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^4
    order: lexicographic order
    interpretations:
      h#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (0,0,2,1)
      g#_A(x1,x2) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,0)) x1 + ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x2 + (0,0,1,1)

The next rules are strictly ordered:

  p1, p2

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