We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  i(0()) -> 0()
       , +(0(), y) -> y
       , +(x, 0()) -> x
       , i(i(x)) -> x
       , +(i(x), x) -> 0()
       , +(x, i(x)) -> 0()
       , i(+(x, y)) -> +(i(x), i(y))
       , +(x, +(y, z)) -> +(+(x, y), z)
       , +(+(x, i(y)), y) -> x
       , +(+(x, y), i(y)) -> x}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  i(0()) -> 0()
       , i(i(x)) -> x
       , +(i(x), x) -> 0()
       , +(x, i(x)) -> 0()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(i) = {}, Uargs(+) = {1, 2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       i(x1) = [1 0] x1 + [1]
               [0 1]      [1]
       0() = [0]
             [0]
       +(x1, x2) = [1 0] x1 + [1 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:
        {  +(0(), y) -> y
         , +(x, 0()) -> x
         , i(+(x, y)) -> +(i(x), i(y))
         , +(x, +(y, z)) -> +(+(x, y), z)
         , +(+(x, i(y)), y) -> x
         , +(+(x, y), i(y)) -> x}
      Weak Trs:
        {  i(0()) -> 0()
         , i(i(x)) -> x
         , +(i(x), x) -> 0()
         , +(x, i(x)) -> 0()}
      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
         , +(x, 0()) -> x
         , +(+(x, i(y)), y) -> x
         , +(+(x, y), i(y)) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(i) = {}, Uargs(+) = {1, 2}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         i(x1) = [0 1] x1 + [0]
                 [1 0]      [0]
         0() = [0]
               [0]
         +(x1, x2) = [1 0] x1 + [1 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:
          {  i(+(x, y)) -> +(i(x), i(y))
           , +(x, +(y, z)) -> +(+(x, y), z)}
        Weak Trs:
          {  +(0(), y) -> y
           , +(x, 0()) -> x
           , +(+(x, i(y)), y) -> x
           , +(+(x, y), i(y)) -> x
           , i(0()) -> 0()
           , i(i(x)) -> x
           , +(i(x), x) -> 0()
           , +(x, i(x)) -> 0()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs:
            {  i(+(x, y)) -> +(i(x), i(y))
             , +(x, +(y, z)) -> +(+(x, y), z)}
          Weak Trs:
            {  +(0(), y) -> y
             , +(x, 0()) -> x
             , +(+(x, i(y)), y) -> x
             , +(+(x, y), i(y)) -> x
             , i(0()) -> 0()
             , i(i(x)) -> x
             , +(i(x), x) -> 0()
             , +(x, i(x)) -> 0()}
          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:
          {  i_0(2) -> 1
           , 0_0() -> 1
           , 0_0() -> 2
           , +_0(2, 2) -> 1}

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