YES

Problem:
 b(w(x)) -> w(w(w(b(x))))
 w(b(x)) -> b(x)
 b(b(x)) -> w(w(w(w(x))))
 w(w(x)) -> w(x)

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [b](x0) = 2x0,
   
   [w](x0) = x0
  orientation:
   b(w(x)) = 2x >= 2x = w(w(w(b(x))))
   
   w(b(x)) = 2x >= 2x = b(x)
   
   b(b(x)) = 4x >= x = w(w(w(w(x))))
   
   w(w(x)) = x >= x = w(x)
  problem:
   b(w(x)) -> w(w(w(b(x))))
   w(b(x)) -> b(x)
   w(w(x)) -> w(x)
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [b](x0) = 3x0,
    
    [w](x0) = x0 + 1
   orientation:
    b(w(x)) = 3x + 3 >= 3x + 3 = w(w(w(b(x))))
    
    w(b(x)) = 3x + 1 >= 3x = b(x)
    
    w(w(x)) = x + 2 >= x + 1 = w(x)
   problem:
    b(w(x)) -> w(w(w(b(x))))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {1}
     transitions:
      f20() -> 2*
      w0(5) -> 1*
      w0(4) -> 5*
      w0(3) -> 4*
      b0(2) -> 3*
      1 -> 3*
    problem:
     
    Qed