YES

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

Proof:
 DP Processor:
  DPs:
   b#(c(x1)) -> b#(x1)
   b#(c(x1)) -> b#(b(x1))
   b#(c(x1)) -> a#(b(b(x1)))
  TRS:
   a(a(x1)) -> x1
   b(b(x1)) -> c(c(c(x1)))
   b(c(x1)) -> a(b(b(x1)))
  TDG Processor:
   DPs:
    b#(c(x1)) -> b#(x1)
    b#(c(x1)) -> b#(b(x1))
    b#(c(x1)) -> a#(b(b(x1)))
   TRS:
    a(a(x1)) -> x1
    b(b(x1)) -> c(c(c(x1)))
    b(c(x1)) -> a(b(b(x1)))
   graph:
    b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> a#(b(b(x1)))
    b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> b#(b(x1))
    b#(c(x1)) -> b#(b(x1)) -> b#(c(x1)) -> b#(x1)
    b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> a#(b(b(x1)))
    b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(b(x1))
    b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
   SCC Processor:
    #sccs: 1
    #rules: 2
    #arcs: 6/9
    DPs:
     b#(c(x1)) -> b#(b(x1))
     b#(c(x1)) -> b#(x1)
    TRS:
     a(a(x1)) -> x1
     b(b(x1)) -> c(c(c(x1)))
     b(c(x1)) -> a(b(b(x1)))
    Arctic Interpretation Processor:
     dimension: 3
     interpretation:
      [b#](x0) = [1  0  -&]x0 + [0],
      
                [0  1  1 ]     [1 ]
      [c](x0) = [-& -& 0 ]x0 + [-2]
                [-& -& 0 ]     [0 ],
      
                [-& -& 0 ]     [0 ]
      [b](x0) = [0  -& -&]x0 + [-&]
                [0  1  1 ]     [1 ],
      
                [-& 0  -&]     [0 ]
      [a](x0) = [0  0  -1]x0 + [1 ]
                [-& 1  -&]     [-&]
     orientation:
      b#(c(x1)) = [1 2 2]x1 + [2] >= [0  -& 1 ]x1 + [1] = b#(b(x1))
      
      b#(c(x1)) = [1 2 2]x1 + [2] >= [1  0  -&]x1 + [0] = b#(x1)
      
                 [0  0  -1]     [1]           
      a(a(x1)) = [0  0  -1]x1 + [1] >= x1 = x1
                 [1  1  0 ]     [2]           
      
                 [0  1  1 ]     [1]    [0  1  1 ]     [1]              
      b(b(x1)) = [-& -& 0 ]x1 + [0] >= [-& -& 0 ]x1 + [0] = c(c(c(x1)))
                 [1  2  2 ]     [2]    [-& -& 0 ]     [0]              
      
                 [-& -& 0 ]     [0]    [-& -& 0 ]     [0]              
      b(c(x1)) = [0  1  1 ]x1 + [1] >= [0  1  1 ]x1 + [1] = a(b(b(x1)))
                 [0  1  1 ]     [1]    [-& -& 1 ]     [1]              
     problem:
      DPs:
       b#(c(x1)) -> b#(x1)
      TRS:
       a(a(x1)) -> x1
       b(b(x1)) -> c(c(c(x1)))
       b(c(x1)) -> a(b(b(x1)))
     EDG Processor:
      DPs:
       b#(c(x1)) -> b#(x1)
      TRS:
       a(a(x1)) -> x1
       b(b(x1)) -> c(c(c(x1)))
       b(c(x1)) -> a(b(b(x1)))
      graph:
       b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1)
      CDG Processor:
       DPs:
        b#(c(x1)) -> b#(x1)
       TRS:
        a(a(x1)) -> x1
        b(b(x1)) -> c(c(c(x1)))
        b(c(x1)) -> a(b(b(x1)))
       graph:
        
       Qed