YES

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

Proof:
 String Reversal Processor:
  a(x1) -> b(x1)
  c(a(x1)) -> b(a(a(c(c(x1)))))
  b(b(x1)) -> x1
  DP Processor:
   DPs:
    a#(x1) -> b#(x1)
    c#(a(x1)) -> c#(x1)
    c#(a(x1)) -> c#(c(x1))
    c#(a(x1)) -> a#(c(c(x1)))
    c#(a(x1)) -> a#(a(c(c(x1))))
    c#(a(x1)) -> b#(a(a(c(c(x1)))))
   TRS:
    a(x1) -> b(x1)
    c(a(x1)) -> b(a(a(c(c(x1)))))
    b(b(x1)) -> x1
   TDG Processor:
    DPs:
     a#(x1) -> b#(x1)
     c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> b#(a(a(c(c(x1)))))
    TRS:
     a(x1) -> b(x1)
     c(a(x1)) -> b(a(a(c(c(x1)))))
     b(b(x1)) -> x1
    graph:
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> b#(a(a(c(c(x1)))))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> c#(c(x1)) -> c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(a(a(c(c(x1)))))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(a(c(c(x1))))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> a#(c(c(x1)))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(c(x1))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> a#(c(c(x1))) -> a#(x1) -> b#(x1)
     c#(a(x1)) -> a#(a(c(c(x1)))) -> a#(x1) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 12/36
     DPs:
      c#(a(x1)) -> c#(c(x1))
      c#(a(x1)) -> c#(x1)
     TRS:
      a(x1) -> b(x1)
      c(a(x1)) -> b(a(a(c(c(x1)))))
      b(b(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 2
      usable rules:
       a(x1) -> b(x1)
       c(a(x1)) -> b(a(a(c(c(x1)))))
       b(b(x1)) -> x1
      interpretation:
                 [0  0 ]     [0]
       [b](x0) = [0  -&]x0 + [0],
       
       [c#](x0) = [0 1]x0 + [0],
       
                 [0 0]     [0]
       [a](x0) = [0 2]x0 + [2],
       
                 [0  2 ]     [0 ]
       [c](x0) = [-& 0 ]x0 + [-&]
      orientation:
       c#(a(x1)) = [1 3]x1 + [3] >= [0 2]x1 + [0] = c#(c(x1))
       
       c#(a(x1)) = [1 3]x1 + [3] >= [0 1]x1 + [0] = c#(x1)
       
               [0 0]     [0]    [0  0 ]     [0]        
       a(x1) = [0 2]x1 + [2] >= [0  -&]x1 + [0] = b(x1)
       
                  [2 4]     [4]    [2 4]     [4]                    
       c(a(x1)) = [0 2]x1 + [2] >= [0 2]x1 + [2] = b(a(a(c(c(x1)))))
       
                  [0 0]     [0]           
       b(b(x1)) = [0 0]x1 + [0] >= x1 = x1
      problem:
       DPs:
        
       TRS:
        a(x1) -> b(x1)
        c(a(x1)) -> b(a(a(c(c(x1)))))
        b(b(x1)) -> x1
      Qed