YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: f#(x,y) -> g#(x,y)
p2: g#(h(x),y) -> f#(x,y)
p3: g#(h(x),y) -> g#(x,y)

and R consists of:

r1: f(x,y) -> g(x,y)
r2: g(h(x),y) -> h(f(x,y))
r3: g(h(x),y) -> h(g(x,y))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: f#(x,y) -> g#(x,y)
p2: g#(h(x),y) -> g#(x,y)
p3: g#(h(x),y) -> f#(x,y)

and R consists of:

r1: f(x,y) -> g(x,y)
r2: g(h(x),y) -> h(f(x,y))
r3: g(h(x),y) -> h(g(x,y))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3

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