YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> c(a(b(b(x1))))
  b(x1) -> x1
  c(b(c(x1))) -> a(x1)
  DP Processor:
   DPs:
    b#(a(x1)) -> b#(x1)
    b#(a(x1)) -> b#(b(x1))
    b#(a(x1)) -> a#(b(b(x1)))
    b#(a(x1)) -> c#(a(b(b(x1))))
    c#(b(c(x1))) -> a#(x1)
   TRS:
    a(x1) -> x1
    b(a(x1)) -> c(a(b(b(x1))))
    b(x1) -> x1
    c(b(c(x1))) -> a(x1)
   TDG Processor:
    DPs:
     b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> b#(b(x1))
     b#(a(x1)) -> a#(b(b(x1)))
     b#(a(x1)) -> c#(a(b(b(x1))))
     c#(b(c(x1))) -> a#(x1)
    TRS:
     a(x1) -> x1
     b(a(x1)) -> c(a(b(b(x1))))
     b(x1) -> x1
     c(b(c(x1))) -> a(x1)
    graph:
     b#(a(x1)) -> c#(a(b(b(x1)))) -> c#(b(c(x1))) -> a#(x1)
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> c#(a(b(b(x1))))
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> a#(b(b(x1)))
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> b#(b(x1))
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(a(b(b(x1))))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> a#(b(b(x1)))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(b(x1))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 9/25
     DPs:
      b#(a(x1)) -> b#(b(x1))
      b#(a(x1)) -> b#(x1)
     TRS:
      a(x1) -> x1
      b(a(x1)) -> c(a(b(b(x1))))
      b(x1) -> x1
      c(b(c(x1))) -> a(x1)
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [b#](x0) = [0 0]x0,
       
                 [0 1]     [0]
       [c](x0) = [0 0]x0 + [0],
       
                 [0 0]  
       [b](x0) = [0 0]x0,
       
                 [1 1]     [0 ]
       [a](x0) = [0 0]x0 + [-&]
      orientation:
       b#(a(x1)) = [1 1]x1 + [0] >= [0 0]x1 = b#(b(x1))
       
       b#(a(x1)) = [1 1]x1 + [0] >= [0 0]x1 = b#(x1)
       
               [1 1]     [0 ]           
       a(x1) = [0 0]x1 + [-&] >= x1 = x1
       
                  [1 1]     [0]    [1 1]     [0]                 
       b(a(x1)) = [1 1]x1 + [0] >= [1 1]x1 + [0] = c(a(b(b(x1))))
       
               [0 0]             
       b(x1) = [0 0]x1 >= x1 = x1
       
                     [1 2]     [1]    [1 1]     [0 ]        
       c(b(c(x1))) = [0 1]x1 + [0] >= [0 0]x1 + [-&] = a(x1)
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(x1)) -> c(a(b(b(x1))))
        b(x1) -> x1
        c(b(c(x1))) -> a(x1)
      Qed