YES

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

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