YES

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