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
     interpretation:
      [a#](x0) = [0 3]x0 + [0],
      
                [0 1]     [0]
      [c](x0) = [0 0]x0 + [1],
      
                [0  1 ]     [0]
      [b](x0) = [-& 0 ]x0 + [0],
      
                [0  1 ]     [0]
      [a](x0) = [-& 0 ]x0 + [0]
     orientation:
      a#(c(b(x1))) = [3 4]x1 + [4] >= [0 3]x1 + [3] = a#(a(x1))
      
      a#(c(b(x1))) = [3 4]x1 + [4] >= [0 3]x1 + [0] = a#(x1)
      
              [0  1 ]     [0]           
      a(x1) = [-& 0 ]x1 + [0] >= x1 = x1
      
              [0  1 ]     [0]    [0  1 ]     [0]        
      a(x1) = [-& 0 ]x1 + [0] >= [-& 0 ]x1 + [0] = b(x1)
      
                    [1 2]     [2]    [1 2]     [2]                 
      a(c(b(x1))) = [0 1]x1 + [1] >= [0 1]x1 + [1] = c(c(a(a(x1))))
      
              [0 1]     [0]    [0  1 ]     [0]        
      c(x1) = [0 0]x1 + [1] >= [-& 0 ]x1 + [0] = 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