YES

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:
 DP Processor:
  DPs:
   5#(9(x1)) -> 5#(x1)
   5#(9(x1)) -> 2#(6(5(x1)))
   3#(5(x1)) -> 7#(x1)
   3#(5(x1)) -> 9#(7(x1))
   3#(5(x1)) -> 8#(9(7(x1)))
   9#(x1) -> 3#(x1)
   9#(x1) -> 2#(3(x1))
   9#(x1) -> 3#(2(3(x1)))
   2#(6(x1)) -> 3#(x1)
   3#(8(x1)) -> 7#(x1)
   3#(8(x1)) -> 2#(7(x1))
   3#(8(x1)) -> 3#(2(7(x1)))
   9#(x1) -> 2#(x1)
   9#(x1) -> 5#(0(2(x1)))
   8#(8(4(x1))) -> 9#(x1)
   7#(1(x1)) -> 9#(x1)
   3#(9(x1)) -> 3#(x1)
   3#(9(x1)) -> 9#(3(x1))
  TRS:
   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))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
    [8#](x0) = [1 1]x0,
    
    [9#](x0) = [4 0]x0 + [3],
    
    [7#](x0) = [0 1]x0 + [3],
    
    [2#](x0) = [4 0]x0 + [1],
    
    [5#](x0) = [6 4]x0,
    
    [3#](x0) = [4 0]x0,
    
              [0]
    [0](x0) = [0],
    
              [0 1]     [1]
    [8](x0) = [1 0]x0 + [4],
    
              [0 1]  
    [7](x0) = [1 0]x0,
    
              [0 0]  
    [2](x0) = [4 0]x0,
    
              [1 0]  
    [6](x0) = [0 0]x0,
    
              [1 1]     [1]
    [5](x0) = [4 4]x0 + [0],
    
                   [1]
    [9](x0) = x0 + [0],
    
              [0 0]  
    [4](x0) = [4 0]x0,
    
              [1 0]  
    [3](x0) = [6 0]x0,
    
              [0 0]     [0]
    [1](x0) = [4 0]x0 + [1]
   orientation:
    5#(9(x1)) = [6 4]x1 + [6] >= [6 4]x1 = 5#(x1)
    
    5#(9(x1)) = [6 4]x1 + [6] >= [4 4]x1 + [5] = 2#(6(5(x1)))
    
    3#(5(x1)) = [4 4]x1 + [4] >= [0 1]x1 + [3] = 7#(x1)
    
    3#(5(x1)) = [4 4]x1 + [4] >= [0 4]x1 + [3] = 9#(7(x1))
    
    3#(5(x1)) = [4 4]x1 + [4] >= [1 1]x1 + [1] = 8#(9(7(x1)))
    
    9#(x1) = [4 0]x1 + [3] >= [4 0]x1 = 3#(x1)
    
    9#(x1) = [4 0]x1 + [3] >= [4 0]x1 + [1] = 2#(3(x1))
    
    9#(x1) = [4 0]x1 + [3] >= [0] = 3#(2(3(x1)))
    
    2#(6(x1)) = [4 0]x1 + [1] >= [4 0]x1 = 3#(x1)
    
    3#(8(x1)) = [0 4]x1 + [4] >= [0 1]x1 + [3] = 7#(x1)
    
    3#(8(x1)) = [0 4]x1 + [4] >= [0 4]x1 + [1] = 2#(7(x1))
    
    3#(8(x1)) = [0 4]x1 + [4] >= [0] = 3#(2(7(x1)))
    
    9#(x1) = [4 0]x1 + [3] >= [4 0]x1 + [1] = 2#(x1)
    
    9#(x1) = [4 0]x1 + [3] >= [0] = 5#(0(2(x1)))
    
    8#(8(4(x1))) = [4 0]x1 + [5] >= [4 0]x1 + [3] = 9#(x1)
    
    7#(1(x1)) = [4 0]x1 + [4] >= [4 0]x1 + [3] = 9#(x1)
    
    3#(9(x1)) = [4 0]x1 + [4] >= [4 0]x1 = 3#(x1)
    
    3#(9(x1)) = [4 0]x1 + [4] >= [4 0]x1 + [3] = 9#(3(x1))
    
               [0]    [0]           
    3(1(x1)) = [0] >= [0] = 4(1(x1))
    
               [1 1]     [2]    [0 0]     [0]              
    5(9(x1)) = [4 4]x1 + [4] >= [4 4]x1 + [4] = 2(6(5(x1)))
    
               [1 1]     [1]         [1]              
    3(5(x1)) = [6 6]x1 + [6] >= x1 + [5] = 8(9(7(x1)))
    
                 [1]    [0]              
    9(x1) = x1 + [0] >= [0] = 3(2(3(x1)))
    
               [4 0]     [1]    [1 0]          
    8(4(x1)) = [0 0]x1 + [4] >= [0 0]x1 = 6(x1)
    
               [0 0]      [0 0]             
    2(6(x1)) = [4 0]x1 >= [4 0]x1 = 4(3(x1))
    
               [0 1]     [1]    [0]              
    3(8(x1)) = [0 6]x1 + [6] >= [0] = 3(2(7(x1)))
    
                 [1]    [1]              
    9(x1) = x1 + [0] >= [0] = 5(0(2(x1)))
    
                  [0 0]     [5]    [0 0]     [0]           
    8(8(4(x1))) = [4 0]x1 + [5] >= [4 0]x1 + [5] = 1(9(x1))
    
               [4 0]     [1]    [1 0]     [1]           
    7(1(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = 6(9(x1))
    
               [1 0]     [1]    [1 0]     [1]           
    3(9(x1)) = [6 0]x1 + [6] >= [6 0]x1 + [0] = 9(3(x1))
    
               [4 4]     [0]    [0]           
    7(5(x1)) = [1 1]x1 + [1] >= [1] = 1(0(x1))
   problem:
    DPs:
     
    TRS:
     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))
   Qed