/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 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 0 0 ⟶ 1 , 1 ⟶ , 1 2 ⟶ 2 2 2 1 0 } The system was reversed. After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 }, it remains to prove termination of the 4-rule system { 0 ⟶ , 0 0 ⟶ 1 , 1 ⟶ , 2 1 ⟶ 0 1 2 2 2 } Applying sparse tiling TRFC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (0,0) ↦ 0, (0,1) ↦ 1, (0,2) ↦ 2, (0,4) ↦ 3, (1,0) ↦ 4, (1,1) ↦ 5, (1,2) ↦ 6, (1,4) ↦ 7, (2,0) ↦ 8, (2,1) ↦ 9, (2,2) ↦ 10, (2,4) ↦ 11, (3,0) ↦ 12, (3,1) ↦ 13, (3,2) ↦ 14, (3,4) ↦ 15 }, it remains to prove termination of the 64-rule system { 0 0 ⟶ 0 , 0 1 ⟶ 1 , 0 2 ⟶ 2 , 0 3 ⟶ 3 , 4 0 ⟶ 4 , 4 1 ⟶ 5 , 4 2 ⟶ 6 , 4 3 ⟶ 7 , 8 0 ⟶ 8 , 8 1 ⟶ 9 , 8 2 ⟶ 10 , 8 3 ⟶ 11 , 12 0 ⟶ 12 , 12 1 ⟶ 13 , 12 2 ⟶ 14 , 12 3 ⟶ 15 , 0 0 0 ⟶ 1 4 , 0 0 1 ⟶ 1 5 , 0 0 2 ⟶ 1 6 , 0 0 3 ⟶ 1 7 , 4 0 0 ⟶ 5 4 , 4 0 1 ⟶ 5 5 , 4 0 2 ⟶ 5 6 , 4 0 3 ⟶ 5 7 , 8 0 0 ⟶ 9 4 , 8 0 1 ⟶ 9 5 , 8 0 2 ⟶ 9 6 , 8 0 3 ⟶ 9 7 , 12 0 0 ⟶ 13 4 , 12 0 1 ⟶ 13 5 , 12 0 2 ⟶ 13 6 , 12 0 3 ⟶ 13 7 , 1 4 ⟶ 0 , 1 5 ⟶ 1 , 1 6 ⟶ 2 , 1 7 ⟶ 3 , 5 4 ⟶ 4 , 5 5 ⟶ 5 , 5 6 ⟶ 6 , 5 7 ⟶ 7 , 9 4 ⟶ 8 , 9 5 ⟶ 9 , 9 6 ⟶ 10 , 9 7 ⟶ 11 , 13 4 ⟶ 12 , 13 5 ⟶ 13 , 13 6 ⟶ 14 , 13 7 ⟶ 15 , 2 9 4 ⟶ 0 1 6 10 10 8 , 2 9 5 ⟶ 0 1 6 10 10 9 , 2 9 6 ⟶ 0 1 6 10 10 10 , 2 9 7 ⟶ 0 1 6 10 10 11 , 6 9 4 ⟶ 4 1 6 10 10 8 , 6 9 5 ⟶ 4 1 6 10 10 9 , 6 9 6 ⟶ 4 1 6 10 10 10 , 6 9 7 ⟶ 4 1 6 10 10 11 , 10 9 4 ⟶ 8 1 6 10 10 8 , 10 9 5 ⟶ 8 1 6 10 10 9 , 10 9 6 ⟶ 8 1 6 10 10 10 , 10 9 7 ⟶ 8 1 6 10 10 11 , 14 9 4 ⟶ 12 1 6 10 10 8 , 14 9 5 ⟶ 12 1 6 10 10 9 , 14 9 6 ⟶ 12 1 6 10 10 10 , 14 9 7 ⟶ 12 1 6 10 10 11 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 9 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 10 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 11 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 12 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 13 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 14 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 15 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 4 ↦ 0, 1 ↦ 1, 5 ↦ 2, 8 ↦ 3, 9 ↦ 4, 0 ↦ 5, 6 ↦ 6, 2 ↦ 7, 10 ↦ 8 }, it remains to prove termination of the 15-rule system { 0 1 ⟶ 2 , 3 1 ⟶ 4 , 0 5 1 ⟶ 2 2 , 3 5 1 ⟶ 4 2 , 1 0 ⟶ 5 , 1 6 ⟶ 7 , 7 4 0 ⟶ 5 1 6 8 8 3 , 7 4 2 ⟶ 5 1 6 8 8 4 , 7 4 6 ⟶ 5 1 6 8 8 8 , 6 4 0 ⟶ 0 1 6 8 8 3 , 6 4 2 ⟶ 0 1 6 8 8 4 , 6 4 6 ⟶ 0 1 6 8 8 8 , 8 4 0 ⟶ 3 1 6 8 8 3 , 8 4 2 ⟶ 3 1 6 8 8 4 , 8 4 6 ⟶ 3 1 6 8 8 8 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (1,↑) ↦ 0, (6,↓) ↦ 1, (7,↑) ↦ 2, (4,↓) ↦ 3, (0,↓) ↦ 4, (8,↓) ↦ 5, (3,↓) ↦ 6, (6,↑) ↦ 7, (8,↑) ↦ 8, (3,↑) ↦ 9, (2,↓) ↦ 10, (0,↑) ↦ 11, (1,↓) ↦ 12, (5,↓) ↦ 13, (7,↓) ↦ 14 }, it remains to prove termination of the 64-rule system { 0 1 ⟶ 2 , 2 3 4 ⟶ 0 1 5 5 6 , 2 3 4 ⟶ 7 5 5 6 , 2 3 4 ⟶ 8 5 6 , 2 3 4 ⟶ 8 6 , 2 3 4 ⟶ 9 , 2 3 10 ⟶ 0 1 5 5 3 , 2 3 10 ⟶ 7 5 5 3 , 2 3 10 ⟶ 8 5 3 , 2 3 10 ⟶ 8 3 , 2 3 1 ⟶ 0 1 5 5 5 , 2 3 1 ⟶ 7 5 5 5 , 2 3 1 ⟶ 8 5 5 , 2 3 1 ⟶ 8 5 , 2 3 1 ⟶ 8 , 7 3 4 ⟶ 11 12 1 5 5 6 , 7 3 4 ⟶ 0 1 5 5 6 , 7 3 4 ⟶ 7 5 5 6 , 7 3 4 ⟶ 8 5 6 , 7 3 4 ⟶ 8 6 , 7 3 4 ⟶ 9 , 7 3 10 ⟶ 11 12 1 5 5 3 , 7 3 10 ⟶ 0 1 5 5 3 , 7 3 10 ⟶ 7 5 5 3 , 7 3 10 ⟶ 8 5 3 , 7 3 10 ⟶ 8 3 , 7 3 1 ⟶ 11 12 1 5 5 5 , 7 3 1 ⟶ 0 1 5 5 5 , 7 3 1 ⟶ 7 5 5 5 , 7 3 1 ⟶ 8 5 5 , 7 3 1 ⟶ 8 5 , 7 3 1 ⟶ 8 , 8 3 4 ⟶ 9 12 1 5 5 6 , 8 3 4 ⟶ 0 1 5 5 6 , 8 3 4 ⟶ 7 5 5 6 , 8 3 4 ⟶ 8 5 6 , 8 3 4 ⟶ 8 6 , 8 3 4 ⟶ 9 , 8 3 10 ⟶ 9 12 1 5 5 3 , 8 3 10 ⟶ 0 1 5 5 3 , 8 3 10 ⟶ 7 5 5 3 , 8 3 10 ⟶ 8 5 3 , 8 3 10 ⟶ 8 3 , 8 3 1 ⟶ 9 12 1 5 5 5 , 8 3 1 ⟶ 0 1 5 5 5 , 8 3 1 ⟶ 7 5 5 5 , 8 3 1 ⟶ 8 5 5 , 8 3 1 ⟶ 8 5 , 8 3 1 ⟶ 8 , 4 12 →= 10 , 6 12 →= 3 , 4 13 12 →= 10 10 , 6 13 12 →= 3 10 , 12 4 →= 13 , 12 1 →= 14 , 14 3 4 →= 13 12 1 5 5 6 , 14 3 10 →= 13 12 1 5 5 3 , 14 3 1 →= 13 12 1 5 5 5 , 1 3 4 →= 4 12 1 5 5 6 , 1 3 10 →= 4 12 1 5 5 3 , 1 3 1 →= 4 12 1 5 5 5 , 5 3 4 →= 6 12 1 5 5 6 , 5 3 10 →= 6 12 1 5 5 3 , 5 3 1 →= 6 12 1 5 5 5 } 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 9 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 10 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 11 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 12 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 13 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 14 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 4 ↦ 0, 12 ↦ 1, 10 ↦ 2, 6 ↦ 3, 3 ↦ 4, 13 ↦ 5, 1 ↦ 6, 14 ↦ 7, 5 ↦ 8 }, it remains to prove termination of the 15-rule system { 0 1 →= 2 , 3 1 →= 4 , 0 5 1 →= 2 2 , 3 5 1 →= 4 2 , 1 0 →= 5 , 1 6 →= 7 , 7 4 0 →= 5 1 6 8 8 3 , 7 4 2 →= 5 1 6 8 8 4 , 7 4 6 →= 5 1 6 8 8 8 , 6 4 0 →= 0 1 6 8 8 3 , 6 4 2 →= 0 1 6 8 8 4 , 6 4 6 →= 0 1 6 8 8 8 , 8 4 0 →= 3 1 6 8 8 3 , 8 4 2 →= 3 1 6 8 8 4 , 8 4 6 →= 3 1 6 8 8 8 } The system is trivially terminating.