YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x1)) -> a#(x1)
   c#(b(x1)) -> c#(x1)
   c#(b(x1)) -> c#(c(x1))
  TRS:
   a(b(x1)) -> b(b(a(x1)))
   c(b(x1)) -> b(c(c(x1)))
  Usable Rule Processor:
   DPs:
    a#(b(x1)) -> a#(x1)
    c#(b(x1)) -> c#(x1)
    c#(b(x1)) -> c#(c(x1))
   TRS:
    c(b(x1)) -> b(c(c(x1)))
   KBO Processor:
    weight function:
     w0 = 1
     w(c#) = w(a#) = w(b) = 1
     w(c) = 0
    precedence:
     c > c# ~ a# ~ b
    problem:
     DPs:
      
     TRS:
      
    Qed