YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
  TRS:
   a(x1) -> x1
   a(a(x1)) -> b(c(x1))
   a(b(b(x1))) -> b(b(a(a(x1))))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
    [a#](x0) = [0 1]x0 + [0],
    
              [-& 0 ]     [3]
    [b](x0) = [1  -&]x0 + [0],
    
              [-& -1]     [2]
    [c](x0) = [-4 -&]x0 + [3],
    
              [0  1 ]     [1]
    [a](x0) = [-& 0 ]x0 + [3]
   orientation:
    a#(b(b(x1))) = [1 2]x1 + [5] >= [0 1]x1 + [0] = a#(x1)
    
    a#(b(b(x1))) = [1 2]x1 + [5] >= [0 1]x1 + [4] = a#(a(x1))
    
            [0  1 ]     [1]           
    a(x1) = [-& 0 ]x1 + [3] >= x1 = x1
    
               [0  1 ]     [4]    [-4 -&]     [3]           
    a(a(x1)) = [-& 0 ]x1 + [3] >= [-& 0 ]x1 + [3] = b(c(x1))
    
                  [1  2 ]     [5]    [1  2 ]     [5]                 
    a(b(b(x1))) = [-& 1 ]x1 + [4] >= [-& 1 ]x1 + [4] = b(b(a(a(x1))))
   problem:
    DPs:
     
    TRS:
     a(x1) -> x1
     a(a(x1)) -> b(c(x1))
     a(b(b(x1))) -> b(b(a(a(x1))))
   Qed