YES After renaming modulo { a->0, b->1, c->2 }, it remains to prove termination of the 1-rule system { 0 0 1 0 2 0 0 0 -> 0 0 0 0 0 1 0 2 0 0 1 0 2 } The system was reversed. After renaming modulo { 0->0, 2->1, 1->2 }, it remains to prove termination of the 1-rule system { 0 0 0 1 0 2 0 0 -> 1 0 2 0 0 1 0 2 0 0 0 0 0 } Applying the dependency pairs transformation. After renaming modulo { (0,true)->0, (0,false)->1, (1,false)->2, (2,false)->3 }, it remains to prove termination of the 10-rule system { 0 1 1 2 1 3 1 1 -> 0 3 1 1 2 1 3 1 1 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 1 2 1 3 1 1 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 2 1 3 1 1 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 3 1 1 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 1 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 1 1 1 , 0 1 1 2 1 3 1 1 -> 0 1 1 , 0 1 1 2 1 3 1 1 -> 0 1 , 0 1 1 2 1 3 1 1 -> 0 , 1 1 1 2 1 3 1 1 ->= 2 1 3 1 1 2 1 3 1 1 1 1 1 } The system was filtered by the following matrix interpretation of type E_J with J = {1,...,2} and dimension 9: 0 is interpreted by / \ | 1 0 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 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 is interpreted by / \ | 1 0 0 0 0 0 0 0 0 | | 0 1 0 0 0 0 0 0 0 | | 0 0 0 2 0 0 0 0 0 | | 0 0 0 0 1 0 0 0 0 | | 0 0 1 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 0 1 1 0 0 0 0 1 | | 0 1 1 0 0 0 0 0 0 | \ / 2 is interpreted by / \ | 1 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 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 0 0 | | 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 | \ / 3 is interpreted by / \ | 1 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 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 0 | | 0 0 0 0 0 0 0 0 0 | | 0 0 0 0 0 0 0 0 0 | \ / After renaming modulo { 1->0, 2->1, 3->2 }, it remains to prove termination of the 1-rule system { 0 0 0 1 0 2 0 0 ->= 1 0 2 0 0 1 0 2 0 0 0 0 0 } The system is trivially terminating.