/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, C ↦ 2, c ↦ 3, A ↦ 4, B ↦ 5 }, it remains to prove termination of the 24-rule system { 0 1 ⟶ 2 , 1 3 ⟶ 4 , 3 0 ⟶ 5 , 4 2 ⟶ 1 , 2 5 ⟶ 0 , 5 4 ⟶ 3 , 0 0 0 0 ⟶ 4 4 4 , 4 4 4 4 ⟶ 0 0 0 , 1 1 1 1 ⟶ 5 5 5 , 5 5 5 5 ⟶ 1 1 1 , 3 3 3 3 ⟶ 2 2 2 , 2 2 2 2 ⟶ 3 3 3 , 5 0 0 0 ⟶ 3 4 4 4 , 4 4 4 1 ⟶ 0 0 0 2 , 2 1 1 1 ⟶ 0 5 5 5 , 5 5 5 3 ⟶ 1 1 1 4 , 4 3 3 3 ⟶ 1 2 2 2 , 2 2 2 0 ⟶ 3 3 3 5 , 0 4 ⟶ , 4 0 ⟶ , 1 5 ⟶ , 5 1 ⟶ , 3 2 ⟶ , 2 3 ⟶ } 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 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 5 ↦ 0, 0 ↦ 1, 3 ↦ 2, 4 ↦ 3, 1 ↦ 4, 2 ↦ 5 }, it remains to prove termination of the 6-rule system { 0 1 1 1 ⟶ 2 3 3 3 , 3 3 3 4 ⟶ 1 1 1 5 , 5 4 4 4 ⟶ 1 0 0 0 , 0 0 0 2 ⟶ 4 4 4 3 , 3 2 2 2 ⟶ 4 5 5 5 , 5 5 5 1 ⟶ 2 2 2 0 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (3,↑) ↦ 2, (3,↓) ↦ 3, (4,↓) ↦ 4, (5,↑) ↦ 5, (0,↓) ↦ 6, (2,↓) ↦ 7, (5,↓) ↦ 8 }, it remains to prove termination of the 18-rule system { 0 1 1 1 ⟶ 2 3 3 , 0 1 1 1 ⟶ 2 3 , 0 1 1 1 ⟶ 2 , 2 3 3 4 ⟶ 5 , 5 4 4 4 ⟶ 0 6 6 , 5 4 4 4 ⟶ 0 6 , 5 4 4 4 ⟶ 0 , 0 6 6 7 ⟶ 2 , 2 7 7 7 ⟶ 5 8 8 , 2 7 7 7 ⟶ 5 8 , 2 7 7 7 ⟶ 5 , 5 8 8 1 ⟶ 0 , 6 1 1 1 →= 7 3 3 3 , 3 3 3 4 →= 1 1 1 8 , 8 4 4 4 →= 1 6 6 6 , 6 6 6 7 →= 4 4 4 3 , 3 7 7 7 →= 4 8 8 8 , 8 8 8 1 →= 7 7 7 6 } 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 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 6 ↦ 0, 1 ↦ 1, 7 ↦ 2, 3 ↦ 3, 4 ↦ 4, 8 ↦ 5 }, it remains to prove termination of the 6-rule system { 0 1 1 1 →= 2 3 3 3 , 3 3 3 4 →= 1 1 1 5 , 5 4 4 4 →= 1 0 0 0 , 0 0 0 2 →= 4 4 4 3 , 3 2 2 2 →= 4 5 5 5 , 5 5 5 1 →= 2 2 2 0 } The system is trivially terminating.