YES

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

Proof:
 String Reversal Processor:
  a(x1) -> b(c(b(b(x1))))
  b(b(c(x1))) -> c(a(x1))
  c(c(x1)) -> x1
  DP Processor:
   DPs:
    a#(x1) -> b#(x1)
    a#(x1) -> b#(b(x1))
    a#(x1) -> c#(b(b(x1)))
    a#(x1) -> b#(c(b(b(x1))))
    b#(b(c(x1))) -> a#(x1)
    b#(b(c(x1))) -> c#(a(x1))
   TRS:
    a(x1) -> b(c(b(b(x1))))
    b(b(c(x1))) -> c(a(x1))
    c(c(x1)) -> x1
   TDG Processor:
    DPs:
     a#(x1) -> b#(x1)
     a#(x1) -> b#(b(x1))
     a#(x1) -> c#(b(b(x1)))
     a#(x1) -> b#(c(b(b(x1))))
     b#(b(c(x1))) -> a#(x1)
     b#(b(c(x1))) -> c#(a(x1))
    TRS:
     a(x1) -> b(c(b(b(x1))))
     b(b(c(x1))) -> c(a(x1))
     c(c(x1)) -> x1
    graph:
     b#(b(c(x1))) -> a#(x1) -> a#(x1) -> b#(c(b(b(x1))))
     b#(b(c(x1))) -> a#(x1) -> a#(x1) -> c#(b(b(x1)))
     b#(b(c(x1))) -> a#(x1) -> a#(x1) -> b#(b(x1))
     b#(b(c(x1))) -> a#(x1) -> a#(x1) -> b#(x1)
     a#(x1) -> b#(c(b(b(x1)))) -> b#(b(c(x1))) -> c#(a(x1))
     a#(x1) -> b#(c(b(b(x1)))) -> b#(b(c(x1))) -> a#(x1)
     a#(x1) -> b#(b(x1)) -> b#(b(c(x1))) -> c#(a(x1))
     a#(x1) -> b#(b(x1)) -> b#(b(c(x1))) -> a#(x1)
     a#(x1) -> b#(x1) -> b#(b(c(x1))) -> c#(a(x1))
     a#(x1) -> b#(x1) -> b#(b(c(x1))) -> a#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 4
     #arcs: 10/36
     DPs:
      b#(b(c(x1))) -> a#(x1)
      a#(x1) -> b#(x1)
      a#(x1) -> b#(b(x1))
      a#(x1) -> b#(c(b(b(x1))))
     TRS:
      a(x1) -> b(c(b(b(x1))))
      b(b(c(x1))) -> c(a(x1))
      c(c(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      interpretation:
       [b#](x0) = [-& 0 ]x0 + [0],
       
       [a#](x0) = [1 0]x0 + [3],
       
                 [1  0 ]     [3]
       [c](x0) = [0  -&]x0 + [2],
       
                 [-& 0 ]     [0]
       [b](x0) = [0  -&]x0 + [1],
       
                 [0  -&]     [2]
       [a](x0) = [1  0 ]x0 + [3]
      orientation:
       b#(b(c(x1))) = [1 0]x1 + [3] >= [1 0]x1 + [3] = a#(x1)
       
       a#(x1) = [1 0]x1 + [3] >= [-& 0 ]x1 + [0] = b#(x1)
       
       a#(x1) = [1 0]x1 + [3] >= [0  -&]x1 + [1] = b#(b(x1))
       
       a#(x1) = [1 0]x1 + [3] >= [0  -&]x1 + [2] = b#(c(b(b(x1))))
       
               [0  -&]     [2]    [0  -&]     [2]                 
       a(x1) = [1  0 ]x1 + [3] >= [1  0 ]x1 + [3] = b(c(b(b(x1))))
       
                     [1  0 ]     [3]    [1  0 ]     [3]           
       b(b(c(x1))) = [0  -&]x1 + [2] >= [0  -&]x1 + [2] = c(a(x1))
       
                  [2 1]     [4]           
       c(c(x1)) = [1 0]x1 + [3] >= x1 = x1
      problem:
       DPs:
        b#(b(c(x1))) -> a#(x1)
        a#(x1) -> b#(x1)
       TRS:
        a(x1) -> b(c(b(b(x1))))
        b(b(c(x1))) -> c(a(x1))
        c(c(x1)) -> x1
      EDG Processor:
       DPs:
        b#(b(c(x1))) -> a#(x1)
        a#(x1) -> b#(x1)
       TRS:
        a(x1) -> b(c(b(b(x1))))
        b(b(c(x1))) -> c(a(x1))
        c(c(x1)) -> x1
       graph:
        b#(b(c(x1))) -> a#(x1) -> a#(x1) -> b#(x1)
        a#(x1) -> b#(x1) -> b#(b(c(x1))) -> a#(x1)
       CDG Processor:
        DPs:
         b#(b(c(x1))) -> a#(x1)
         a#(x1) -> b#(x1)
        TRS:
         a(x1) -> b(c(b(b(x1))))
         b(b(c(x1))) -> c(a(x1))
         c(c(x1)) -> x1
        graph:
         b#(b(c(x1))) -> a#(x1) -> a#(x1) -> b#(x1)
        SCC Processor:
         #sccs: 0
         #rules: 0
         #arcs: 1/4