We consider the following Problem:

  Strict Trs:
    {  f(a, empty()) -> g(a, empty())
     , f(a, cons(x, k)) -> f(cons(x, a), k)
     , g(empty(), d) -> d
     , g(cons(x, k), d) -> g(k, cons(x, d))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(a, empty()) -> g(a, empty())
       , f(a, cons(x, k)) -> f(cons(x, a), k)
       , g(empty(), d) -> d
       , g(cons(x, k), d) -> g(k, cons(x, d))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {f(a, empty()) -> g(a, empty())}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(g) = {}, Uargs(cons) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       empty() = [0]
                 [0]
       g(x1, x2) = [0 0] x1 + [1 2] x2 + [0]
                   [0 0]      [0 0]      [1]
       cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                      [0 0]      [0 0]      [2]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(a, cons(x, k)) -> f(cons(x, a), k)
         , g(empty(), d) -> d
         , g(cons(x, k), d) -> g(k, cons(x, d))}
      Weak Trs: {f(a, empty()) -> g(a, empty())}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {g(empty(), d) -> d}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {}, Uargs(g) = {}, Uargs(cons) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                     [0 0]      [0 0]      [1]
         empty() = [0]
                   [0]
         g(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
                     [0 0]      [0 1]      [1]
         cons(x1, x2) = [0 0] x1 + [0 0] x2 + [3]
                        [0 0]      [0 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  f(a, cons(x, k)) -> f(cons(x, a), k)
           , g(cons(x, k), d) -> g(k, cons(x, d))}
        Weak Trs:
          {  g(empty(), d) -> d
           , f(a, empty()) -> g(a, empty())}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {g(cons(x, k), d) -> g(k, cons(x, d))}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(f) = {}, Uargs(g) = {}, Uargs(cons) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           f(x1, x2) = [0 1] x1 + [0 2] x2 + [0]
                       [0 1]      [0 0]      [3]
           empty() = [0]
                     [2]
           g(x1, x2) = [0 1] x1 + [1 0] x2 + [0]
                       [0 1]      [0 1]      [1]
           cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                          [0 0]      [0 1]      [1]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {f(a, cons(x, k)) -> f(cons(x, a), k)}
          Weak Trs:
            {  g(cons(x, k), d) -> g(k, cons(x, d))
             , g(empty(), d) -> d
             , f(a, empty()) -> g(a, empty())}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The weightgap principle applies, where following rules are oriented strictly:
          
          TRS Component: {f(a, cons(x, k)) -> f(cons(x, a), k)}
          
          Interpretation of nonconstant growth:
          -------------------------------------
            The following argument positions are usable:
              Uargs(f) = {}, Uargs(g) = {}, Uargs(cons) = {}
            We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
            Interpretation Functions:
             f(x1, x2) = [0 0] x1 + [0 1] x2 + [0]
                         [0 1]      [0 1]      [3]
             empty() = [0]
                       [0]
             g(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                         [0 1]      [0 1]      [3]
             cons(x1, x2) = [0 0] x1 + [0 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(a, cons(x, k)) -> f(cons(x, a), k)
               , g(cons(x, k), d) -> g(k, cons(x, d))
               , g(empty(), d) -> d
               , f(a, empty()) -> g(a, empty())}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Weak Trs:
                {  f(a, cons(x, k)) -> f(cons(x, a), k)
                 , g(cons(x, k), d) -> g(k, cons(x, d))
                 , g(empty(), d) -> d
                 , f(a, empty()) -> g(a, empty())}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              Empty rules are trivially bounded

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