YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

r1: g(f(x),y) -> f(h(x,y))
r2: h(x,y) -> g(x,f(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: g#(f(x),y) -> h#(x,y)
p2: h#(x,y) -> g#(x,f(y))

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1,x2) = x1
        f_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        h#_A(x1,x2) = ((1,1),(1,1)) x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1,x2) = ((0,1),(1,0)) x1 + (1,1)
        f_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        h#_A(x1,x2) = x1
    

The next rules are strictly ordered:

  p1, p2

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