We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(x, y) -> x
       , g(a()) -> h(a(), b(), a())
       , i(x) -> f(x, x)
       , h(x, x, y) -> 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(x, y) -> x}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(g) = {}, Uargs(h) = {}, Uargs(i) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [1 0] x1 + [0 1] x2 + [2]
                   [0 1]      [1 0]      [0]
       g(x1) = [0 0] x1 + [1]
               [0 0]      [1]
       a() = [0]
             [0]
       h(x1, x2, x3) = [0 1] x1 + [1 0] x2 + [0 0] x3 + [1]
                       [1 0]      [0 1]      [0 0]      [1]
       b() = [0]
             [0]
       i(x1) = [1 1] x1 + [0]
               [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  g(a()) -> h(a(), b(), a())
         , i(x) -> f(x, x)
         , h(x, x, y) -> g(x)}
      Weak Trs: {f(x, y) -> 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(a()) -> h(a(), b(), a())}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(g) = {}, Uargs(h) = {}, Uargs(i) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
                     [0 1]      [0 1]      [0]
         g(x1) = [0 2] x1 + [1]
                 [0 0]      [1]
         a() = [0]
               [1]
         h(x1, x2, x3) = [0 0] x1 + [0 2] x2 + [0 0] x3 + [1]
                         [0 0]      [0 1]      [0 0]      [1]
         b() = [0]
               [0]
         i(x1) = [1 2] x1 + [0]
                 [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  i(x) -> f(x, x)
           , h(x, x, y) -> g(x)}
        Weak Trs:
          {  g(a()) -> h(a(), b(), a())
           , f(x, y) -> x}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {h(x, x, y) -> g(x)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {}, Uargs(g) = {}, Uargs(h) = {}, Uargs(i) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1, x2) = [1 0] x1 + [0 1] x2 + [0]
                       [0 1]      [1 0]      [0]
           g(x1) = [1 0] x1 + [0]
                   [0 0]      [1]
           a() = [1]
                 [0]
           h(x1, x2, x3) = [0 1] x1 + [1 0] x2 + [0 0] x3 + [1]
                           [0 0]      [0 0]      [0 0]      [1]
           b() = [0]
                 [0]
           i(x1) = [1 1] x1 + [0]
                   [0 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {i(x) -> f(x, x)}
          Weak Trs:
            {  h(x, x, y) -> g(x)
             , g(a()) -> h(a(), b(), a())
             , f(x, y) -> x}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {i(x) -> f(x, x)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {}, Uargs(g) = {}, Uargs(h) = {}, Uargs(i) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
                         [0 1]      [0 0]      [0]
             g(x1) = [0 0] x1 + [1]
                     [0 0]      [1]
             a() = [0]
                   [0]
             h(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
                             [0 1]      [1 0]      [0 0]      [1]
             b() = [0]
                   [0]
             i(x1) = [1 0] x1 + [2]
                     [0 1]      [0]
          
          The strictly oriented rules are moved into the weak component.
          
          We consider the following Problem:
          
            Weak Trs:
              {  i(x) -> f(x, x)
               , h(x, x, y) -> g(x)
               , g(a()) -> h(a(), b(), a())
               , f(x, y) -> x}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  i(x) -> f(x, x)
                 , h(x, x, y) -> g(x)
                 , g(a()) -> h(a(), b(), a())
                 , f(x, y) -> 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))