YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   a#(b(x1)) -> a#(a(x1))
   a#(b(x1)) -> b#(a(a(x1)))
   a#(b(x1)) -> b#(b(a(a(x1))))
   a#(b(x1)) -> c#(b(b(a(a(x1)))))
  TRS:
   a(x1) -> x1
   a(b(x1)) -> c(b(b(a(a(x1)))))
   b(b(x1)) -> x1
   c(c(x1)) -> x1
  TDG Processor:
   DPs:
    a#(b(x1)) -> a#(x1)
    a#(b(x1)) -> a#(a(x1))
    a#(b(x1)) -> b#(a(a(x1)))
    a#(b(x1)) -> b#(b(a(a(x1))))
    a#(b(x1)) -> c#(b(b(a(a(x1)))))
   TRS:
    a(x1) -> x1
    a(b(x1)) -> c(b(b(a(a(x1)))))
    b(b(x1)) -> x1
    c(c(x1)) -> x1
   graph:
    a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> c#(b(b(a(a(x1)))))
    a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> b#(b(a(a(x1))))
    a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> b#(a(a(x1)))
    a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(a(x1))
    a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(x1)
    a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> c#(b(b(a(a(x1)))))
    a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> b#(b(a(a(x1))))
    a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> b#(a(a(x1)))
    a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(a(x1))
    a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(x1)
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 10/25
    DPs:
     a#(b(x1)) -> a#(a(x1))
     a#(b(x1)) -> a#(x1)
    TRS:
     a(x1) -> x1
     a(b(x1)) -> c(b(b(a(a(x1)))))
     b(b(x1)) -> x1
     c(c(x1)) -> x1
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [a#](x0) = [1 0]x0 + [-4],
      
                [-& 0 ]     [0]
      [c](x0) = [0  1 ]x0 + [0],
      
                [1  0 ]     [0 ]
      [b](x0) = [-& 0 ]x0 + [-&],
      
                [0  -3]     [-&]
      [a](x0) = [1  0 ]x0 + [-4]
     orientation:
      a#(b(x1)) = [2 1]x1 + [1] >= [1 0]x1 + [-4] = a#(a(x1))
      
      a#(b(x1)) = [2 1]x1 + [1] >= [1 0]x1 + [-4] = a#(x1)
      
              [0  -3]     [-&]           
      a(x1) = [1  0 ]x1 + [-4] >= x1 = x1
      
                 [1 0]     [0]    [1 0]     [0]                    
      a(b(x1)) = [2 1]x1 + [1] >= [2 1]x1 + [1] = c(b(b(a(a(x1)))))
      
                 [2  1 ]     [1 ]           
      b(b(x1)) = [-& 0 ]x1 + [-&] >= x1 = x1
      
                 [0 1]     [0]           
      c(c(x1)) = [1 2]x1 + [1] >= x1 = x1
     problem:
      DPs:
       
      TRS:
       a(x1) -> x1
       a(b(x1)) -> c(b(b(a(a(x1)))))
       b(b(x1)) -> x1
       c(c(x1)) -> x1
     Qed