Consider the TRS R consisting of the rewrite rules

1: f(x,y) -> g1(x,x,y)
2: f(x,y) -> g1(y,x,x)
3: f(x,y) -> g2(x,y,y)
4: f(x,y) -> g2(y,y,x)
5: g1(x,x,y) -> h(x,y)
6: g1(y,x,x) -> h(x,y)
7: g2(x,y,y) -> h(x,y)
8: g2(y,y,x) -> h(x,y)
9: h(x,x) -> x

There are 8 dependency pairs:

10: F(x,y) -> G1(x,x,y)
11: F(x,y) -> G1(y,x,x)
12: F(x,y) -> G2(x,y,y)
13: F(x,y) -> G2(y,y,x)
14: G1(x,x,y) -> H(x,y)
15: G1(y,x,x) -> H(x,y)
16: G2(x,y,y) -> H(x,y)
17: G2(y,y,x) -> H(x,y)

The approximated dependency graph contains no SCCs
and hence the TRS is trivially terminating.