YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = max{x1 - 1, x2}
      g_A(x1,x2) = max{x1 + 3, x2 + 2}
      f_A(x1,x2) = max{x1 + 3, x2 + 3}
    
    2. max/plus interpretations on natural numbers:
    
      f#_A(x1,x2) = 0
      g_A(x1,x2) = 0
      f_A(x1,x2) = 0
    

The next rules are strictly ordered:

  p1

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