YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(a(),f(x,a())) -> f#(x,f(f(f(a(),a()),a()),a()))
p2: f#(a(),f(x,a())) -> f#(f(f(a(),a()),a()),a())
p3: f#(a(),f(x,a())) -> f#(f(a(),a()),a())
p4: f#(a(),f(x,a())) -> f#(a(),a())

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = max{x1, x2 + 6}
      a_A = 7
      f_A(x1,x2) = max{0, x1 - 5, x2 - 14}
    
    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.