YES

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

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