YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> c(a(b(x1)))
  b(b(x1)) -> x1
  c(c(x1)) -> a(b(x1))
  DP Processor:
   DPs:
    b#(a(x1)) -> b#(x1)
    b#(a(x1)) -> a#(b(x1))
    b#(a(x1)) -> c#(a(b(x1)))
    c#(c(x1)) -> b#(x1)
    c#(c(x1)) -> a#(b(x1))
   TRS:
    a(x1) -> x1
    b(a(x1)) -> c(a(b(x1)))
    b(b(x1)) -> x1
    c(c(x1)) -> a(b(x1))
   TDG Processor:
    DPs:
     b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> a#(b(x1))
     b#(a(x1)) -> c#(a(b(x1)))
     c#(c(x1)) -> b#(x1)
     c#(c(x1)) -> a#(b(x1))
    TRS:
     a(x1) -> x1
     b(a(x1)) -> c(a(b(x1)))
     b(b(x1)) -> x1
     c(c(x1)) -> a(b(x1))
    graph:
     c#(c(x1)) -> b#(x1) -> b#(a(x1)) -> c#(a(b(x1)))
     c#(c(x1)) -> b#(x1) -> b#(a(x1)) -> a#(b(x1))
     c#(c(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> c#(a(b(x1))) -> c#(c(x1)) -> a#(b(x1))
     b#(a(x1)) -> c#(a(b(x1))) -> c#(c(x1)) -> b#(x1)
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(a(b(x1)))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> a#(b(x1))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 3
     #arcs: 8/25
     DPs:
      c#(c(x1)) -> b#(x1)
      b#(a(x1)) -> b#(x1)
      b#(a(x1)) -> c#(a(b(x1)))
     TRS:
      a(x1) -> x1
      b(a(x1)) -> c(a(b(x1)))
      b(b(x1)) -> x1
      c(c(x1)) -> a(b(x1))
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [c#](x0) = [-& 0 ]x0 + [0],
       
       [b#](x0) = [-& 0 ]x0 + [0],
       
                 [0 2]     [-&]
       [c](x0) = [0 2]x0 + [2 ],
       
                 [0 0]     [-&]
       [b](x0) = [0 0]x0 + [0 ],
       
                 [2 2]     [2]
       [a](x0) = [0 0]x0 + [0]
      orientation:
       c#(c(x1)) = [0 2]x1 + [2] >= [-& 0 ]x1 + [0] = b#(x1)
       
       b#(a(x1)) = [0 0]x1 + [0] >= [-& 0 ]x1 + [0] = b#(x1)
       
       b#(a(x1)) = [0 0]x1 + [0] >= [0 0]x1 + [0] = c#(a(b(x1)))
       
               [2 2]     [2]           
       a(x1) = [0 0]x1 + [0] >= x1 = x1
       
                  [2 2]     [2]    [2 2]     [2]              
       b(a(x1)) = [2 2]x1 + [2] >= [2 2]x1 + [2] = c(a(b(x1)))
       
                  [0 0]     [0]           
       b(b(x1)) = [0 0]x1 + [0] >= x1 = x1
       
                  [2 4]     [4]    [2 2]     [2]           
       c(c(x1)) = [2 4]x1 + [4] >= [0 0]x1 + [0] = a(b(x1))
      problem:
       DPs:
        b#(a(x1)) -> b#(x1)
        b#(a(x1)) -> c#(a(b(x1)))
       TRS:
        a(x1) -> x1
        b(a(x1)) -> c(a(b(x1)))
        b(b(x1)) -> x1
        c(c(x1)) -> a(b(x1))
      EDG Processor:
       DPs:
        b#(a(x1)) -> b#(x1)
        b#(a(x1)) -> c#(a(b(x1)))
       TRS:
        a(x1) -> x1
        b(a(x1)) -> c(a(b(x1)))
        b(b(x1)) -> x1
        c(c(x1)) -> a(b(x1))
       graph:
        b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1)
        b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(a(b(x1)))
       CDG Processor:
        DPs:
         b#(a(x1)) -> b#(x1)
         b#(a(x1)) -> c#(a(b(x1)))
        TRS:
         a(x1) -> x1
         b(a(x1)) -> c(a(b(x1)))
         b(b(x1)) -> x1
         c(c(x1)) -> a(b(x1))
        graph:
         
        Qed