YES

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

Proof:
 DP Processor:
  DPs:
   a#(x1) -> b#(x1)
   a#(x1) -> b#(b(x1))
   b#(b(b(x1))) -> a#(x1)
  TRS:
   a(x1) -> b(b(x1))
   b(b(b(x1))) -> a(x1)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [b#](x0) = 1x0,
    
    [a#](x0) = 8x0,
    
    [b](x0) = 5x0,
    
    [a](x0) = 14x0
   orientation:
    a#(x1) = 8x1 >= 1x1 = b#(x1)
    
    a#(x1) = 8x1 >= 6x1 = b#(b(x1))
    
    b#(b(b(x1))) = 11x1 >= 8x1 = a#(x1)
    
    a(x1) = 14x1 >= 10x1 = b(b(x1))
    
    b(b(b(x1))) = 15x1 >= 14x1 = a(x1)
   problem:
    DPs:
     
    TRS:
     a(x1) -> b(b(x1))
     b(b(b(x1))) -> a(x1)
   Qed