YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> x1
  a(b(x1)) -> c(b(a(a(x1))))
  c(c(x1)) -> b(x1)
  DP Processor:
   DPs:
    a#(b(x1)) -> a#(x1)
    a#(b(x1)) -> a#(a(x1))
    a#(b(x1)) -> b#(a(a(x1)))
    a#(b(x1)) -> c#(b(a(a(x1))))
    c#(c(x1)) -> b#(x1)
   TRS:
    a(x1) -> x1
    b(a(x1)) -> x1
    a(b(x1)) -> c(b(a(a(x1))))
    c(c(x1)) -> b(x1)
   TDG Processor:
    DPs:
     a#(b(x1)) -> a#(x1)
     a#(b(x1)) -> a#(a(x1))
     a#(b(x1)) -> b#(a(a(x1)))
     a#(b(x1)) -> c#(b(a(a(x1))))
     c#(c(x1)) -> b#(x1)
    TRS:
     a(x1) -> x1
     b(a(x1)) -> x1
     a(b(x1)) -> c(b(a(a(x1))))
     c(c(x1)) -> b(x1)
    graph:
     a#(b(x1)) -> c#(b(a(a(x1)))) -> c#(c(x1)) -> b#(x1)
     a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> c#(b(a(a(x1))))
     a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> b#(a(a(x1)))
     a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(a(x1))
     a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(x1)
     a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> c#(b(a(a(x1))))
     a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> b#(a(a(x1)))
     a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(a(x1))
     a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 9/25
     DPs:
      a#(b(x1)) -> a#(a(x1))
      a#(b(x1)) -> a#(x1)
     TRS:
      a(x1) -> x1
      b(a(x1)) -> x1
      a(b(x1)) -> c(b(a(a(x1))))
      c(c(x1)) -> b(x1)
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [a#](x0) = [0 0]x0,
       
                 [-& 0 ]     [1]
       [c](x0) = [2  0 ]x0 + [1],
       
                 [0 0]     [0]
       [b](x0) = [2 2]x0 + [3],
       
                 [0 0]     [0]
       [a](x0) = [0 0]x0 + [1]
      orientation:
       a#(b(x1)) = [2 2]x1 + [3] >= [0 0]x1 + [1] = a#(a(x1))
       
       a#(b(x1)) = [2 2]x1 + [3] >= [0 0]x1 = a#(x1)
       
               [0 0]     [0]           
       a(x1) = [0 0]x1 + [1] >= x1 = x1
       
                  [0 0]     [1]           
       b(a(x1)) = [2 2]x1 + [3] >= x1 = x1
       
                  [2 2]     [3]    [2 2]     [3]                 
       a(b(x1)) = [2 2]x1 + [3] >= [2 2]x1 + [3] = c(b(a(a(x1))))
       
                  [2 0]     [1]    [0 0]     [0]        
       c(c(x1)) = [2 2]x1 + [3] >= [2 2]x1 + [3] = b(x1)
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(x1)) -> x1
        a(b(x1)) -> c(b(a(a(x1))))
        c(c(x1)) -> b(x1)
      Qed