YES

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

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