We consider the following Problem:

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

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

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

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