YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

and R consists of:

r1: f(g(x),y,y) -> g(f(x,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,x3) = ((0,1),(0,1)) x1 + (2,2)
          g_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      g > f#
    
    partial status:
  
      pi(f#) = []
      pi(g) = []

The next rules are strictly ordered:

  p1

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