YES

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

Proof:
 String Reversal Processor:
  a(x1) -> x1
  a(x1) -> c(b(x1))
  b(b(x1)) -> a(a(x1))
  c(c(c(x1))) -> b(x1)
  DP Processor:
   DPs:
    a#(x1) -> b#(x1)
    a#(x1) -> c#(b(x1))
    b#(b(x1)) -> a#(x1)
    b#(b(x1)) -> a#(a(x1))
    c#(c(c(x1))) -> b#(x1)
   TRS:
    a(x1) -> x1
    a(x1) -> c(b(x1))
    b(b(x1)) -> a(a(x1))
    c(c(c(x1))) -> b(x1)
   TDG Processor:
    DPs:
     a#(x1) -> b#(x1)
     a#(x1) -> c#(b(x1))
     b#(b(x1)) -> a#(x1)
     b#(b(x1)) -> a#(a(x1))
     c#(c(c(x1))) -> b#(x1)
    TRS:
     a(x1) -> x1
     a(x1) -> c(b(x1))
     b(b(x1)) -> a(a(x1))
     c(c(c(x1))) -> b(x1)
    graph:
     c#(c(c(x1))) -> b#(x1) -> b#(b(x1)) -> a#(a(x1))
     c#(c(c(x1))) -> b#(x1) -> b#(b(x1)) -> a#(x1)
     b#(b(x1)) -> a#(a(x1)) -> a#(x1) -> c#(b(x1))
     b#(b(x1)) -> a#(a(x1)) -> a#(x1) -> b#(x1)
     b#(b(x1)) -> a#(x1) -> a#(x1) -> c#(b(x1))
     b#(b(x1)) -> a#(x1) -> a#(x1) -> b#(x1)
     a#(x1) -> c#(b(x1)) -> c#(c(c(x1))) -> b#(x1)
     a#(x1) -> b#(x1) -> b#(b(x1)) -> a#(a(x1))
     a#(x1) -> b#(x1) -> b#(b(x1)) -> a#(x1)
    Arctic Interpretation Processor:
     dimension: 2
     interpretation:
      [c#](x0) = [-& 0 ]x0 + [0],
      
      [b#](x0) = [-& 0 ]x0 + [0],
      
      [a#](x0) = [0 2]x0 + [2],
      
                [-& 0 ]     [0]
      [b](x0) = [0  2 ]x0 + [2],
      
                [2  0 ]     [0]
      [c](x0) = [0  -&]x0 + [0],
      
                [0  2 ]     [2]
      [a](x0) = [-& 0 ]x0 + [0]
     orientation:
      a#(x1) = [0 2]x1 + [2] >= [-& 0 ]x1 + [0] = b#(x1)
      
      a#(x1) = [0 2]x1 + [2] >= [0 2]x1 + [2] = c#(b(x1))
      
      b#(b(x1)) = [0 2]x1 + [2] >= [0 2]x1 + [2] = a#(x1)
      
      b#(b(x1)) = [0 2]x1 + [2] >= [0 2]x1 + [2] = a#(a(x1))
      
      c#(c(c(x1))) = [2 0]x1 + [0] >= [-& 0 ]x1 + [0] = b#(x1)
      
              [0  2 ]     [2]           
      a(x1) = [-& 0 ]x1 + [0] >= x1 = x1
      
              [0  2 ]     [2]    [0  2 ]     [2]           
      a(x1) = [-& 0 ]x1 + [0] >= [-& 0 ]x1 + [0] = c(b(x1))
      
                 [0 2]     [2]    [0  2 ]     [2]           
      b(b(x1)) = [2 4]x1 + [4] >= [-& 0 ]x1 + [0] = a(a(x1))
      
