YES

Problem:
 a(b(c(x1))) -> a(a(b(x1)))
 a(b(c(x1))) -> b(c(b(c(x1))))
 a(b(c(x1))) -> c(b(c(a(x1))))

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