We consider the following Problem:

  Strict Trs:
    {  f(s(x)) -> s(s(f(p(s(x)))))
     , f(0()) -> 0()
     , p(s(x)) -> x}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(s(x)) -> s(s(f(p(s(x)))))
       , f(0()) -> 0()
       , p(s(x)) -> x}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component: {f(0()) -> 0()}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {1}, Uargs(s) = {1}, Uargs(p) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1) = [1 0] x1 + [1]
               [1 0]      [1]
       s(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       p(x1) = [1 0] x1 + [0]
               [0 0]      [1]
       0() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(s(x)) -> s(s(f(p(s(x)))))
         , p(s(x)) -> x}
      Weak Trs: {f(0()) -> 0()}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The weightgap principle applies, where following rules are oriented strictly:
      
      TRS Component: {p(s(x)) -> x}
      
      Interpretation of nonconstant growth:
      -------------------------------------
        The following argument positions are usable:
          Uargs(f) = {1}, Uargs(s) = {1}, Uargs(p) = {}
        We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
        Interpretation Functions:
         f(x1) = [1 0] x1 + [2]
                 [1 0]      [0]
         s(x1) = [1 0] x1 + [1]
                 [0 1]      [0]
         p(x1) = [1 0] x1 + [3]
                 [0 1]      [1]
         0() = [0]
               [0]
      
      The strictly oriented rules are moved into the weak component.
      
      We consider the following Problem:
      
        Strict Trs: {f(s(x)) -> s(s(f(p(s(x)))))}
        Weak Trs:
          {  p(s(x)) -> x
           , f(0()) -> 0()}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(?,O(n^1))
      
      Proof:
        We consider the following Problem:
        
          Strict Trs: {f(s(x)) -> s(s(f(p(s(x)))))}
          Weak Trs:
            {  p(s(x)) -> x
             , f(0()) -> 0()}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(?,O(n^1))
        
        Proof:
          The problem is match-bounded by 1.
          The enriched problem is compatible with the following automaton:
          {  f_0(2) -> 1
           , f_1(5) -> 4
           , s_0(2) -> 1
           , s_0(2) -> 2
           , s_0(2) -> 5
           , s_1(2) -> 6
           , s_1(3) -> 1
           , s_1(3) -> 4
           , s_1(4) -> 3
           , p_0(2) -> 1
           , p_1(6) -> 5
           , 0_0() -> 1
           , 0_0() -> 2
           , 0_0() -> 5
           , 0_1() -> 4}

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