YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> c(a(a(b(b(x1)))))
  b(x1) -> x1
  c(c(x1)) -> 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)) -> a#(a(b(b(x1))))
    b#(a(x1)) -> c#(a(a(b(b(x1)))))
   TRS:
    a(x1) -> x1
    b(a(x1)) -> c(a(a(b(b(x1)))))
    b(x1) -> x1
    c(c(x1)) -> 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)) -> a#(a(b(b(x1))))
     b#(a(x1)) -> c#(a(a(b(b(x1)))))
    TRS:
     a(x1) -> x1
     b(a(x1)) -> c(a(a(b(b(x1)))))
     b(x1) -> x1
     c(c(x1)) -> x1
    graph:
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> c#(a(a(b(b(x1)))))
     b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> a#(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(a(b(b(x1)))))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> a#(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: 10/25
     DPs:
      b#(a(x1)) -> b#(b(x1))
      b#(a(x1)) -> b#(x1)
     TRS:
      a(x1) -> x1
      b(a(x1)) -> c(a(a(b(b(x1)))))
      b(x1) -> x1
      c(c(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(x1) -> x1
       b(a(x1)) -> c(a(a(b(b(x1)))))
       b(x1) -> x1
       c(c(x1)) -> x1
      interpretation:
       [b#](x0) = [0 2]x0 + [0],
       
                 [0  2 ]     [0 ]
       [b](x0) = [-& 0 ]x0 + [-&],
       
                 [0 0]     [0]
       [a](x0) = [0 2]x0 + [0],
       
                 [0  0 ]     [2]
       [c](x0) = [0  -&]x0 + [0]
      orientation:
       b#(a(x1)) = [2 4]x1 + [2] >= [0 2]x1 + [0] = b#(b(x1))
       
       b#(a(x1)) = [2 4]x1 + [2] >= [0 2]x1 + [0] = b#(x1)
       
               [0 0]     [0]           
       a(x1) = [0 2]x1 + [0] >= x1 = x1
       
                  [2 4]     [2]    [2 4]     [2]                    
       b(a(x1)) = [0 2]x1 + [0] >= [0 2]x1 + [0] = c(a(a(b(b(x1)))))
       
               [0  2 ]     [0 ]           
       b(x1) = [-& 0 ]x1 + [-&] >= x1 = x1
       
                  [0 0]     [2]           
       c(c(x1)) = [0 0]x1 + [2] >= x1 = x1
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(x1)) -> c(a(a(b(b(x1)))))
        b(x1) -> x1
        c(c(x1)) -> x1
      Qed