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

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

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

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 1 ⎟
    ⎜ 0 0 2 0 0 ⎟
    ⎝           ⎠

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

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

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

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 1 ⎟
    ⎜ 0 1 0 ⎟
    ⎜ 0 1 0 ⎟
    ⎝       ⎠

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

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

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

The system is trivially terminating.