YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> x1
  b(a(x1)) -> c(c(x1))
  b(c(x1)) -> a(b(b(x1)))
  DP Processor:
   DPs:
    b#(c(x1)) -> b#(x1)
    b#(c(x1)) -> b#(b(x1))
    b#(c(x1)) -> a#(b(b(x1)))
   TRS:
    a(a(x1)) -> x1
    b(a(x1)) -> c(c(x1))
    b(c(x1)) -> a(b(b(x1)))
   TDG Processor:
    DPs:
     b#(c(x1)) -> b#(x1)
     b#(c(x1)) -> b#(b(x1))
     b#(c(x1)) -> a#(b(b(x1)))
    TRS:
     a(a(x1)) -> x1
     b(a(x1)) -> c(c(x1))
     b(c(x1)) -> a(b(b(x1)))
    graph:
     b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> a#(b(b(x1)))
     b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> b#(b(x1))
     b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> b#(x1)
     b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> a#(b(b(x1)))
     b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(b(x1))
     b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 6/9
     DPs:
      b#(c(x1)) -> b#(b(x1))
      b#(c(x1)) -> b#(x1)
     TRS:
      a(a(x1)) -> x1
      b(a(x1)) -> c(c(x1))
      b(c(x1)) -> a(b(b(x1)))
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(a(x1)) -> x1
       b(a(x1)) -> c(c(x1))
       b(c(x1)) -> a(b(b(x1)))
      interpretation:
       [b#](x0) = [0 1]x0 + [0],
       
                 [-& 0 ]     [0 ]
       [b](x0) = [0  -&]x0 + [-&],
       
                 [1  0 ]     [1 ]
       [a](x0) = [0  -&]x0 + [-&],
       
                 [0  -&]     [0]
       [c](x0) = [1  0 ]x0 + [1]
      orientation:
       b#(c(x1)) = [2 1]x1 + [2] >= [1 0]x1 + [0] = b#(b(x1))
       
       b#(c(x1)) = [2 1]x1 + [2] >= [0 1]x1 + [0] = b#(x1)
       
                  [2 1]     [2]           
       a(a(x1)) = [1 0]x1 + [1] >= x1 = x1
       
                  [0  -&]     [0]    [0  -&]     [0]           
       b(a(x1)) = [1  0 ]x1 + [1] >= [1  0 ]x1 + [1] = c(c(x1))
       
                  [1  0 ]     [1]    [1  0 ]     [1]              
       b(c(x1)) = [0  -&]x1 + [0] >= [0  -&]x1 + [0] = a(b(b(x1)))
      problem:
       DPs:
        b#(c(x1)) -> b#(x1)
       TRS:
        a(a(x1)) -> x1
        b(a(x1)) -> c(c(x1))
        b(c(x1)) -> a(b(b(x1)))
      Restore Modifier:
       DPs:
        b#(c(x1)) -> b#(x1)
       TRS:
        a(a(x1)) -> x1
        b(a(x1)) -> c(c(x1))
        b(c(x1)) -> a(b(b(x1)))
       EDG Processor:
        DPs:
         b#(c(x1)) -> b#(x1)
        TRS:
         a(a(x1)) -> x1
         b(a(x1)) -> c(c(x1))
         b(c(x1)) -> a(b(b(x1)))
        graph:
         b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
        Usable Rule Processor:
         DPs:
          b#(c(x1)) -> b#(x1)
         TRS:
          
         Arctic Interpretation Processor:
          dimension: 4
          usable rules:
           
          interpretation:
           [b#](x0) = [0  -& -& -&]x0,
           
                     [1 0 0 0]     [0]
                     [0 0 0 0]     [0]
           [c](x0) = [1 0 0 0]x0 + [0]
                     [0 0 0 0]     [0]
          orientation:
           b#(c(x1)) = [1 0 0 0]x1 + [0] >= [0  -& -& -&]x1 = b#(x1)
          problem:
           DPs:
            
           TRS:
            
          Qed