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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((1,0),(1,0)) x2 + (2,3)
          g_A(x1,x2) = x1 + (4,2)
          f_A(x1,x2) = ((1,1),(1,1)) x1 + (1,1)
  
    precedence:
    
      f > f# = g
    
    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.