YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   a#(b(x1)) -> a#(a(x1))
   a#(b(x1)) -> b#(a(a(x1)))
   b#(x1) -> a#(c(x1))
   a#(a(x1)) -> a#(c(a(x1)))
  TRS:
   a(b(x1)) -> b(a(a(x1)))
   b(x1) -> c(a(c(x1)))
   a(a(x1)) -> a(c(a(x1)))
  KBO Processor:
   argument filtering:
    pi(b) = [0]
    pi(a) = [0]
    pi(c) = []
    pi(a#) = 0
    pi(b#) = [0]
   usable rules:
    a(b(x1)) -> b(a(a(x1)))
    b(x1) -> c(a(c(x1)))
    a(a(x1)) -> a(c(a(x1)))
   weight function:
    w0 = 1
    w(c) = w(b) = 1
    w(b#) = w(a#) = w(a) = 0
   precedence:
    b# ~ a > a# ~ c ~ b
   problem:
    DPs:
     
    TRS:
     a(b(x1)) -> b(a(a(x1)))
     b(x1) -> c(a(c(x1)))
     a(a(x1)) -> a(c(a(x1)))
   Qed