YES

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

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