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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = x2 + 7
          g_A(x1,x2) = max{8, x1 + 4, x2 + 6}
          f_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        f# = g = f
      
      partial status:
    
        pi(f#) = [2]
        pi(g) = [2]
        pi(f) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{2, x2 + 1}
          g_A(x1,x2) = max{3, x2 + 1}
          f_A(x1,x2) = max{x1, x2}
    
      precedence:
      
        f# = g = f
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
        pi(f) = [2]
    

The next rules are strictly ordered:

  p1

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