YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = max{0, x1 - 8, x2 - 14}
    f_A(x1,x2) = max{10, x1 - 8, x2 - 5}
    a_A = 17

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

  max/plus interpretations on natural numbers:
  
    f#_A(x1,x2) = max{x1 + 5, x2 - 10}
    f_A(x1,x2) = max{0, x1 - 8}
    a_A = 9

The next rules are strictly ordered:

  p1

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