YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  b(a(a(x1))) -> a(a(b(b(x1))))
  b(c(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#(c(x1)) -> a#(c(x1))
   TRS:
    a(x1) -> x1
    b(a(a(x1))) -> a(a(b(b(x1))))
    b(c(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#(c(x1)) -> a#(c(x1))
    TRS:
     a(x1) -> x1
     b(a(a(x1))) -> a(a(b(b(x1))))
     b(c(x1)) -> a(c(x1))
    graph:
     b#(a(a(x1))) -> b#(b(x1)) -> b#(c(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#(c(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(c(x1)) -> a(c(x1))
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [b#](x0) = [-1 2 ]x0 + [0],
       
                 [-4 -&]     [0]
       [c](x0) = [-4 3 ]x0 + [3],
       
                 [-& 1 ]     [2 ]
       [b](x0) = [-& 0 ]x0 + [-3],
       
                 [0 1]     [3]
       [a](x0) = [0 0]x0 + [3]
      orientation:
       b#(a(a(x1))) = [2 3]x1 + [5] >= [-& 2 ]x1 + [1] = b#(b(x1))
       
       b#(a(a(x1))) = [2 3]x1 + [5] >= [-1 2 ]x1 + [0] = b#(x1)
       
               [0 1]     [3]           
       a(x1) = [0 0]x1 + [3] >= x1 = x1
       
                     [1 2]     [4]    [-& 2 ]     [4]                 
       b(a(a(x1))) = [0 1]x1 + [3] >= [-& 1 ]x1 + [3] = a(a(b(b(x1))))
       
                  [-3 4 ]     [4]    [-3 4 ]     [4]           
       b(c(x1)) = [-4 3 ]x1 + [3] >= [-4 3 ]x1 + [3] = a(c(x1))
      problem:
       DPs:
        
       TRS:
        a(x1) -> x1
        b(a(a(x1))) -> a(a(b(b(x1))))
        b(c(x1)) -> a(c(x1))
      Qed