YES

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

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