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:

  lexicographic path order with precedence:
  
    precedence:
    
      g > f#
    
    argument filter:
  
      pi(f#) = 1
      pi(g) = [1]

The next rules are strictly ordered:

  p1

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