/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, l ↦ 1, r ↦ 2, b ↦ 3 },
it remains to prove termination of the 4-rule system
{ 0 1 ⟶ 1 0 ,
  2 0 0 ⟶ 0 0 2 ,
  3 1 ⟶ 3 0 2 ,
  2 3 ⟶ 1 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 1 ⎟
    ⎜ 0 1 0 0 ⎟
    ⎜ 0 0 0 1 ⎟
    ⎜ 0 0 1 0 ⎟
    ⎝         ⎠

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

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

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

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

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

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

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

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

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

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

After renaming modulo the bijection {  },
it remains to prove termination of the 0-rule system
{ }

The system is trivially terminating.