YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: h#(f(x),y) -> g#(x,y)
p2: g#(x,y) -> h#(x,y)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: h#(f(x),y) -> g#(x,y)
p2: g#(x,y) -> h#(x,y)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      h#_A(x1,x2) = x1
      f_A(x1) = x1 + 2
      g#_A(x1,x2) = x1 + 1

The next rules are strictly ordered:

  p1, p2

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