YES(?,O(n^3))

Problem:
 3(1(x1)) -> 4(1(x1))
 5(9(x1)) -> 2(6(5(x1)))
 3(5(x1)) -> 8(9(7(x1)))
 9(x1) -> 3(2(3(x1)))
 8(4(x1)) -> 6(x1)
 2(6(x1)) -> 4(3(x1))
 3(8(x1)) -> 3(2(7(x1)))
 9(x1) -> 5(0(2(x1)))
 8(8(4(x1))) -> 1(9(x1))
 7(1(x1)) -> 6(9(x1))
 3(9(x1)) -> 9(3(x1))
 7(5(x1)) -> 1(0(x1))

Proof:
 Matrix Interpretation Processor:
  dimension: 3
  interpretation:
             [1 0 2]  
   [0](x0) = [0 0 1]x0
             [0 0 0]  ,
   
             [1 5 0]     [2]
   [8](x0) = [0 1 0]x0 + [2]
             [0 0 1]     [2],
   
             [1 0 1]     [0]
   [7](x0) = [0 1 0]x0 + [4]
             [0 0 1]     [0],
   
             [1 0 0]     [0]
   [2](x0) = [0 1 4]x0 + [4]
             [0 0 0]     [0],
   
             [1 0 1]     [1]
   [6](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [0],
   
             [1 0 2]     [0]
   [5](x0) = [0 0 2]x0 + [0]
             [0 0 1]     [4],
   
             [1 0 1]     [1]
   [9](x0) = [0 0 0]x0 + [0]
             [0 0 1]     [4],
   
             [1 0 0]  
   [4](x0) = [0 0 1]x0
             [0 0 0]  ,
   
             [1 0 1]     [0]
   [3](x0) = [0 0 1]x0 + [0]
             [0 0 1]     [2],
   
             [1 1 2]     [3]
   [1](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [4]
  orientation:
              [1 1 2]     [7]    [1 1 2]     [3]           
   3(1(x1)) = [0 0 0]x1 + [4] >= [0 0 0]x1 + [4] = 4(1(x1))
              [0 0 0]     [6]    [0 0 0]     [0]           
   
              [1 0 3]     [9]    [1 0 3]     [5]              
   5(9(x1)) = [0 0 2]x1 + [8] >= [0 0 1]x1 + [8] = 2(6(5(x1)))
              [0 0 1]     [8]    [0 0 0]     [0]              
   
              [1 0 3]     [4]    [1 0 2]     [3]              
   3(5(x1)) = [0 0 1]x1 + [4] >= [0 0 0]x1 + [2] = 8(9(7(x1)))
              [0 0 1]     [6]    [0 0 1]     [6]              
   
           [1 0 1]     [1]    [1 0 1]     [0]              
   9(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 3(2(3(x1)))
           [0 0 1]     [4]    [0 0 0]     [2]              
   
              [1 0 5]     [2]    [1 0 1]     [1]        
   8(4(x1)) = [0 0 1]x1 + [2] >= [0 0 1]x1 + [0] = 6(x1)
              [0 0 0]     [2]    [0 0 0]     [0]        
   
              [1 0 1]     [1]    [1 0 1]     [0]           
   2(6(x1)) = [0 0 1]x1 + [4] >= [0 0 1]x1 + [2] = 4(3(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
   
              [1 5 1]     [4]    [1 0 1]     [0]              
   3(8(x1)) = [0 0 1]x1 + [2] >= [0 0 0]x1 + [0] = 3(2(7(x1)))
              [0 0 1]     [4]    [0 0 0]     [2]              
   
           [1 0 1]     [1]    [1 0 0]     [0]              
   9(x1) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = 5(0(2(x1)))
           [0 0 1]     [4]    [0 0 0]     [4]              
   
                 [1  0  10]     [14]    [1 0 3]     [12]           
   8(8(4(x1))) = [0  0  1 ]x1 + [4 ] >= [0 0 1]x1 + [4 ] = 1(9(x1))
                 [0  0  0 ]     [4 ]    [0 0 0]     [4 ]           
   
              [1 1 2]     [7]    [1 0 2]     [6]           
   7(1(x1)) = [0 0 1]x1 + [4] >= [0 0 1]x1 + [4] = 6(9(x1))
              [0 0 0]     [4]    [0 0 0]     [0]           
   
              [1 0 2]     [5]    [1 0 2]     [3]           
   3(9(x1)) = [0 0 1]x1 + [4] >= [0 0 0]x1 + [0] = 9(3(x1))
              [0 0 1]     [6]    [0 0 1]     [6]           
   
              [1 0 3]     [4]    [1 0 3]     [3]           
   7(5(x1)) = [0 0 2]x1 + [4] >= [0 0 0]x1 + [0] = 1(0(x1))
              [0 0 1]     [4]    [0 0 0]     [4]           
  problem:
   
  Qed