We consider the following Problem:

  Strict Trs:
    {  c() -> f()
     , f() -> g()}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  c() -> f()
       , f() -> g()}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {f() -> g()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       c() = [0]
             [0]
       f() = [2]
             [0]
       g() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {c() -> f()}
      Weak Trs: {f() -> g()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {c() -> f()}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         c() = [1]
               [0]
         f() = [0]
               [0]
         g() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Weak Trs:
          {  c() -> f()
           , f() -> g()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        We consider the following Problem:
        
          Weak Trs:
            {  c() -> f()
             , f() -> g()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          Empty rules are trivially bounded

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