We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))