YES

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

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