/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

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

The system was reversed.

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

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

After renaming modulo the bijection { (0,↑) ↦ 0, (0,↓) ↦ 1, (1,↓) ↦ 2 },
it remains to prove termination of the 11-rule system
{ 0 1 1 1 1 1 2 1 ⟶ 0 1 1 2 1 2 1 1 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 2 1 2 1 1 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 2 1 2 1 1 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 2 1 1 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 1 ,
  0 1 1 1 1 1 2 1 ⟶ 0 1 ,
  1 1 1 1 1 1 2 1 →= 2 1 1 1 2 1 2 1 1 1 1 1 1 1 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 9:

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

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

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

The system is trivially terminating.