/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 3-rule system { 0 ⟶ , 0 1 ⟶ 2 0 , 2 2 ⟶ 2 1 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 3-rule system { 0 ⟶ , 1 0 ⟶ 0 2 , 2 2 ⟶ 0 1 2 1 2 } Applying the dependency pairs transformation. Here, ↑ marks so-called defined symbols. After renaming modulo the bijection { (1,↑) ↦ 0, (0,↓) ↦ 1, (0,↑) ↦ 2, (2,↓) ↦ 3, (2,↑) ↦ 4, (1,↓) ↦ 5 }, it remains to prove termination of the 9-rule system { 0 1 ⟶ 2 3 , 0 1 ⟶ 4 , 4 3 ⟶ 2 5 3 5 3 , 4 3 ⟶ 0 3 5 3 , 4 3 ⟶ 4 5 3 , 4 3 ⟶ 0 3 , 1 →= , 5 1 →= 1 3 , 3 3 →= 1 5 3 5 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 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 4 ↦ 2, 3 ↦ 3, 5 ↦ 4 }, it remains to prove termination of the 7-rule system { 0 1 ⟶ 2 , 2 3 ⟶ 0 3 4 3 , 2 3 ⟶ 2 4 3 , 2 3 ⟶ 0 3 , 1 →= , 4 1 →= 1 3 , 3 3 →= 1 4 3 4 3 } Applying sparse tiling TROC(2) [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { (5,0) ↦ 0, (0,1) ↦ 1, (1,1) ↦ 2, (5,2) ↦ 3, (2,1) ↦ 4, (1,3) ↦ 5, (2,3) ↦ 6, (1,4) ↦ 7, (2,4) ↦ 8, (3,1) ↦ 9, (0,3) ↦ 10, (3,4) ↦ 11, (4,3) ↦ 12, (3,3) ↦ 13, (3,6) ↦ 14, (0,4) ↦ 15, (4,1) ↦ 16, (4,4) ↦ 17, (5,1) ↦ 18, (5,3) ↦ 19, (5,4) ↦ 20 }, it remains to prove termination of the 75-rule system { 0 1 2 ⟶ 3 4 , 0 1 5 ⟶ 3 6 , 0 1 7 ⟶ 3 8 , 3 6 9 ⟶ 0 10 11 12 9 , 3 6 13 ⟶ 0 10 11 12 13 , 3 6 11 ⟶ 0 10 11 12 11 , 3 6 14 ⟶ 0 10 11 12 14 , 3 6 9 ⟶ 3 8 12 9 , 3 6 13 ⟶ 3 8 12 13 , 3 6 11 ⟶ 3 8 12 11 , 3 6 14 ⟶ 3 8 12 14 , 3 6 9 ⟶ 0 10 9 , 3 6 13 ⟶ 0 10 13 , 3 6 11 ⟶ 0 10 11 , 3 6 14 ⟶ 0 10 14 , 1 2 →= 1 , 1 5 →= 10 , 1 7 →= 15 , 2 2 →= 2 , 2 5 →= 5 , 2 7 →= 7 , 4 2 →= 4 , 4 5 →= 6 , 4 7 →= 8 , 9 2 →= 9 , 9 5 →= 13 , 9 7 →= 11 , 16 2 →= 16 , 16 5 →= 12 , 16 7 →= 17 , 18 2 →= 18 , 18 5 →= 19 , 18 7 →= 20 , 15 16 2 →= 1 5 9 , 15 16 5 →= 1 5 13 , 15 16 7 →= 1 5 11 , 7 16 2 →= 2 5 9 , 7 16 5 →= 2 5 13 , 7 16 7 →= 2 5 11 , 8 16 2 →= 4 5 9 , 8 16 5 →= 4 5 13 , 8 16 7 →= 4 5 11 , 11 16 2 →= 9 5 9 , 11 16 5 →= 9 5 13 , 11 16 7 →= 9 5 11 , 17 16 2 →= 16 5 9 , 17 16 5 →= 16 5 13 , 17 16 7 →= 16 5 11 , 20 16 2 →= 18 5 9 , 20 16 5 →= 18 5 13 , 20 16 7 →= 18 5 11 , 10 13 9 →= 1 7 12 11 12 9 , 10 13 13 →= 1 7 12 11 12 13 , 10 13 11 →= 1 7 12 11 12 11 , 10 13 14 →= 1 7 12 11 12 14 , 5 13 9 →= 2 7 12 11 12 9 , 5 13 13 →= 2 7 12 11 12 13 , 5 13 11 →= 2 7 12 11 12 11 , 5 13 14 →= 2 7 12 11 12 14 , 6 13 9 →= 4 7 12 11 12 9 , 6 13 13 →= 4 7 12 11 12 13 , 6 13 11 →= 4 7 12 11 12 11 , 6 13 14 →= 4 7 12 11 12 14 , 13 13 9 →= 9 7 12 11 12 9 , 13 13 13 →= 9 7 12 11 12 13 , 13 13 11 →= 9 7 12 11 12 11 , 13 13 14 →= 9 7 12 11 12 14 , 12 13 9 →= 16 7 12 11 12 9 , 12 13 13 →= 16 7 12 11 12 13 , 12 13 11 →= 16 7 12 11 12 11 , 12 13 14 →= 16 7 12 11 12 14 , 19 13 9 →= 18 7 12 11 12 9 , 19 13 13 →= 18 7 12 11 12 13 , 19 13 11 →= 18 7 12 11 12 11 , 19 13 14 →= 18 7 12 11 12 14 } Applying sparse untiling TROCU(2) after reversal [Geser/Hofbauer/Waldmann, FSCD 2019]. After renaming modulo the bijection { 2 ↦ 0, 1 ↦ 1, 0 ↦ 2, 4 ↦ 3, 3 ↦ 4, 5 ↦ 5, 6 ↦ 6, 7 ↦ 7, 8 ↦ 8, 9 ↦ 9, 12 ↦ 10, 11 ↦ 11, 10 ↦ 12, 13 ↦ 13, 14 ↦ 14, 15 ↦ 15, 16 ↦ 16, 17 ↦ 17, 18 ↦ 18, 19 ↦ 19 }, it remains to prove termination of the 66-rule system { 0 1 2 ⟶ 3 4 , 0 1 5 ⟶ 3 6 , 0 1 7 ⟶ 3 8 , 3 6 9 ⟶ 0 10 11 12 9 , 3 6 13 ⟶ 0 10 11 12 13 , 3 6 11 ⟶ 0 10 11 12 11 , 3 6 9 ⟶ 3 8 12 9 , 3 6 13 ⟶ 3 8 12 13 , 3 6 11 ⟶ 3 8 12 11 , 3 6 9 ⟶ 0 10 9 , 3 6 13 ⟶ 0 10 13 , 3 6 11 ⟶ 0 10 11 , 1 2 →= 1 , 1 5 →= 10 , 1 7 →= 14 , 2 2 →= 2 , 2 5 →= 5 , 2 7 →= 7 , 4 2 →= 4 , 4 5 →= 6 , 4 7 →= 8 , 9 2 →= 9 , 9 5 →= 13 , 9 7 →= 11 , 15 2 →= 15 , 15 5 →= 12 , 15 7 →= 16 , 17 2 →= 17 , 17 5 →= 18 , 17 7 →= 19 , 14 15 2 →= 1 5 9 , 14 15 5 →= 1 5 13 , 14 15 7 →= 1 5 11 , 7 15 2 →= 2 5 9 , 7 15 5 →= 2 5 13 , 7 15 7 →= 2 5 11 , 8 15 2 →= 4 5 9 , 8 15 5 →= 4 5 13 , 8 15 7 →= 4 5 11 , 11 15 2 →= 9 5 9 , 11 15 5 →= 9 5 13 , 11 15 7 →= 9 5 11 , 16 15 2 →= 15 5 9 , 16 15 5 →= 15 5 13 , 16 15 7 →= 15 5 11 , 19 15 2 →= 17 5 9 , 19 15 5 →= 17 5 13 , 19 15 7 →= 17 5 11 , 10 13 9 →= 1 7 12 11 12 9 , 10 13 13 →= 1 7 12 11 12 13 , 10 13 11 →= 1 7 12 11 12 11 , 5 13 9 →= 2 7 12 11 12 9 , 5 13 13 →= 2 7 12 11 12 13 , 5 13 11 →= 2 7 12 11 12 11 , 6 13 9 →= 4 7 12 11 12 9 , 6 13 13 →= 4 7 12 11 12 13 , 6 13 11 →= 4 7 12 11 12 11 , 13 13 9 →= 9 7 12 11 12 9 , 13 13 13 →= 9 7 12 11 12 13 , 13 13 11 →= 9 7 12 11 12 11 , 12 13 9 →= 15 7 12 11 12 9 , 12 13 13 →= 15 7 12 11 12 13 , 12 13 11 →= 15 7 12 11 12 11 , 18 13 9 →= 17 7 12 11 12 9 , 18 13 13 →= 17 7 12 11 12 13 , 18 13 11 →= 17 7 12 11 12 11 } 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 2 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 3 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 4 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 5 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 6 ↦ ⎛ ⎞ ⎜ 1 1 ⎟ ⎜ 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 1 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 15 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 16 ↦ ⎛ ⎞ ⎜ 1 2 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 17 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 18 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ 19 ↦ ⎛ ⎞ ⎜ 1 0 ⎟ ⎜ 0 1 ⎟ ⎝ ⎠ After renaming modulo the bijection { 1 ↦ 0, 2 ↦ 1, 7 ↦ 2, 14 ↦ 3, 5 ↦ 4, 4 ↦ 5, 8 ↦ 6, 9 ↦ 7, 13 ↦ 8, 11 ↦ 9, 15 ↦ 10, 16 ↦ 11, 17 ↦ 12, 19 ↦ 13, 12 ↦ 14 }, it remains to prove termination of the 35-rule system { 0 1 →= 0 , 0 2 →= 3 , 1 1 →= 1 , 1 4 →= 4 , 1 2 →= 2 , 5 1 →= 5 , 5 2 →= 6 , 7 1 →= 7 , 7 4 →= 8 , 7 2 →= 9 , 10 1 →= 10 , 10 2 →= 11 , 12 1 →= 12 , 12 2 →= 13 , 3 10 1 →= 0 4 7 , 3 10 4 →= 0 4 8 , 3 10 2 →= 0 4 9 , 2 10 1 →= 1 4 7 , 2 10 4 →= 1 4 8 , 2 10 2 →= 1 4 9 , 6 10 1 →= 5 4 7 , 6 10 4 →= 5 4 8 , 6 10 2 →= 5 4 9 , 9 10 1 →= 7 4 7 , 9 10 4 →= 7 4 8 , 9 10 2 →= 7 4 9 , 11 10 1 →= 10 4 7 , 11 10 4 →= 10 4 8 , 11 10 2 →= 10 4 9 , 13 10 1 →= 12 4 7 , 13 10 4 →= 12 4 8 , 13 10 2 →= 12 4 9 , 14 8 7 →= 10 2 14 9 14 7 , 14 8 8 →= 10 2 14 9 14 8 , 14 8 9 →= 10 2 14 9 14 9 } The system is trivially terminating.