We consider the following Problem:

  Strict Trs:
    {  f(c(a(), z, x)) -> b(a(), z)
     , b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y)))
     , b(y, z) -> z}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  Arguments of following rules are not normal-forms:
  {b(x, b(z, y)) -> f(b(f(f(z)), c(x, z, y)))}
  
  All above mentioned rules can be savely removed.
  
  We consider the following Problem:
  
    Strict Trs:
      {  f(c(a(), z, x)) -> b(a(), z)
       , b(y, z) -> z}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {b(y, z) -> z}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(c) = {}, Uargs(b) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [1]
               [1 0]      [1]
       c(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                       [0 1]      [0 0]      [0 0]      [0]
       a() = [0]
             [0]
       b(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                   [0 0]      [0 1]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {f(c(a(), z, x)) -> b(a(), z)}
      Weak Trs: {b(y, z) -> z}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {f(c(a(), z, x)) -> b(a(), z)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(c) = {}, Uargs(b) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [1 0] x1 + [1]
                 [0 1]      [1]
         c(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                         [0 0]      [0 1]      [0 0]      [0]
         a() = [0]
               [0]
         b(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                     [0 0]      [0 1]      [1]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Weak Trs:
          {  f(c(a(), z, x)) -> b(a(), z)
           , b(y, z) -> z}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        We consider the following Problem:
        
          Weak Trs:
            {  f(c(a(), z, x)) -> b(a(), z)
             , b(y, z) -> z}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          Empty rules are trivially bounded

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