We consider the following Problem:

  Strict Trs:
    {  .(1(), x) -> x
     , .(x, 1()) -> x
     , .(i(x), x) -> 1()
     , .(x, i(x)) -> 1()
     , i(1()) -> 1()
     , i(i(x)) -> x
     , .(i(y), .(y, z)) -> z
     , .(y, .(i(y), z)) -> z
     , .(.(x, y), z) -> .(x, .(y, z))
     , i(.(x, y)) -> .(i(y), i(x))}
  StartTerms: basic terms
  Strategy: innermost

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

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

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