We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(1()) -> f(g(1()))
       , f(f(x)) -> f(x)
       , g(0()) -> g(f(0()))
       , g(g(x)) -> g(x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {f(f(x)) -> f(x)}
    
    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) = [1 0] x1 + [1]
               [0 1]      [1]
       1() = [0]
             [0]
       g(x1) = [1 0] x1 + [0]
               [0 0]      [0]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(1()) -> f(g(1()))
         , g(0()) -> g(f(0()))
         , g(g(x)) -> g(x)}
      Weak Trs: {f(f(x)) -> f(x)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {g(g(x)) -> g(x)}
      
      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) = [0 0] x1 + [1]
                 [0 0]      [1]
         1() = [0]
               [0]
         g(x1) = [1 0] x1 + [1]
                 [0 0]      [1]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  f(1()) -> f(g(1()))
           , g(0()) -> g(f(0()))}
        Weak Trs:
          {  g(g(x)) -> g(x)
           , f(f(x)) -> f(x)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {g(0()) -> g(f(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) = [0 0] x1 + [1]
                   [0 0]      [1]
           1() = [0]
                 [0]
           g(x1) = [0 1] x1 + [1]
                   [0 1]      [1]
           0() = [0]
                 [2]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {f(1()) -> f(g(1()))}
          Weak Trs:
            {  g(0()) -> g(f(0()))
             , g(g(x)) -> g(x)
             , f(f(x)) -> f(x)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {f(1()) -> f(g(1()))}
          
          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) = [1 1] x1 + [1]
                     [0 0]      [1]
             1() = [0]
                   [2]
             g(x1) = [0 0] x1 + [0]
                     [0 0]      [0]
             0() = [0]
                   [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  f(1()) -> f(g(1()))
               , g(0()) -> g(f(0()))
               , g(g(x)) -> g(x)
               , f(f(x)) -> f(x)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  f(1()) -> f(g(1()))
                 , g(0()) -> g(f(0()))
                 , g(g(x)) -> g(x)
                 , f(f(x)) -> f(x)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

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