We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  +(0(), y) -> y
       , +(s(x), 0()) -> s(x)
       , +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {+(s(x), 0()) -> s(x)}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(+) = {2}, 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]
                   [1 0]      [0 0]      [1]
       0() = [0]
             [0]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  +(0(), y) -> y
         , +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
      Weak Trs: {+(s(x), 0()) -> s(x)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {+(0(), y) -> y}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(+) = {2}, 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 + [2]
                     [0 1]      [0 1]      [1]
         0() = [0]
               [2]
         s(x1) = [1 0] x1 + [3]
                 [0 0]      [3]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {+(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
        Weak Trs:
          {  +(0(), y) -> y
           , +(s(x), 0()) -> s(x)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {+(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
          Weak Trs:
            {  +(0(), y) -> y
             , +(s(x), 0()) -> s(x)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The problem is match-bounded by 2.
          The enriched problem is compatible with the following automaton:
          {  +_0(2, 2) -> 1
           , +_1(2, 6) -> 5
           , +_1(4, 5) -> 3
           , +_2(2, 10) -> 9
           , +_2(8, 9) -> 7
           , 0_0() -> 1
           , 0_0() -> 2
           , 0_1() -> 5
           , 0_1() -> 6
           , 0_2() -> 9
           , 0_2() -> 10
           , s_0(2) -> 1
           , s_0(2) -> 2
           , s_1(2) -> 3
           , s_1(2) -> 4
           , s_1(2) -> 5
           , s_1(2) -> 9
           , s_1(3) -> 1
           , s_2(2) -> 7
           , s_2(2) -> 8
           , s_2(7) -> 3
           , s_2(7) -> 7}

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