YES

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

Proof:
 DP Processor:
  DPs:
   a#(x1) -> b#(c(x1))
   a#(b(b(x1))) -> a#(x1)
   a#(b(b(x1))) -> a#(a(x1))
   a#(b(b(x1))) -> b#(a(a(x1)))
   a#(b(b(x1))) -> b#(b(a(a(x1))))
  TRS:
   a(x1) -> b(c(x1))
   a(b(b(x1))) -> b(b(a(a(x1))))
   b(a(x1)) -> x1
  TDG Processor:
   DPs:
    a#(x1) -> b#(c(x1))
    a#(b(b(x1))) -> a#(x1)
    a#(b(b(x1))) -> a#(a(x1))
    a#(b(b(x1))) -> b#(a(a(x1)))
    a#(b(b(x1))) -> b#(b(a(a(x1))))
   TRS:
    a(x1) -> b(c(x1))
    a(b(b(x1))) -> b(b(a(a(x1))))
    b(a(x1)) -> x1
   graph:
    a#(b(b(x1))) -> a#(a(x1)) -> a#(b(b(x1))) -> b#(b(a(a(x1))))
    a#(b(b(x1))) -> a#(a(x1)) -> a#(b(b(x1))) -> b#(a(a(x1)))
    a#(b(b(x1))) -> a#(a(x1)) -> a#(b(b(x1))) -> a#(a(x1))
    a#(b(b(x1))) -> a#(a(x1)) -> a#(b(b(x1))) -> a#(x1)
    a#(b(b(x1))) -> a#(a(x1)) -> a#(x1) -> b#(c(x1))
    a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> b#(b(a(a(x1))))
    a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> b#(a(a(x1)))
    a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> a#(a(x1))
    a#(b(b(x1))) -> a#(x1) -> a#(b(b(x1))) -> a#(x1)
    a#(b(b(x1))) -> a#(x1) -> a#(x1) -> b#(c(x1))
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 10/25
    DPs:
     a#(b(b(x1))) -> a#(a(x1))
     a#(b(b(x1))) -> a#(x1)
    TRS:
     a(x1) -> b(c(x1))
     a(b(b(x1))) -> b(b(a(a(x1))))
     b(a(x1)) -> x1
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      a(x1) -> b(c(x1))
      a(b(b(x1))) -> b(b(a(a(x1))))
      b(a(x1)) -> x1
     interpretation:
      [a#](x0) = [0  -&]x0,
      
                [0 3]     [0]
      [b](x0) = [0 0]x0 + [2],
      
                [0 ]
      [c](x0) = [-&],
      
                [0 0]     [0]
      [a](x0) = [0 0]x0 + [2]
     orientation:
      a#(b(b(x1))) = [3 3]x1 + [5] >= [0 0]x1 + [0] = a#(a(x1))
      
      a#(b(b(x1))) = [3 3]x1 + [5] >= [0  -&]x1 = a#(x1)
      
              [0 0]     [0]    [0]           
      a(x1) = [0 0]x1 + [2] >= [2] = b(c(x1))
      
                    [3 3]     [5]    [3 3]     [5]                 
      a(b(b(x1))) = [3 3]x1 + [5] >= [3 3]x1 + [5] = b(b(a(a(x1))))
      
                 [3 3]     [5]           
      b(a(x1)) = [0 0]x1 + [2] >= x1 = x1
     problem:
      DPs:
       
      TRS:
       a(x1) -> b(c(x1))
       a(b(b(x1))) -> b(b(a(a(x1))))
       b(a(x1)) -> x1
     Qed