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:
              [0  -&]  
    [c](x0) = [-& -&]x0,
    
    [a#](x0) = [0  -&]x0 + [0],
    
              [0  -&]     [-&]
    [a](x0) = [3  0 ]x0 + [2 ],
    
              [0 1]     [-&]
    [b](x0) = [3 0]x0 + [2 ]
   orientation:
    a#(b(b(x1))) = [4 1]x1 + [3] >= [0  -&]x1 + [0] = a#(x1)
    
    a#(b(b(x1))) = [4 1]x1 + [3] >= [0  -&]x1 + [0] = a#(a(x1))
    
            [0  -&]     [-&]    [0  -&]     [-&]           
    a(x1) = [3  0 ]x1 + [2 ] >= [3  -&]x1 + [2 ] = b(c(x1))
    
               [0  -&]     [-&]           
    a(a(x1)) = [3  0 ]x1 + [2 ] >= x1 = x1
    
                  [4 1]     [3]    [4 1]     [3]                 
    a(b(b(x1))) = [7 4]x1 + [6] >= [7 4]x1 + [6] = 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