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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((0,1),(0,0)) x1 + (3,4)
          g#_A(x1,x2) = ((0,1),(0,0)) x1 + (1,4)
          h_A(x1) = ((0,0),(0,1)) x1 + (2,3)
  
    precedence:
    
      f# = g# = h
    
    partial status:
  
      pi(f#) = []
      pi(g#) = []
      pi(h) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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)

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}


-- 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) -> 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          f#_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (2,2)
          g#_A(x1,x2) = ((1,1),(0,0)) x1 + x2 + (1,2)
          h_A(x1) = ((0,0),(1,1)) x1 + (1,1)
  
    precedence:
    
      h > f# = g#
    
    partial status:
  
      pi(f#) = []
      pi(g#) = []
      pi(h) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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 estimated dependency graph contains the following SCCs:

  (no SCCs)