We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
       , times(X, s(Y)) -> plus(X, times(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: {times(X, s(Y)) -> plus(X, times(Y, X))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(plus) = {2}, Uargs(times) = {}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       plus(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
                      [0 1]      [0 0]      [0]
       times(x1, x2) = [0 1] x1 + [0 1] x2 + [1]
                       [0 1]      [0 0]      [1]
       s(x1) = [0 0] x1 + [0]
               [0 1]      [3]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs: {plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))}
      Weak Trs: {times(X, s(Y)) -> plus(X, times(Y, X))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs: {plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))}
        Weak Trs: {times(X, s(Y)) -> plus(X, times(Y, X))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The following argument positions are usable:
          Uargs(plus) = {2}, Uargs(times) = {}, Uargs(s) = {}
        We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         plus(x1, x2) = [3 0] x1 + [1 0] x2 + [1]
                        [0 0]      [0 1]      [0]
         times(x1, x2) = [3 3] x1 + [0 3] x2 + [0]
                         [1 1]      [1 1]      [0]
         s(x1) = [0 0] x1 + [0]
                 [1 1]      [1]

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