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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = x1 + ((0,0),(1,0)) x2
      f_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,0)
      a_A() = (2,1)

The next rules are strictly ordered:

  p1, p2

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