YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(x1) -> b(b(x1))
  c(a(b(x1))) -> a(a(c(c(x1))))
  c(x1) -> x1
  DP Processor:
   DPs:
    c#(a(b(x1))) -> c#(x1)
    c#(a(b(x1))) -> c#(c(x1))
    c#(a(b(x1))) -> a#(c(c(x1)))
    c#(a(b(x1))) -> a#(a(c(c(x1))))
   TRS:
    a(x1) -> x1
    a(x1) -> b(b(x1))
    c(a(b(x1))) -> a(a(c(c(x1))))
    c(x1) -> x1
   TDG Processor:
    DPs:
     c#(a(b(x1))) -> c#(x1)
     c#(a(b(x1))) -> c#(c(x1))
     c#(a(b(x1))) -> a#(c(c(x1)))
     c#(a(b(x1))) -> a#(a(c(c(x1))))
    TRS:
     a(x1) -> x1
     a(x1) -> b(b(x1))
     c(a(b(x1))) -> a(a(c(c(x1))))
     c(x1) -> x1
    graph:
     c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> a#(a(c(c(x1))))
     c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> a#(c(c(x1)))
     c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> c#(c(x1))
     c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> c#(x1)
     c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> a#(a(c(c(x1))))
     c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> a#(c(c(x1)))
     c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> c#(c(x1))
     c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> c#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 8/16
     DPs:
      c#(a(b(x1))) -> c#(c(x1))
      c#(a(b(x1))) -> c#(x1)
     TRS:
      a(x1) -> x1
      a(x1) -> b(b(x1))
      c(a(b(x1))) -> a(a(c(c(x1))))
      c(x1) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [c#](x0) = [-4 -3]x0 + [0],
       
                 [0  1 ]  
       [c](x0) = [-& 0 ]x0,
       
                 [0  1 ]     [2 ]
       [b](x0) = [-1 0 ]x0 + [-1],
       
                 [0 1]     [2]
       [a](x0) = [3 0]x0 + [1]
      orientation:
       c#(a(b(x1))) = [0 1]x1 + [2] >= [-4 -3]x1 + [0] = c#(c(x1))
       
       c#(a(b(x1))) = [0 1]x1 + [2] >= [-4 -3]x1 + [0] = c#(x1)
       
               [0 1]     [2]           
       a(x1) = [3 0]x1 + [1] >= x1 = x1
       
               [0 1]     [2]    [0  1 ]     [2]           
       a(x1) = [3 0]x1 + [1] >= [-1 0 ]x1 + [1] = b(b(x1))
       
                     [4 5]     [6]    [4 5]     [2]                 
       c(a(b(x1))) = [3 4]x1 + [5] >= [3 4]x1 + [5] = a(a(c(c(x1))))
       
               [0  1 ]             
       c(x1) = [-& 0 ]x1 >= x1 = x1
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        a(x1) -> b(b(x1))
        c(a(b(x1))) -> a(a(c(c(x1))))
        c(x1) -> x1
      Qed