YES

We show the termination of the TRS R:

  g(x,y) -> x
  g(x,y) -> y
  f(s(x),y,y) -> f(y,x,s(x))

-- SCC decomposition.

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

p1: f#(s(x),y,y) -> f#(y,x,s(x))

and R consists of:

r1: g(x,y) -> x
r2: g(x,y) -> y
r3: f(s(x),y,y) -> f(y,x,s(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#(s(x),y,y) -> f#(y,x,s(x))

and R consists of:

r1: g(x,y) -> x
r2: g(x,y) -> y
r3: f(s(x),y,y) -> f(y,x,s(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,x3) = max{x1 + 2, x2 + 3, x3 + 1}
        s_A(x1) = max{5, x1 + 2}
  
    precedence:
    
      f# = s
    
    partial status:
  
      pi(f#) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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