YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: f(f(x,a()),a()) -> f(f(f(a(),a()),f(x,a())),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(x,a()),a()) -> f#(f(f(a(),a()),f(x,a())),a())
p2: f#(f(x,a()),a()) -> f#(f(a(),a()),f(x,a()))

and R consists of:

r1: f(f(x,a()),a()) -> f(f(f(a(),a()),f(x,a())),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 + 4, x2 + 1}
    f_A(x1,x2) = max{0, x1 - 2, x2 - 2}
    a_A = 7

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: f(f(x,a()),a()) -> f(f(f(a(),a()),f(x,a())),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 - 3, x2 - 6}
    f_A(x1,x2) = max{2, x1 - 7, x2 - 4}
    a_A = 8

The next rules are strictly ordered:

  p1

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