We consider the following Problem:

  Strict Trs:
    {  f(a(), x) -> f(b(), f(c(), x))
     , f(a(), f(b(), x)) -> f(b(), f(a(), x))
     , f(d(), f(c(), x)) -> f(d(), f(a(), x))
     , f(a(), f(c(), x)) -> f(c(), f(a(), x))}
  StartTerms: basic terms
  Strategy: innermost

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(a(), x) -> f(b(), f(c(), x))
       , f(a(), f(b(), x)) -> f(b(), f(a(), x))
       , f(d(), f(c(), x)) -> f(d(), f(a(), x))
       , f(a(), f(c(), x)) -> f(c(), f(a(), 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(d(), f(c(), x)) -> f(d(), f(a(), x))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {2}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [0 3] x1 + [1 0] x2 + [1]
                   [0 0]      [0 1]      [1]
       a() = [0]
             [0]
       b() = [0]
             [0]
       c() = [0]
             [1]
       d() = [0]
             [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Strict Trs:
        {  f(a(), x) -> f(b(), f(c(), x))
         , f(a(), f(b(), x)) -> f(b(), f(a(), x))
         , f(a(), f(c(), x)) -> f(c(), f(a(), x))}
      Weak Trs: {f(d(), f(c(), x)) -> f(d(), f(a(), x))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      We consider the following Problem:
      
        Strict Trs:
          {  f(a(), x) -> f(b(), f(c(), x))
           , f(a(), f(b(), x)) -> f(b(), f(a(), x))
           , f(a(), f(c(), x)) -> f(c(), f(a(), x))}
        Weak Trs: {f(d(), f(c(), x)) -> f(d(), f(a(), x))}
        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, 2) -> 1
         , f_1(3, 4) -> 1
         , f_1(5, 2) -> 4
         , a_0() -> 2
         , b_0() -> 2
         , b_1() -> 3
         , c_0() -> 2
         , c_1() -> 5
         , d_0() -> 2}

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