YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> x1
  c(a(x1)) -> a(c(c(b(x1))))
  c(b(x1)) -> b(a(x1))
  DP Processor:
   DPs:
    c#(a(x1)) -> b#(x1)
    c#(a(x1)) -> c#(b(x1))
    c#(a(x1)) -> c#(c(b(x1)))
    c#(b(x1)) -> b#(a(x1))
   TRS:
    b(a(x1)) -> x1
    c(a(x1)) -> a(c(c(b(x1))))
    c(b(x1)) -> b(a(x1))
   TDG Processor:
    DPs:
     c#(a(x1)) -> b#(x1)
     c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(c(b(x1)))
     c#(b(x1)) -> b#(a(x1))
    TRS:
     b(a(x1)) -> x1
     c(a(x1)) -> a(c(c(b(x1))))
     c(b(x1)) -> b(a(x1))
    graph:
     c#(a(x1)) -> c#(c(b(x1))) -> c#(b(x1)) -> b#(a(x1))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(c(b(x1)))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> b#(x1)
     c#(a(x1)) -> c#(b(x1)) -> c#(b(x1)) -> b#(a(x1))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(c(b(x1)))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 8/16
     DPs:
      c#(a(x1)) -> c#(c(b(x1)))
      c#(a(x1)) -> c#(b(x1))
     TRS:
      b(a(x1)) -> x1
      c(a(x1)) -> a(c(c(b(x1))))
      c(b(x1)) -> b(a(x1))
     Arctic Interpretation Processor:
      dimension: 1
      interpretation:
       [c#](x0) = x0 + 0,
       
       [c](x0) = 2x0 + 0,
       
       [a](x0) = 2x0 + 7,
       
       [b](x0) = -2x0 + 3
      orientation:
       c#(a(x1)) = 2x1 + 7 >= x1 + 5 = c#(c(b(x1)))
       
       c#(a(x1)) = 2x1 + 7 >= -2x1 + 3 = c#(b(x1))
       
       b(a(x1)) = x1 + 5 >= x1 = x1
       
       c(a(x1)) = 4x1 + 9 >= 4x1 + 9 = a(c(c(b(x1))))
       
       c(b(x1)) = x1 + 5 >= x1 + 5 = b(a(x1))
      problem:
       DPs:
        
       TRS:
        b(a(x1)) -> x1
        c(a(x1)) -> a(c(c(b(x1))))
        c(b(x1)) -> b(a(x1))
      Qed