/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 2-rule system
{ 0 0 1 ⟶ 1 1 0 0 ,
  1 0 1 ⟶ 0 0 0 0 }

The system was reversed.

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

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 4-rule system
{ 0 1 1 ⟶ 0 2 ,
  0 1 1 ⟶ 0 ,
  2 1 1 →= 1 1 2 2 ,
  2 1 2 →= 1 1 1 1 }

The system was filtered by the following matrix interpretation
of type E_J with J = {1,...,2} and dimension 6:

0 ↦ ⎛             ⎞
    ⎜ 1 0 1 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 ⎟
    ⎝             ⎠

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

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

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

The system is trivially terminating.