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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = ((0,0),(1,0)) x2 + (2,1)
            g_A(x1,x2) = (1,5)
            f_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(0,0)) x2 + (3,6)
    
      precedence:
      
        f# = g = f
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1,x2) = (1,1)
            g_A(x1,x2) = (1,1)
            f_A(x1,x2) = ((1,0),(1,1)) x1 + (2,2)
    
      precedence:
      
        f# > g = f
      
      partial status:
    
        pi(f#) = []
        pi(g) = []
        pi(f) = [1]
    

The next rules are strictly ordered:

  p1

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