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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{22, x1 + 8, x2 + 6}
        a_A = 13
        f_A(x1,x2) = max{20, x1 + 7, x2 + 10}
  
    precedence:
    
      f# = a = f
    
    partial status:
  
      pi(f#) = [2]
      pi(a) = []
      pi(f) = [2]

The next rules are strictly ordered:

  p1

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