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
     usable rules:
      a(x1) -> x1
      a(x1) -> c(b(x1))
      b(b(x1)) -> a(a(x1))
      c(c(c(x1))) -> b(x1)
     interpretation:
      [b#](x0) = [2 0]x0 + [0],
      
                [-& 0 ]     [0]
      [c](x0) = [0  1 ]x0 + [1],
      
      [a#](x0) = [2 0]x0 + [3],
      
      [c#](x0) = [-& 2 ]x0 + [0],
      
                [0  -&]     [0]
      [a](x0) = [1  0 ]x0 + [2],
      
                [1  0 ]     [2]
      [b](x0) = [0  -&]x0 + [0]
     orientation:
      a#(x1) = [2 0]x1 + [3] >= [2 0]x1 + [0] = b#(x1)
      
      a#(x1) = [2 0]x1 + [3] >= [2  -&]x1 + [2] = c#(b(x1))
      
      b#(b(x1)) = [3 2]x1 + [4] >= [2 0]x1 + [3] = a#(x1)
      
      b#(b(x1)) = [3 2]x1 + [4] >= [2 0]x1 + [3] = a#(a(x1))
      
      c#(c(c(x1))) = [3 4]x1 + [4] >= [2 0]x1 + [0] = b#(x1)
      
              [0  -&]     [0]           
      a(x1) = [1  0 ]x1 + [2] >= x1 = x1
      
              [0  -&]     [0]    [0  -&]     [0]           
      a(x1) = [1  0 ]x1 + [2] >= [1  0 ]x1 + [2] = c(b(x1))
      
                 [2 1]     [3]    [0  -&]     [0]           
      b(b(x1)) = [1 0]x1 + [2] >= [1  0 ]x1 + [2] = a(a(x1))
      
                    [1 2]     [2]    [1  0 ]     [2]        
      c(c(c(x1))) = [2 3]x1 + [3] >= [0  -&]x1 + [0] = b(x1)
     problem:
      DPs:
       a#(x1) -> b#(x1)
       a#(x1) -> c#(b(x1))
      TRS:
       a(x1) -> x1
       a(x1) -> c(b(x1))
       b(b(x1)) -> a(a(x1))
       c(c(c(x1))) -> b(x1)
     Restore Modifier:
      DPs:
       a#(x1) -> b#(x1)
       a#(x1) -> c#(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) -> b#(x1)
        a#(x1) -> c#(b(x1))
       TRS:
        a(x1) -> x1
        a(x1) -> c(b(x1))
        b(b(x1)) -> a(a(x1))
        c(c(c(x1))) -> b(x1)
       graph:
        
       SCC Processor:
        #sccs: 0
        #rules: 0
        #arcs: 0/4