YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(a(x1)) -> b(x1)
  b(x1) -> x1
  c(b(x1)) -> a(b(c(c(c(x1)))))
  DP Processor:
   DPs:
    a#(a(x1)) -> b#(x1)
    c#(b(x1)) -> c#(x1)
    c#(b(x1)) -> c#(c(x1))
    c#(b(x1)) -> c#(c(c(x1)))
    c#(b(x1)) -> b#(c(c(c(x1))))
    c#(b(x1)) -> a#(b(c(c(c(x1)))))
   TRS:
    a(x1) -> x1
    a(a(x1)) -> b(x1)
    b(x1) -> x1
    c(b(x1)) -> a(b(c(c(c(x1)))))
   TDG Processor:
    DPs:
     a#(a(x1)) -> b#(x1)
     c#(b(x1)) -> c#(x1)
     c#(b(x1)) -> c#(c(x1))
     c#(b(x1)) -> c#(c(c(x1)))
     c#(b(x1)) -> b#(c(c(c(x1))))
     c#(b(x1)) -> a#(b(c(c(c(x1)))))
    TRS:
     a(x1) -> x1
     a(a(x1)) -> b(x1)
     b(x1) -> x1
     c(b(x1)) -> a(b(c(c(c(x1)))))
    graph:
     c#(b(x1)) -> c#(c(c(x1))) -> c#(b(x1)) -> a#(b(c(c(c(x1)))))
     c#(b(x1)) -> c#(c(c(x1))) -> c#(b(x1)) -> b#(c(c(c(x1))))
     c#(b(x1)) -> c#(c(c(x1))) -> c#(b(x1)) -> c#(c(c(x1)))
     c#(b(x1)) -> c#(c(c(x1))) -> c#(b(x1)) -> c#(c(x1))
     c#(b(x1)) -> c#(c(c(x1))) -> c#(b(x1)) -> c#(x1)
     c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> a#(b(c(c(c(x1)))))
     c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> b#(c(c(c(x1))))
     c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> c#(c(c(x1)))
     c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> c#(c(x1))
     c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> c#(x1)
     c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> a#(b(c(c(c(x1)))))
     c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> b#(c(c(c(x1))))
     c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> c#(c(c(x1)))
     c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> c#(c(x1))
     c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> c#(x1)
     c#(b(x1)) -> a#(b(c(c(c(x1))))) -> a#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 3
     #arcs: 16/36
     DPs:
      c#(b(x1)) -> c#(c(c(x1)))
      c#(b(x1)) -> c#(x1)
      c#(b(x1)) -> c#(c(x1))
     TRS:
      a(x1) -> x1
      a(a(x1)) -> b(x1)
      b(x1) -> x1
      c(b(x1)) -> a(b(c(c(c(x1)))))
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(x1) -> x1
       a(a(x1)) -> b(x1)
       b(x1) -> x1
       c(b(x1)) -> a(b(c(c(c(x1)))))
      interpretation:
                 [0 0]     [0]
       [b](x0) = [2 2]x0 + [2],
       
       [c#](x0) = [-& 0 ]x0 + [0],
       
                 [2 0]     [0]
       [a](x0) = [2 0]x0 + [0],
       
                 [0  0 ]     [0]
       [c](x0) = [-& 0 ]x0 + [0]
      orientation:
       c#(b(x1)) = [2 2]x1 + [2] >= [-& 0 ]x1 + [0] = c#(c(c(x1)))
       
       c#(b(x1)) = [2 2]x1 + [2] >= [-& 0 ]x1 + [0] = c#(x1)
       
       c#(b(x1)) = [2 2]x1 + [2] >= [-& 0 ]x1 + [0] = c#(c(x1))
       
               [2 0]     [0]           
       a(x1) = [2 0]x1 + [0] >= x1 = x1
       
                  [4 2]     [2]    [0 0]     [0]        
       a(a(x1)) = [4 2]x1 + [2] >= [2 2]x1 + [2] = b(x1)
       
               [0 0]     [0]           
       b(x1) = [2 2]x1 + [2] >= x1 = x1
       
                  [2 2]     [2]    [2 2]     [2]                    
       c(b(x1)) = [2 2]x1 + [2] >= [2 2]x1 + [2] = a(b(c(c(c(x1)))))
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        a(a(x1)) -> b(x1)
        b(x1) -> x1
        c(b(x1)) -> a(b(c(c(c(x1)))))
      Qed