YES

1 decompositions
 #1 -----------
   4: f(h(x7),h(x8)) -> f(x8,x7)
   5: f(x9,x10) -> f(x10,x9)
   6: g(x11) -> h(x11)
   7: h(x12) -> g(x12)

@Rule Labeling
--- R
   4: f(h(x7),h(x8)) -> f(x8,x7)
   5: f(x9,x10) -> f(x10,x9)
   6: g(x11) -> h(x11)
   7: h(x12) -> g(x12)

--- S
   4: f(h(x7),h(x8)) -> f(x8,x7)
   5: f(x9,x10) -> f(x10,x9)
   6: g(x11) -> h(x11)
   7: h(x12) -> g(x12)


NOTE: input TRS is reduced

original is
   1: f(g(x1),g(x2)) -> f(g(x1),h(x2))
   2: f(h(x3),g(x4)) -> f(g(x3),g(x4))
   3: f(g(x5),h(x6)) -> f(x5,x6)
   4: f(h(x7),h(x8)) -> f(x8,x7)
   5: f(x9,x10) -> f(x10,x9)
   6: g(x11) -> h(x11)
   7: h(x12) -> g(x12)

reduced to
   4: f(h(x7),h(x8)) -> f(x8,x7)
   5: f(x9,x10) -> f(x10,x9)
   6: g(x11) -> h(x11)
   7: h(x12) -> g(x12)