We consider the following Problem:

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

Certificate: YES(?,O(n^2))

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

Hurray, we answered YES(?,O(n^2))