YES

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

Proof:
 DP Processor:
  DPs:
   a#(c(x1)) -> b#(c(x1))
   b#(c(x1)) -> b#(x1)
   b#(c(x1)) -> a#(b(x1))
  TRS:
   a(b(x1)) -> x1
   a(c(x1)) -> c(c(b(c(x1))))
   b(c(x1)) -> a(b(x1))
  TDG Processor:
   DPs:
    a#(c(x1)) -> b#(c(x1))
    b#(c(x1)) -> b#(x1)
    b#(c(x1)) -> a#(b(x1))
   TRS:
    a(b(x1)) -> x1
    a(c(x1)) -> c(c(b(c(x1))))
    b(c(x1)) -> a(b(x1))
   graph:
    b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> a#(b(x1))
    b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
    b#(c(x1)) -> a#(b(x1)) -> a#(c(x1)) -> b#(c(x1))
    a#(c(x1)) -> b#(c(x1)) -> b#(c(x1)) -> a#(b(x1))
    a#(c(x1)) -> b#(c(x1)) -> b#(c(x1)) -> b#(x1)
   Arctic Interpretation Processor:
    dimension: 1
    interpretation:
     [b#](x0) = -4x0 + 8,
     
     [a#](x0) = x0 + 0,
     
     [c](x0) = 4x0 + 12,
     
     [a](x0) = 4x0 + 0,
     
     [b](x0) = -4x0 + 1
    orientation:
     a#(c(x1)) = 4x1 + 12 >= x1 + 8 = b#(c(x1))
     
     b#(c(x1)) = x1 + 8 >= -4x1 + 8 = b#(x1)
     
     b#(c(x1)) = x1 + 8 >= -4x1 + 1 = a#(b(x1))
     
     a(b(x1)) = x1 + 5 >= x1 = x1
     
     a(c(x1)) = 8x1 + 16 >= 8x1 + 16 = c(c(b(c(x1))))
     
     b(c(x1)) = x1 + 8 >= x1 + 5 = a(b(x1))
    problem:
     DPs:
      b#(c(x1)) -> b#(x1)
     TRS:
      a(b(x1)) -> x1
      a(c(x1)) -> c(c(b(c(x1))))
      b(c(x1)) -> a(b(x1))
    EDG Processor:
     DPs:
      b#(c(x1)) -> b#(x1)
     TRS:
      a(b(x1)) -> x1
      a(c(x1)) -> c(c(b(c(x1))))
      b(c(x1)) -> a(b(x1))
     graph:
      b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
     CDG Processor:
      DPs:
       b#(c(x1)) -> b#(x1)
      TRS:
       a(b(x1)) -> x1
       a(c(x1)) -> c(c(b(c(x1))))
       b(c(x1)) -> a(b(x1))
      graph:
       
      Qed