                    [6 4]     [4]    [-& 0 ]     [0]        
      c(c(c(x1))) = [4 2]x1 + [2] >= [0  2 ]x1 + [2] = b(x1)
     problem:
      DPs:
       a#(x1) -> c#(b(x1))
       b#(b(x1)) -> a#(x1)
       b#(b(x1)) -> a#(a(x1))
       c#(c(c(x1))) -> b#(x1)
      TRS:
       a(x1) -> x1
       a(x1) -> c(b(x1))
       b(b(x1)) -> a(a(x1))
       c(c(c(x1))) -> b(x1)
     EDG Processor:
      DPs:
       a#(x1) -> c#(b(x1))
       b#(b(x1)) -> a#(x1)
       b#(b(x1)) -> a#(a(x1))
       c#(c(c(x1))) -> b#(x1)
      TRS:
       a(x1) -> x1
       a(x1) -> c(b(x1))
       b(b(x1)) -> a(a(x1))
       c(c(c(x1))) -> b(x1)
      graph:
       c#(c(c(x1))) -> b#(x1) -> b#(b(x1)) -> a#(x1)
       c#(c(c(x1))) -> b#(x1) -> b#(b(x1)) -> a#(a(x1))
       b#(b(x1)) -> a#(a(x1)) -> a#(x1) -> c#(b(x1))
       b#(b(x1)) -> a#(x1) -> a#(x1) -> c#(b(x1))
       a#(x1) -> c#(b(x1)) -> c#(c(c(x1))) -> b#(x1)
      Arctic Interpretation Processor:
       dimension: 2
       interpretation:
        [c#](x0) = [0 0]x0 + [0],
        
        [b#](x0) = [2 1]x0 + [3],
        
        [a#](x0) = [0 1]x0 + [2],
        
                  [-& 0 ]     [0]
        [b](x0) = [0  1 ]x0 + [2],
        
                  [1  0 ]     [2]
        [c](x0) = [0  -&]x0 + [0],
        
                  [0  1 ]     [2]
        [a](x0) = [-& 0 ]x0 + [0]
       orientation:
        a#(x1) = [0 1]x1 + [2] >= [0 1]x1 + [2] = c#(b(x1))
        
        b#(b(x1)) = [1 2]x1 + [3] >= [0 1]x1 + [2] = a#(x1)
        
        b#(b(x1)) = [1 2]x1 + [3] >= [0 1]x1 + [2] = a#(a(x1))
        
        c#(c(c(x1))) = [2 1]x1 + [3] >= [2 1]x1 + [3] = b#(x1)
        
                [0  1 ]     [2]           
        a(x1) = [-& 0 ]x1 + [0] >= x1 = x1
        
                [0  1 ]     [2]    [0  1 ]     [2]           
        a(x1) = [-& 0 ]x1 + [0] >= [-& 0 ]x1 + [0] = c(b(x1))
        
                   [0 1]     [2]    [0  1 ]     [2]           
        b(b(x1)) = [1 2]x1 + [3] >= [-& 0 ]x1 + [0] = a(a(x1))
        
                      [3 2]     [4]    [-& 0 ]     [0]        
        c(c(c(x1))) = [2 1]x1 + [3] >= [0  1 ]x1 + [2] = b(x1)
       problem:
        DPs:
         a#(x1) -> c#(b(x1))
         c#(c(c(x1))) -> b#(x1)
        TRS:
         a(x1) -> x1
         a(x1) -> c(b(x1))
         b(b(x1)) -> a(a(x1))
         c(c(c(x1))) -> b(x1)
       EDG Processor:
        DPs:
         a#(x1) -> c#(b(x1))
         c#(c(c(x1))) -> b#(x1)
        TRS:
         a(x1) -> x1
         a(x1) -> c(b(x1))
         b(b(x1)) -> a(a(x1))
         c(c(c(x1))) -> b(x1)
        graph:
         a#(x1) -> c#(b(x1)) -> c#(c(c(x1))) -> b#(x1)
        SCC Processor:
         #sccs: 0
         #rules: 0
         #arcs: 1/4