YES

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

Proof:
 DP Processor:
  DPs:
   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))))
   c#(c(c(x1))) -> b#(x1)
  TRS:
   a(x1) -> x1
   a(b(x1)) -> b(a(c(a(x1))))
   b(x1) -> x1
   c(c(c(x1))) -> b(x1)
  TDG Processor:
   DPs:
    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))))
    c#(c(c(x1))) -> b#(x1)
   TRS:
    a(x1) -> x1
    a(b(x1)) -> b(a(c(a(x1))))
    b(x1) -> x1
    c(c(c(x1))) -> b(x1)
   graph:
    a#(b(x1)) -> c#(a(x1)) -> c#(c(c(x1))) -> b#(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#(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)
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 9/25
    DPs:
     a#(b(x1)) -> a#(c(a(x1)))
     a#(b(x1)) -> a#(x1)
    TRS:
     a(x1) -> x1
     a(b(x1)) -> b(a(c(a(x1))))
     b(x1) -> x1
     c(c(c(x1))) -> b(x1)
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      a(x1) -> x1
      a(b(x1)) -> b(a(c(a(x1))))
      b(x1) -> x1
      c(c(c(x1))) -> b(x1)
     interpretation:
                [1 1]     [1 ]
      [b](x0) = [0 0]x0 + [-&],
      
      [a#](x0) = [1  -&]x0 + [0],
      
                [1 0]     [1 ]
      [a](x0) = [0 0]x0 + [-&],
      
                [-& 0 ]     [0]
      [c](x0) = [0  1 ]x0 + [0]
     orientation:
      a#(b(x1)) = [2 2]x1 + [2] >= [1 1]x1 + [1] = a#(c(a(x1)))
      
      a#(b(x1)) = [2 2]x1 + [2] >= [1  -&]x1 + [0] = a#(x1)
      
              [1 0]     [1 ]           
      a(x1) = [0 0]x1 + [-&] >= x1 = x1
      
                 [2 2]     [2]    [2 2]     [2]                 
      a(b(x1)) = [1 1]x1 + [1] >= [1 1]x1 + [1] = b(a(c(a(x1))))
      
              [1 1]     [1 ]           
      b(x1) = [0 0]x1 + [-&] >= x1 = x1
      
                    [1 2]     [1]    [1 1]     [1 ]        
      c(c(c(x1))) = [2 3]x1 + [2] >= [0 0]x1 + [-&] = b(x1)
     problem:
      DPs:
       
      TRS:
       a(x1) -> x1
       a(b(x1)) -> b(a(c(a(x1))))
       b(x1) -> x1
       c(c(c(x1))) -> b(x1)
     Qed