YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: f(x,g(x)) -> x
r2: f(x,h(y)) -> f(h(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#(x,h(y)) -> f#(h(x),y)

and R consists of:

r1: f(x,g(x)) -> x
r2: f(x,h(y)) -> f(h(x),y)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1,x2) = x1 + ((1,0),(1,1)) x2
      h_A(x1) = x1 + (1,1)

The next rules are strictly ordered:

  p1
  r1, r2

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