YES

We show the termination of the TRS R:

  f(f(a(),a()),x) -> f(x,f(a(),f(a(),f(a(),a()))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),f(a(),a()))))
p2: f#(f(a(),a()),x) -> f#(a(),f(a(),f(a(),a())))
p3: f#(f(a(),a()),x) -> f#(a(),f(a(),a()))

and R consists of:

r1: f(f(a(),a()),x) -> f(x,f(a(),f(a(),f(a(),a()))))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(f(a(),a()),x) -> f#(x,f(a(),f(a(),f(a(),a()))))

and R consists of:

r1: f(f(a(),a()),x) -> f(x,f(a(),f(a(),f(a(),a()))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 15, x2 + 16}
          f_A(x1,x2) = max{0, x1 - 11, x2 - 6}
          a_A = 8
    
      precedence:
      
        f# > f = a
      
      partial status:
    
        pi(f#) = [2]
        pi(f) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = x2 + 6
          f_A(x1,x2) = 5
          a_A = 4
    
      precedence:
      
        f# = f = a
      
      partial status:
    
        pi(f#) = [2]
        pi(f) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.