YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1,x2) = max{x1 + 14, x2 + 11}
        f_A(x1,x2) = max{2, x1 - 6, x2 - 4}
        a_A = 7
  
    precedence:
    
      f# > f = a
    
    partial status:
  
      pi(f#) = []
      pi(f) = []
      pi(a) = []

The next rules are strictly ordered:

  p1

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