We consider the following Problem:

  Strict Trs:
    {  p(m, n, s(r)) -> p(m, r, n)
     , p(m, s(n), 0()) -> p(0(), n, m)
     , p(m, 0(), 0()) -> m}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  p(m, n, s(r)) -> p(m, r, n)
       , p(m, s(n), 0()) -> p(0(), n, m)
       , p(m, 0(), 0()) -> m}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {p(m, 0(), 0()) -> m}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(p) = {}, Uargs(s) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       p(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1]
                       [0 1]      [0 0]      [0 0]      [1]
       s(x1) = [0 0] x1 + [0]
               [0 0]      [0]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  p(m, n, s(r)) -> p(m, r, n)
         , p(m, s(n), 0()) -> p(0(), n, m)}
      Weak Trs: {p(m, 0(), 0()) -> m}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {p(m, s(n), 0()) -> p(0(), n, m)}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(p) = {}, Uargs(s) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         p(x1, x2, x3) = [1 0] x1 + [1 3] x2 + [1 0] x3 + [0]
                         [0 1]      [0 0]      [0 0]      [1]
         s(x1) = [1 3] x1 + [0]
                 [0 0]      [3]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {p(m, n, s(r)) -> p(m, r, n)}
        Weak Trs:
          {  p(m, s(n), 0()) -> p(0(), n, m)
           , p(m, 0(), 0()) -> m}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        The weightgap principle applies, where following rules are oriented strictly:
        
        TRS Component: {p(m, n, s(r)) -> p(m, r, n)}
        
        Interpretation of nonconstant growth:
        -------------------------------------
          The following argument positions are usable:
            Uargs(p) = {}, Uargs(s) = {}
          We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
          Interpretation Functions:
           p(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [3]
                           [1 1]      [1 0]      [1 0]      [0]
           s(x1) = [1 0] x1 + [1]
                   [0 0]      [0]
           0() = [0]
                 [0]
        
        The strictly oriented rules are moved into the weak component.
        
        We consider the following Problem:
        
          Weak Trs:
            {  p(m, n, s(r)) -> p(m, r, n)
             , p(m, s(n), 0()) -> p(0(), n, m)
             , p(m, 0(), 0()) -> m}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Weak Trs:
              {  p(m, n, s(r)) -> p(m, r, n)
               , p(m, s(n), 0()) -> p(0(), n, m)
               , p(m, 0(), 0()) -> m}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            Empty rules are trivially bounded

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