YES

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