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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        t#_A(x1,x2,x3) = x1 + x2 + x3
        f_A(x1) = x1 + 1
        g_A(x1) = x1 + 1
        t_A(x1,x2,x3) = 0
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g > t > f > t#
      
      argument filter:
    
        pi(t#) = [1, 2, 3]
        pi(f) = 1
        pi(g) = 1
        pi(t) = []
    

The next rules are strictly ordered:

  p1, p2

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