YES

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

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(c(b(a(x1))))
  c(b(x1)) -> x1
  b(c(x1)) -> c(a(x1))
  DP Processor:
   DPs:
    a#(a(x1)) -> b#(a(x1))
    a#(a(x1)) -> c#(b(a(x1)))
    a#(a(x1)) -> b#(c(b(a(x1))))
    b#(c(x1)) -> a#(x1)
    b#(c(x1)) -> c#(a(x1))
   TRS:
    a(a(x1)) -> b(c(b(a(x1))))
    c(b(x1)) -> x1
    b(c(x1)) -> c(a(x1))
   TDG Processor:
    DPs:
     a#(a(x1)) -> b#(a(x1))
     a#(a(x1)) -> c#(b(a(x1)))
     a#(a(x1)) -> b#(c(b(a(x1))))
     b#(c(x1)) -> a#(x1)
     b#(c(x1)) -> c#(a(x1))
    TRS:
     a(a(x1)) -> b(c(b(a(x1))))
     c(b(x1)) -> x1
     b(c(x1)) -> c(a(x1))
    graph:
     b#(c(x1)) -> a#(x1) -> a#(a(x1)) -> b#(c(b(a(x1))))
     b#(c(x1)) -> a#(x1) -> a#(a(x1)) -> c#(b(a(x1)))
     b#(c(x1)) -> a#(x1) -> a#(a(x1)) -> b#(a(x1))
     a#(a(x1)) -> b#(c(b(a(x1)))) -> b#(c(x1)) -> c#(a(x1))
     a#(a(x1)) -> b#(c(b(a(x1)))) -> b#(c(x1)) -> a#(x1)
     a#(a(x1)) -> b#(a(x1)) -> b#(c(x1)) -> c#(a(x1))
     a#(a(x1)) -> b#(a(x1)) -> b#(c(x1)) -> a#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 3
     #arcs: 7/25
     DPs:
      b#(c(x1)) -> a#(x1)
      a#(a(x1)) -> b#(a(x1))
      a#(a(x1)) -> b#(c(b(a(x1))))
     TRS:
      a(a(x1)) -> b(c(b(a(x1))))
      c(b(x1)) -> x1
      b(c(x1)) -> c(a(x1))
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(a(x1)) -> b(c(b(a(x1))))
       c(b(x1)) -> x1
       b(c(x1)) -> c(a(x1))
      interpretation:
       [b#](x0) = [0 1]x0 + [0],
       
                 [-& 0 ]     [0]
       [b](x0) = [0  1 ]x0 + [1],
       
       [a#](x0) = [0  -&]x0,
       
                 [1  0 ]     [1]
       [a](x0) = [0  -&]x0 + [0],
       
                 [-& 0 ]     [0 ]
       [c](x0) = [0  -&]x0 + [-&]
      orientation:
       b#(c(x1)) = [1 0]x1 + [0] >= [0  -&]x1 = a#(x1)
       
       a#(a(x1)) = [1 0]x1 + [1] >= [1 0]x1 + [1] = b#(a(x1))
       
       a#(a(x1)) = [1 0]x1 + [1] >= [1 0]x1 + [1] = b#(c(b(a(x1))))
       
                  [2 1]     [2]    [0  -&]     [0]                 
       a(a(x1)) = [1 0]x1 + [1] >= [1  0 ]x1 + [1] = b(c(b(a(x1))))
       
                  [0  1 ]     [1]           
       c(b(x1)) = [-& 0 ]x1 + [0] >= x1 = x1
       
                  [0  -&]     [0]    [0  -&]     [0]           
       b(c(x1)) = [1  0 ]x1 + [1] >= [1  0 ]x1 + [1] = c(a(x1))
      problem:
       DPs:
        a#(a(x1)) -> b#(a(x1))
        a#(a(x1)) -> b#(c(b(a(x1))))
       TRS:
        a(a(x1)) -> b(c(b(a(x1))))
        c(b(x1)) -> x1
        b(c(x1)) -> c(a(x1))
      Restore Modifier:
       DPs:
        a#(a(x1)) -> b#(a(x1))
        a#(a(x1)) -> b#(c(b(a(x1))))
       TRS:
        a(a(x1)) -> b(c(b(a(x1))))
        c(b(x1)) -> x1
        b(c(x1)) -> c(a(x1))
       EDG Processor:
        DPs:
         a#(a(x1)) -> b#(a(x1))
         a#(a(x1)) -> b#(c(b(a(x1))))
        TRS:
         a(a(x1)) -> b(c(b(a(x1))))
         c(b(x1)) -> x1
         b(c(x1)) -> c(a(x1))
        graph:
         
        SCC Processor:
         #sccs: 0
         #rules: 0
         #arcs: 0/4