/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 ⟶ 1 1 , 0 1 ⟶ 0 2 0 2 , 2 2 ⟶ } 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 ⟶ 1 1 , 1 0 ⟶ 2 0 2 0 , 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 ⟶ 1 5 4 , 0 1 ⟶ 1 5 5 , 0 2 ⟶ 1 5 6 , 0 3 ⟶ 1 5 7 , 4 0 ⟶ 5 5 4 , 4 1 ⟶ 5 5 5 , 4 2 ⟶ 5 5 6 , 4 3 ⟶ 5 5 7 , 8 0 ⟶ 9 5 4 , 8 1 ⟶ 9 5 5 , 8 2 ⟶ 9 5 6 , 8 3 ⟶ 9 5 7 , 12 0 ⟶ 13 5 4 , 12 1 ⟶ 13 5 5 , 12 2 ⟶ 13 5 6 , 12 3 ⟶ 13 5 7 , 1 4 0 ⟶ 2 8 2 8 0 , 1 4 1 ⟶ 2 8 2 8 1 , 1 4 2 ⟶ 2 8 2 8 2 , 1 4 3 ⟶ 2 8 2 8 3 , 5 4 0 ⟶ 6 8 2 8 0 , 5 4 1 ⟶ 6 8 2 8 1 , 5 4 2 ⟶ 6 8 2 8 2 , 5 4 3 ⟶ 6 8 2 8 3 , 9 4 0 ⟶ 10 8 2 8 0 , 9 4 1 ⟶ 10 8 2 8 1 , 9 4 2 ⟶ 10 8 2 8 2 , 9 4 3 ⟶ 10 8 2 8 3 , 13 4 0 ⟶ 14 8 2 8 0 , 13 4 1 ⟶ 14 8 2 8 1 , 13 4 2 ⟶ 14 8 2 8 2 , 13 4 3 ⟶ 14 8 2 8 3 , 2 10 8 ⟶ 0 , 2 10 9 ⟶ 1 , 2 10 10 ⟶ 2 , 2 10 11 ⟶ 3 , 6 10 8 ⟶ 4 , 6 10 9 ⟶ 5 , 6 10 10 ⟶ 6 , 6 10 11 ⟶ 7 , 10 10 8 ⟶ 8 , 10 10 9 ⟶ 9 , 10 10 10 ⟶ 10 , 10 10 11 ⟶ 11 , 14 10 8 ⟶ 12 , 14 10 9 ⟶ 13 , 14 10 10 ⟶ 14 , 14 10 11 ⟶ 15 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 2: 0 ↦ ⎛ ⎞ ⎜ 1 4 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 1 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 3 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 9 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 10 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 11 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 12 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 13 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 14 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 15 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 8 ↦ 0, 2 ↦ 1, 10 ↦ 2, 0 ↦ 3, 9 ↦ 4, 5 ↦ 5, 4 ↦ 6, 6 ↦ 7, 1 ↦ 8, 3 ↦ 9 }, it remains to prove termination of the 13-rule system { 0 1 ⟶ 2 , 0 3 ⟶ 4 5 6 , 0 1 ⟶ 4 5 7 , 5 6 3 ⟶ 7 0 1 0 3 , 5 6 8 ⟶ 7 0 1 0 8 , 5 6 1 ⟶ 7 0 1 0 1 , 5 6 9 ⟶ 7 0 1 0 9 , 4 6 3 ⟶ 2 0 1 0 3 , 4 6 8 ⟶ 2 0 1 0 8 , 4 6 1 ⟶ 2 0 1 0 1 , 4 6 9 ⟶ 2 0 1 0 9 , 1 2 0 ⟶ 3 , 7 2 0 ⟶ 6 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (0,↑) ↦ 0, (3,↓) ↦ 1, (4,↑) ↦ 2, (5,↓) ↦ 3, (6,↓) ↦ 4, (5,↑) ↦ 5, (1,↓) ↦ 6, (7,↓) ↦ 7, (7,↑) ↦ 8, (0,↓) ↦ 9, (1,↑) ↦ 10, (8,↓) ↦ 11, (9,↓) ↦ 12, (2,↓) ↦ 13, (4,↓) ↦ 14 }, it remains to prove termination of the 46-rule system { 0 1 ⟶ 2 3 4 , 0 1 ⟶ 5 4 , 0 6 ⟶ 2 3 7 , 0 6 ⟶ 5 7 , 0 6 ⟶ 8 , 5 4 1 ⟶ 8 9 6 9 1 , 5 4 1 ⟶ 0 6 9 1 , 5 4 1 ⟶ 10 9 1 , 5 4 1 ⟶ 0 1 , 5 4 11 ⟶ 8 9 6 9 11 , 5 4 11 ⟶ 0 6 9 11 , 5 4 11 ⟶ 10 9 11 , 5 4 11 ⟶ 0 11 , 5 4 6 ⟶ 8 9 6 9 6 , 5 4 6 ⟶ 0 6 9 6 , 5 4 6 ⟶ 10 9 6 , 5 4 6 ⟶ 0 6 , 5 4 12 ⟶ 8 9 6 9 12 , 5 4 12 ⟶ 0 6 9 12 , 5 4 12 ⟶ 10 9 12 , 5 4 12 ⟶ 0 12 , 2 4 1 ⟶ 0 6 9 1 , 2 4 1 ⟶ 10 9 1 , 2 4 1 ⟶ 0 1 , 2 4 11 ⟶ 0 6 9 11 , 2 4 11 ⟶ 10 9 11 , 2 4 11 ⟶ 0 11 , 2 4 6 ⟶ 0 6 9 6 , 2 4 6 ⟶ 10 9 6 , 2 4 6 ⟶ 0 6 , 2 4 12 ⟶ 0 6 9 12 , 2 4 12 ⟶ 10 9 12 , 2 4 12 ⟶ 0 12 , 9 6 →= 13 , 9 1 →= 14 3 4 , 9 6 →= 14 3 7 , 3 4 1 →= 7 9 6 9 1 , 3 4 11 →= 7 9 6 9 11 , 3 4 6 →= 7 9 6 9 6 , 3 4 12 →= 7 9 6 9 12 , 14 4 1 →= 13 9 6 9 1 , 14 4 11 →= 13 9 6 9 11 , 14 4 6 →= 13 9 6 9 6 , 14 4 12 →= 13 9 6 9 12 , 6 13 9 →= 1 , 7 13 9 →= 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 4 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 7 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 8 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 9 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 10 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 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 { 9 ↦ 0, 6 ↦ 1, 13 ↦ 2, 1 ↦ 3, 14 ↦ 4, 3 ↦ 5, 4 ↦ 6, 7 ↦ 7, 11 ↦ 8, 12 ↦ 9 }, it remains to prove termination of the 13-rule system { 0 1 →= 2 , 0 3 →= 4 5 6 , 0 1 →= 4 5 7 , 5 6 3 →= 7 0 1 0 3 , 5 6 8 →= 7 0 1 0 8 , 5 6 1 →= 7 0 1 0 1 , 5 6 9 →= 7 0 1 0 9 , 4 6 3 →= 2 0 1 0 3 , 4 6 8 →= 2 0 1 0 8 , 4 6 1 →= 2 0 1 0 1 , 4 6 9 →= 2 0 1 0 9 , 1 2 0 →= 3 , 7 2 0 →= 6 } The system is trivially terminating.