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

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

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

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

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

1 ↦ ⎛         ⎞
    ⎜ 1 0 0 0 ⎟
    ⎜ 0 1 0 0 ⎟
    ⎜ 0 1 0 4 ⎟
    ⎜ 0 1 2 0 ⎟
    ⎝         ⎠

2 ↦ ⎛             ⎞
    ⎜  1  0  0  0 ⎟
    ⎜  0  1  0  0 ⎟
    ⎜  0 11 16  2 ⎟
    ⎜  0  0  1  0 ⎟
    ⎝             ⎠

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

The system is trivially terminating.