YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  c(b(a(x1))) -> a(c(c(b(x1))))
  c(x1) -> a(b(x1))
  DP Processor:
   DPs:
    c#(b(a(x1))) -> c#(b(x1))
    c#(b(a(x1))) -> c#(c(b(x1)))
    c#(b(a(x1))) -> a#(c(c(b(x1))))
    c#(x1) -> a#(b(x1))
   TRS:
    a(x1) -> x1
    c(b(a(x1))) -> a(c(c(b(x1))))
    c(x1) -> a(b(x1))
   TDG Processor:
    DPs:
     c#(b(a(x1))) -> c#(b(x1))
     c#(b(a(x1))) -> c#(c(b(x1)))
     c#(b(a(x1))) -> a#(c(c(b(x1))))
     c#(x1) -> a#(b(x1))
    TRS:
     a(x1) -> x1
     c(b(a(x1))) -> a(c(c(b(x1))))
     c(x1) -> a(b(x1))
    graph:
     c#(b(a(x1))) -> c#(b(x1)) -> c#(x1) -> a#(b(x1))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> a#(c(c(b(x1))))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> c#(c(b(x1)))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> c#(b(x1))
     c#(b(a(x1))) -> c#(c(b(x1))) -> c#(x1) -> a#(b(x1))
     c#(b(a(x1))) -> c#(c(b(x1))) -> c#(b(a(x1))) -> a#(c(c(b(x1))))
     c#(b(a(x1))) -> c#(c(b(x1))) -> c#(b(a(x1))) -> c#(c(b(x1)))
     c#(b(a(x1))) -> c#(c(b(x1))) -> c#(b(a(x1))) -> c#(b(x1))
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 8/16
     DPs:
      c#(b(a(x1))) -> c#(b(x1))
      c#(b(a(x1))) -> c#(c(b(x1)))
     TRS:
      a(x1) -> x1
      c(b(a(x1))) -> a(c(c(b(x1))))
      c(x1) -> a(b(x1))
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [c#](x0) = [0  -&]x0 + [3],
       
                 [-& 1 ]     [0]
       [b](x0) = [-& -&]x0 + [1],
       
                 [0 1]     [0]
       [c](x0) = [0 0]x0 + [3],
       
                 [0  -&]     [0]
       [a](x0) = [-& 1 ]x0 + [3]
      orientation:
       c#(b(a(x1))) = [-& 2 ]x1 + [4] >= [-& 1 ]x1 + [3] = c#(b(x1))
       
       c#(b(a(x1))) = [-& 2 ]x1 + [4] >= [-& 1 ]x1 + [3] = c#(c(b(x1)))
       
               [0  -&]     [0]           
       a(x1) = [-& 1 ]x1 + [3] >= x1 = x1
       
                     [-& 2 ]     [4]    [-& 2 ]     [4]                 
       c(b(a(x1))) = [-& 2 ]x1 + [4] >= [-& 2 ]x1 + [4] = a(c(c(b(x1))))
       
               [0 1]     [0]    [-& 1 ]     [0]           
       c(x1) = [0 0]x1 + [3] >= [-& -&]x1 + [3] = a(b(x1))
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        c(b(a(x1))) -> a(c(c(b(x1))))
        c(x1) -> a(b(x1))
      Qed