YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> x1
  c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
  DP Processor:
   DPs:
    c#(b(a(x1))) -> b#(x1)
    c#(b(a(x1))) -> c#(b(x1))
    c#(b(a(x1))) -> b#(c(b(x1)))
    c#(b(a(x1))) -> c#(b(c(b(x1))))
    c#(b(a(x1))) -> b#(a(a(c(b(c(b(x1)))))))
   TRS:
    b(a(x1)) -> x1
    c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
   TDG Processor:
    DPs:
     c#(b(a(x1))) -> b#(x1)
     c#(b(a(x1))) -> c#(b(x1))
     c#(b(a(x1))) -> b#(c(b(x1)))
     c#(b(a(x1))) -> c#(b(c(b(x1))))
     c#(b(a(x1))) -> b#(a(a(c(b(c(b(x1)))))))
    TRS:
     b(a(x1)) -> x1
     c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
    graph:
     c#(b(a(x1))) -> c#(b(c(b(x1)))) ->
     c#(b(a(x1))) -> b#(a(a(c(b(c(b(x1)))))))
     c#(b(a(x1))) -> c#(b(c(b(x1)))) ->
     c#(b(a(x1))) -> c#(b(c(b(x1))))
     c#(b(a(x1))) -> c#(b(c(b(x1)))) -> c#(b(a(x1))) -> b#(c(b(x1)))
     c#(b(a(x1))) -> c#(b(c(b(x1)))) -> c#(b(a(x1))) -> c#(b(x1))
     c#(b(a(x1))) -> c#(b(c(b(x1)))) -> c#(b(a(x1))) -> b#(x1)
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> b#(a(a(c(b(c(b(x1)))))))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> c#(b(c(b(x1))))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> b#(c(b(x1)))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> c#(b(x1))
     c#(b(a(x1))) -> c#(b(x1)) -> c#(b(a(x1))) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 10/25
     DPs:
      c#(b(a(x1))) -> c#(b(c(b(x1))))
      c#(b(a(x1))) -> c#(b(x1))
     TRS:
      b(a(x1)) -> x1
      c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
     Arctic Interpretation Processor:
      dimension: 1
      usable rules:
       b(a(x1)) -> x1
       c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
      interpretation:
       [c#](x0) = -8x0 + 0,
       
       [c](x0) = 3x0 + 0,
       
       [a](x0) = 3x0 + 13,
       
       [b](x0) = -3x0 + 0
      orientation:
       c#(b(a(x1))) = -8x1 + 2 >= -11x1 + 0 = c#(b(c(b(x1))))
       
       c#(b(a(x1))) = -8x1 + 2 >= -11x1 + 0 = c#(b(x1))
       
       b(a(x1)) = x1 + 10 >= x1 = x1
       
       c(b(a(x1))) = 3x1 + 13 >= 3x1 + 13 = b(a(a(c(b(c(b(x1)))))))
      problem:
       DPs:
        
       TRS:
        b(a(x1)) -> x1
        c(b(a(x1))) -> b(a(a(c(b(c(b(x1)))))))
      Qed