YES

Problem:
 0(1(2(3(4(5(1(x1))))))) -> 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))))))))))
 0(1(2(3(4(5(1(x1))))))) ->
 1(2(3(4(5(1(1(0(1(2(3(4(5(0(1(2(3(4(5(0(1(2(3(4(5(x1)))))))))))))))))))))))))

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