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

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

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

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

The system is trivially terminating.