We consider the following Problem:

  Strict Trs:
    {  w(r(x)) -> r(w(x))
     , b(r(x)) -> r(b(x))
     , b(w(x)) -> w(b(x))}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  w(r(x)) -> r(w(x))
       , b(r(x)) -> r(b(x))
       , b(w(x)) -> w(b(x))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {b(w(x)) -> w(b(x))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(w) = {1}, Uargs(r) = {1}, Uargs(b) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       w(x1) = [1 0] x1 + [0]
               [0 1]      [2]
       r(x1) = [1 0] x1 + [0]
               [0 1]      [0]
       b(x1) = [0 1] x1 + [0]
               [0 1]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  w(r(x)) -> r(w(x))
         , b(r(x)) -> r(b(x))}
      Weak Trs: {b(w(x)) -> w(b(x))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {b(r(x)) -> r(b(x))}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(w) = {1}, Uargs(r) = {1}, Uargs(b) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         w(x1) = [1 0] x1 + [0]
                 [0 1]      [0]
         r(x1) = [1 0] x1 + [0]
                 [0 1]      [1]
         b(x1) = [0 1] x1 + [1]
                 [0 1]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {w(r(x)) -> r(w(x))}
        Weak Trs:
          {  b(r(x)) -> r(b(x))
           , b(w(x)) -> w(b(x))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {w(r(x)) -> r(w(x))}
          Weak Trs:
            {  b(r(x)) -> r(b(x))
             , b(w(x)) -> w(b(x))}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The problem is match-bounded by 1.
          The enriched problem is compatible with the following automaton:
          {  w_0(2) -> 1
           , w_1(2) -> 3
           , r_0(1) -> 1
           , r_0(2) -> 2
           , r_1(3) -> 1
           , r_1(3) -> 3
           , b_0(2) -> 1}

Hurray, we answered YES(?,O(n^1))