YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

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

and R consists of:

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{0, x1 - 3}
        f_A(x1,x2) = max{4, x1 - 5, x2 + 2}
        h_A(x1,x2) = max{3, x1 + 1, x2}
  
    precedence:
    
      g# = f = h
    
    partial status:
  
      pi(g#) = []
      pi(f) = []
      pi(h) = [2]

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

and R consists of:

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

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#(x,h(y,z)) -> g#(x,y)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{x1 + 1, x2 + 1}
        h_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      g# = h
    
    partial status:
  
      pi(g#) = [2]
      pi(h) = [1, 2]

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1,x2) = max{0, x1 - 2}
        h_A(x1,x2) = max{3, x1 + 1, x2 + 1}
        f_A(x1,x2) = max{x1 + 4, x2 - 3}
  
    precedence:
    
      g# = h = f
    
    partial status:
  
      pi(g#) = []
      pi(h) = [2]
      pi(f) = []

The next rules are strictly ordered:

  p1

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