We consider the following Problem:

  Strict Trs:
    {  f(X, g(X)) -> f(1(), g(X))
     , g(1()) -> g(0())}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(X, g(X)) -> f(1(), g(X))
       , g(1()) -> g(0())}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {g(1()) -> g(0())}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(g) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [1 1] x1 + [0 0] x2 + [1]
                   [0 0]      [0 0]      [0]
       g(x1) = [0 1] x1 + [0]
               [0 0]      [1]
       1() = [0]
             [2]
       0() = [0]
             [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {f(X, g(X)) -> f(1(), g(X))}
      Weak Trs: {g(1()) -> g(0())}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {f(X, g(X)) -> f(1(), g(X))}
        Weak Trs: {g(1()) -> g(0())}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The problem is match-bounded by 0.
        The enriched problem is compatible with the following automaton:
        {  f_0(2, 2) -> 1
         , g_0(2) -> 1
         , 1_0() -> 2
         , 0_0() -> 2}

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