YES

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

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