YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(x1)) -> x1
  c(a(x1)) -> b(a(a(c(c(x1)))))
  b(b(x1)) -> x1
  DP Processor:
   DPs:
    c#(a(x1)) -> c#(x1)
    c#(a(x1)) -> c#(c(x1))
    c#(a(x1)) -> a#(c(c(x1)))
    c#(a(x1)) -> a#(a(c(c(x1))))
    c#(a(x1)) -> b#(a(a(c(c(x1)))))
   TRS:
    a(x1) -> x1
    b(a(x1)) -> x1
    c(a(x1)) -> b(a(a(c(c(x1)))))
    b(b(x1)) -> x1
   TDG Processor:
    DPs:
     c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> b#(a(a(c(c(x1)))))
    TRS:
     a(x1) -> x1
     b(a(x1)) -> x1
     c(a(x1)) -> b(a(a(c(c(x1)))))
     b(b(x1)) -> x1
    graph:
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> b#(a(a(c(c(x1)))))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(a(a(c(c(x1)))))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 10/25
     DPs:
      c#(a(x1)) -> c#(c(x1))
      c#(a(x1)) -> c#(x1)
     TRS:
      a(x1) -> x1
      b(a(x1)) -> x1
      c(a(x1)) -> b(a(a(c(c(x1)))))
      b(b(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [c#](x0) = [-& 1 ]x0 + [0],
       
                 [0  3 ]  
       [c](x0) = [-& 0 ]x0,
       
                 [-& 1 ]     [0 ]
       [b](x0) = [-1 -2]x0 + [-&],
       
                 [0  -3]     [-&]
       [a](x0) = [-1 2 ]x0 + [1 ]
      orientation:
       c#(a(x1)) = [0 3]x1 + [2] >= [-& 1 ]x1 + [0] = c#(c(x1))
       
       c#(a(x1)) = [0 3]x1 + [2] >= [-& 1 ]x1 + [0] = c#(x1)
       
               [0  -3]     [-&]           
       a(x1) = [-1 2 ]x1 + [1 ] >= x1 = x1
       
                  [0  3 ]     [2 ]           
       b(a(x1)) = [-1 0 ]x1 + [-1] >= x1 = x1
       
                  [2  5 ]     [4]    [2  5 ]     [4]                    
       c(a(x1)) = [-1 2 ]x1 + [1] >= [-1 2 ]x1 + [1] = b(a(a(c(c(x1)))))
       
                  [0  -1]     [0 ]           
       b(b(x1)) = [-3 0 ]x1 + [-1] >= x1 = x1
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(x1)) -> x1
        c(a(x1)) -> b(a(a(c(c(x1)))))
        b(b(x1)) -> x1
      Qed