YES

We show the termination of the relative TRS R/S:

  R:
  t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
  t(g(x),g(y),f(z)) -> t(f(y),f(z),x)

  S:
  f(g(x)) -> g(f(x))
  g(f(x)) -> f(g(x))
  f(f(x)) -> g(g(x))
  g(g(x)) -> f(f(x))

-- SCC decomposition.

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

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

and R consists of:

r1: t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
r2: t(g(x),g(y),f(z)) -> t(f(y),f(z),x)
r3: f(g(x)) -> g(f(x))
r4: g(f(x)) -> f(g(x))
r5: f(f(x)) -> g(g(x))
r6: g(g(x)) -> f(f(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
r2: t(g(x),g(y),f(z)) -> t(f(y),f(z),x)
r3: f(g(x)) -> g(f(x))
r4: g(f(x)) -> f(g(x))
r5: f(f(x)) -> g(g(x))
r6: g(g(x)) -> f(f(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            t#_A(x1,x2,x3) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + ((0,1),(0,0)) x3 + (0,3)
            f_A(x1) = ((0,0),(0,1)) x1 + (5,2)
            g_A(x1) = ((0,0),(0,1)) x1 + (5,2)
            t_A(x1,x2,x3) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + ((0,1),(0,0)) x3 + (1,3)
    
      precedence:
      
        t# = f = g = t
      
      partial status:
    
        pi(t#) = []
        pi(f) = []
        pi(g) = []
        pi(t) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            t#_A(x1,x2,x3) = (1,1)
            f_A(x1) = (1,1)
            g_A(x1) = (1,1)
            t_A(x1,x2,x3) = ((1,0),(0,0)) x2 + (1,2)
    
      precedence:
      
        t# > f = g = t
      
      partial status:
    
        pi(t#) = []
        pi(f) = []
        pi(g) = []
        pi(t) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
r2: t(g(x),g(y),f(z)) -> t(f(y),f(z),x)
r3: f(g(x)) -> g(f(x))
r4: g(f(x)) -> f(g(x))
r5: f(f(x)) -> g(g(x))
r6: g(g(x)) -> f(f(x))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

r1: t(f(x),g(y),f(z)) -> t(z,g(x),g(y))
r2: t(g(x),g(y),f(z)) -> t(f(y),f(z),x)
r3: f(g(x)) -> g(f(x))
r4: g(f(x)) -> f(g(x))
r5: f(f(x)) -> g(g(x))
r6: g(g(x)) -> f(f(x))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          t#_A(x1,x2,x3) = max{x1 + 1, x2 - 3, x3 - 5}
          f_A(x1) = x1 + 7
          g_A(x1) = x1 + 7
          t_A(x1,x2,x3) = 7
    
      precedence:
      
        t# = f = g = t
      
      partial status:
    
        pi(t#) = []
        pi(f) = []
        pi(g) = []
        pi(t) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          t#_A(x1,x2,x3) = 4
          f_A(x1) = max{4, x1}
          g_A(x1) = max{4, x1}
          t_A(x1,x2,x3) = 0
    
      precedence:
      
        t# = f = g = t
      
      partial status:
    
        pi(t#) = []
        pi(f) = [1]
        pi(g) = [1]
        pi(t) = []
    

The next rules are strictly ordered:

  p1

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