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

The length-preserving system was inverted.

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

The system was reversed.

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2, (1,↑) ↦ 3 },
it remains to prove termination of the 15-rule system
{ 0 1 2 2 ⟶ 0 2 2 1 ,
  0 1 2 2 ⟶ 3 2 1 ,
  0 1 2 2 ⟶ 3 1 ,
  0 1 2 2 ⟶ 0 ,
  0 1 1 1 ⟶ 0 2 2 2 ,
  0 1 1 1 ⟶ 3 2 2 ,
  0 1 1 1 ⟶ 3 2 ,
  0 1 1 1 ⟶ 3 ,
  3 2 1 2 ⟶ 0 1 2 1 ,
  3 2 1 2 ⟶ 0 2 1 ,
  3 2 1 2 ⟶ 3 1 ,
  3 2 1 2 ⟶ 0 ,
  1 1 2 2 →= 1 2 2 1 ,
  1 1 1 1 →= 1 2 2 2 ,
  2 2 1 2 →= 1 1 2 1 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

2 ↦ ⎛     ⎞
    ⎜ 1 1 ⎟
    ⎜ 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 6-rule system
{ 0 1 2 2 ⟶ 0 2 2 1 ,
  0 1 1 1 ⟶ 0 2 2 2 ,
  3 2 1 2 ⟶ 0 1 2 1 ,
  1 1 2 2 →= 1 2 2 1 ,
  1 1 1 1 →= 1 2 2 2 ,
  2 2 1 2 →= 1 1 2 1 }

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 1 ⎟
    ⎜ 0 1 ⎟
    ⎝     ⎠

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

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

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

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

The system is trivially terminating.