YES

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

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