We consider the following Problem:

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

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

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  c(c(z, y, a()), a(), a()) -> b(z, y)
       , f(c(x, y, z)) -> c(z, f(b(y, z)), a())
       , b(z, b(c(a(), y, a()), f(f(x)))) -> c(c(y, a(), z), z, x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    We consider the following Problem:
    
      Strict Trs:
        {  c(c(z, y, a()), a(), a()) -> b(z, y)
         , f(c(x, y, z)) -> c(z, f(b(y, z)), a())
         , b(z, b(c(a(), y, a()), f(f(x)))) -> c(c(y, a(), z), z, x)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(?,O(n^1))
    
    Proof:
      The following argument positions are usable:
        Uargs(c) = {1, 2}, Uargs(b) = {}, Uargs(f) = {1}
      We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       c(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
                       [1 0]      [0 0]      [2 1]      [2]
       a() = [1]
             [0]
       b(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                   [0 0]      [1 0]      [0]
       f(x1) = [1 2] x1 + [1]
               [1 1]      [2]

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