YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = max{0, x1 - 6, x2 - 4}
      a_A = 9
      f_A(x1,x2) = max{x1 + 5, x2 + 3}
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = 0
      a_A = 0
      f_A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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