We consider the following Problem:

  Strict Trs:
    {  +(x, 0()) -> x
     , +(x, s(y)) -> s(+(x, y))
     , +(0(), s(y)) -> s(y)
     , s(+(0(), y)) -> s(y)}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  +(x, 0()) -> x
       , +(x, s(y)) -> s(+(x, y))
       , +(0(), s(y)) -> s(y)
       , s(+(0(), y)) -> s(y)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {s(+(0(), y)) -> s(y)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(+) = {}, Uargs(s) = {1}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                   [0 0]      [0 0]      [1]
       0() = [0]
             [0]
       s(x1) = [1 0] x1 + [0]
               [1 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  +(x, 0()) -> x
         , +(x, s(y)) -> s(+(x, y))
         , +(0(), s(y)) -> s(y)}
      Weak Trs: {s(+(0(), y)) -> s(y)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {+(x, 0()) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(+) = {}, Uargs(s) = {1}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                     [0 1]      [0 0]      [1]
         0() = [0]
               [0]
         s(x1) = [1 0] x1 + [0]
                 [1 0]      [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs:
          {  +(x, s(y)) -> s(+(x, y))
           , +(0(), s(y)) -> s(y)}
        Weak Trs:
          {  +(x, 0()) -> x
           , s(+(0(), y)) -> s(y)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {+(0(), s(y)) -> s(y)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(+) = {}, Uargs(s) = {1}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           +(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
                       [0 1]      [1 0]      [1]
           0() = [0]
                 [0]
           s(x1) = [1 0] x1 + [0]
                   [1 0]      [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Strict Trs: {+(x, s(y)) -> s(+(x, y))}
          Weak Trs:
            {  +(0(), s(y)) -> s(y)
             , +(x, 0()) -> x
             , s(+(0(), y)) -> s(y)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          We consider the following Problem:
          
            Strict Trs: {+(x, s(y)) -> s(+(x, y))}
            Weak Trs:
              {  +(0(), s(y)) -> s(y)
               , +(x, 0()) -> x
               , s(+(0(), y)) -> s(y)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(?,O(n^1))
          
          Proof:
            The problem is match-bounded by 0.
            The enriched problem is compatible with the following automaton:
            {  +_0(2, 2) -> 1
             , 0_0() -> 1
             , 0_0() -> 2
             , s_0(2) -> 1}

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