/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, s ↦ 1, b ↦ 2, c ↦ 3 }, it remains to prove termination of the 5-rule system { 0 1 ⟶ 1 0 , 2 0 2 1 ⟶ 0 2 1 0 , 2 0 2 2 ⟶ 3 1 , 3 1 ⟶ 0 2 0 2 , 0 2 0 0 ⟶ 2 0 2 0 } The system was reversed. After renaming modulo the bijection { 1 ↦ 0, 0 ↦ 1, 2 ↦ 2, 3 ↦ 3 }, it remains to prove termination of the 5-rule system { 0 1 ⟶ 1 0 , 0 2 1 2 ⟶ 1 0 2 1 , 2 2 1 2 ⟶ 0 3 , 0 3 ⟶ 2 1 2 1 , 1 1 2 1 ⟶ 1 2 1 2 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (1,↓) ↦ 1, (1,↑) ↦ 2, (0,↓) ↦ 3, (2,↓) ↦ 4, (2,↑) ↦ 5, (3,↓) ↦ 6 }, it remains to prove termination of the 20-rule system { 0 1 ⟶ 2 3 , 0 1 ⟶ 0 , 0 4 1 4 ⟶ 2 3 4 1 , 0 4 1 4 ⟶ 0 4 1 , 0 4 1 4 ⟶ 5 1 , 0 4 1 4 ⟶ 2 , 5 4 1 4 ⟶ 0 6 , 0 6 ⟶ 5 1 4 1 , 0 6 ⟶ 2 4 1 , 0 6 ⟶ 5 1 , 0 6 ⟶ 2 , 2 1 4 1 ⟶ 2 4 1 4 , 2 1 4 1 ⟶ 5 1 4 , 2 1 4 1 ⟶ 2 4 , 2 1 4 1 ⟶ 5 , 3 1 →= 1 3 , 3 4 1 4 →= 1 3 4 1 , 4 4 1 4 →= 3 6 , 3 6 →= 4 1 4 1 , 1 1 4 1 →= 1 4 1 4 } 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 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 7 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 5 ↦ 0, 4 ↦ 1, 1 ↦ 2, 0 ↦ 3, 6 ↦ 4, 2 ↦ 5, 3 ↦ 6 }, it remains to prove termination of the 8-rule system { 0 1 2 1 ⟶ 3 4 , 3 4 ⟶ 0 2 1 2 , 5 2 1 2 ⟶ 5 1 2 1 , 6 2 →= 2 6 , 6 1 2 1 →= 2 6 1 2 , 1 1 2 1 →= 6 4 , 6 4 →= 1 2 1 2 , 2 2 1 2 →= 2 1 2 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 7: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 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 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 1 1 0 0 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 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 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 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 1 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 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 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 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 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 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 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 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 1 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 3 ↦ 0, 4 ↦ 1, 0 ↦ 2, 2 ↦ 3, 1 ↦ 4, 5 ↦ 5, 6 ↦ 6 }, it remains to prove termination of the 7-rule system { 0 1 ⟶ 2 3 4 3 , 5 3 4 3 ⟶ 5 4 3 4 , 6 3 →= 3 6 , 6 4 3 4 →= 3 6 4 3 , 4 4 3 4 →= 6 1 , 6 1 →= 4 3 4 3 , 3 3 4 3 →= 3 4 3 4 } 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 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 5 ↦ 0, 3 ↦ 1, 4 ↦ 2, 6 ↦ 3, 1 ↦ 4 }, it remains to prove termination of the 6-rule system { 0 1 2 1 ⟶ 0 2 1 2 , 3 1 →= 1 3 , 3 2 1 2 →= 1 3 2 1 , 2 2 1 2 →= 3 4 , 3 4 →= 2 1 2 1 , 1 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 7: 0 ↦ ⎛ ⎞ ⎜ 1 0 1 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 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 1 1 0 0 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 1 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 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 1 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 0 0 0 0 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 1 ⎟ ⎜ 0 0 0 0 0 1 0 ⎟ ⎜ 0 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 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 1 0 0 0 0 0 ⎟ ⎜ 0 0 0 0 0 0 0 ⎟ ⎝ ⎠ After renaming modulo the bijection { 3 ↦ 0, 1 ↦ 1, 2 ↦ 2, 4 ↦ 3 }, it remains to prove termination of the 5-rule system { 0 1 →= 1 0 , 0 2 1 2 →= 1 0 2 1 , 2 2 1 2 →= 0 3 , 0 3 →= 2 1 2 1 , 1 1 2 1 →= 1 2 1 2 } The system is trivially terminating.