YES

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

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