YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  r1

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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