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:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > g# > h#
      
      argument filter:
    
        pi(h#) = 1
        pi(f) = 1
        pi(g#) = 1
    
    2. matrix interpretations:
    
      carrier: N^4
      order: lexicographic order
      interpretations:
        h#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
        f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (2,1,1,1)
        g#_A(x1,x2) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
    

The next rules are strictly ordered:

  p1, p2

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