YES

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

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