YES

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

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