YES

We show the termination of the TRS R:

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x2 + (2,4)
          a_A() = (2,3)
          f_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,0)
  
    precedence:
    
      a > f# = f
    
    partial status:
  
      pi(f#) = []
      pi(a) = []
      pi(f) = []

The next rules are strictly ordered:

  p1

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