YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: f(a(),f(a(),x)) -> f(a(),f(f(a(),x),f(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 - 2}
    a_A = 10
    f_A(x1,x2) = 1

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

and R consists of:

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

and R consists of:

r1: f(a(),f(a(),x)) -> f(a(),f(f(a(),x),f(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 - 3, x2 + 6}
    a_A = 23
    f_A(x1,x2) = max{1, x1 - 8, x2 - 5}

The next rules are strictly ordered:

  p1

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