YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: f(x,f(a(),a())) -> f(f(f(a(),a()),a()),f(x,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 + 10, x2 + 12}
          f_A(x1,x2) = max{1, x1 - 5, x2 - 6}
          a_A = 8
    
      precedence:
      
        f# = f = a
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = 0
          f_A(x1,x2) = 2
          a_A = 1
    
      precedence:
      
        f > f# = a
      
      partial status:
    
        pi(f#) = []
        pi(f) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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