YES

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

Proof:
 DP Processor:
  DPs:
   a#(c(x1)) -> b#(x1)
   a#(x1) -> b#(x1)
   a#(x1) -> b#(b(x1))
   a#(x1) -> b#(b(b(x1)))
   b#(c(b(x1))) -> a#(c(x1))
  TRS:
   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))
  TDG Processor:
   DPs:
    a#(c(x1)) -> b#(x1)
    a#(x1) -> b#(x1)
    a#(x1) -> b#(b(x1))
    a#(x1) -> b#(b(b(x1)))
    b#(c(b(x1))) -> a#(c(x1))
   TRS:
    a(c(x1)) -> c(b(x1))
    a(x1) -> b(b(b(x1)))
    b(c(b(x1))) -> a(c(x1))
   graph:
    b#(c(b(x1))) -> a#(c(x1)) -> a#(x1) -> b#(b(b(x1)))
    b#(c(b(x1))) -> a#(c(x1)) -> a#(x1) -> b#(b(x1))
    b#(c(b(x1))) -> a#(c(x1)) -> a#(x1) -> b#(x1)
    b#(c(b(x1))) -> a#(c(x1)) -> a#(c(x1)) -> b#(x1)
    a#(c(x1)) -> b#(x1) -> b#(c(b(x1))) -> a#(c(x1))
    a#(x1) -> b#(b(b(x1))) -> b#(c(b(x1))) -> a#(c(x1))
    a#(x1) -> b#(b(x1)) -> b#(c(b(x1))) -> a#(c(x1))
    a#(x1) -> b#(x1) -> b#(c(b(x1))) -> a#(c(x1))
   Matrix Interpretation Processor: dim=2
    
    interpretation:
     [b#](x0) = [1 0]x0,
     
     [a#](x0) = [1 0]x0 + [3],
     
                    [1]
     [b](x0) = x0 + [0],
     
                    [5]
     [a](x0) = x0 + [4],
     
               [4 0]     [0]
     [c](x0) = [4 0]x0 + [2]
    orientation:
     a#(c(x1)) = [4 0]x1 + [3] >= [1 0]x1 = b#(x1)
     
     a#(x1) = [1 0]x1 + [3] >= [1 0]x1 = b#(x1)
     
     a#(x1) = [1 0]x1 + [3] >= [1 0]x1 + [1] = b#(b(x1))
     
     a#(x1) = [1 0]x1 + [3] >= [1 0]x1 + [2] = b#(b(b(x1)))
     
     b#(c(b(x1))) = [4 0]x1 + [4] >= [4 0]x1 + [3] = a#(c(x1))
     
                [4 0]     [5]    [4 0]     [4]           
     a(c(x1)) = [4 0]x1 + [6] >= [4 0]x1 + [6] = c(b(x1))
     
                  [5]         [3]              
     a(x1) = x1 + [4] >= x1 + [0] = b(b(b(x1)))
     
                   [4 0]     [5]    [4 0]     [5]           
     b(c(b(x1))) = [4 0]x1 + [6] >= [4 0]x1 + [6] = a(c(x1))
    problem:
     DPs:
      
     TRS:
      a(c(x1)) -> c(b(x1))
      a(x1) -> b(b(b(x1)))
      b(c(b(x1))) -> a(c(x1))
    Qed