We consider the following Problem:

  Strict Trs:
    {  plus(s(X), plus(Y, Z)) -> plus(X, plus(s(s(Y)), Z))
     , plus(s(X1), plus(X2, plus(X3, X4))) ->
       plus(X1, plus(X3, plus(X2, X4)))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  plus(s(X), plus(Y, Z)) -> plus(X, plus(s(s(Y)), Z))
       , plus(s(X1), plus(X2, plus(X3, X4))) ->
         plus(X1, plus(X3, plus(X2, X4)))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {plus(s(X1), plus(X2, plus(X3, X4))) ->
       plus(X1, plus(X3, plus(X2, X4)))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(plus) = {2}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       plus(x1, x2) = [1 1] x1 + [1 0] x2 + [0]
                      [0 0]      [0 0]      [0]
       s(x1) = [1 1] x1 + [1]
               [0 0]      [1]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {plus(s(X), plus(Y, Z)) -> plus(X, plus(s(s(Y)), Z))}
      Weak Trs:
        {plus(s(X1), plus(X2, plus(X3, X4))) ->
         plus(X1, plus(X3, plus(X2, X4)))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {plus(s(X), plus(Y, Z)) -> plus(X, plus(s(s(Y)), Z))}
        Weak Trs:
          {plus(s(X1), plus(X2, plus(X3, X4))) ->
           plus(X1, plus(X3, plus(X2, X4)))}
        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:
        {  plus_0(2, 2) -> 1
         , s_0(2) -> 2}

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