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:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = max{6, x1 - 1, x2}
    f_A(x1,x2) = max{5, x1 - 16, x2 - 6}
    a_A = 19

The next rules are strictly ordered:

  p1

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