Consider the TRS R consisting of the rewrite rules

1: half(0) -> 0
2: half(s(0)) -> 0
3: half(s(s(x))) -> s(half(x))
4: lastbit(0) -> 0
5: lastbit(s(0)) -> s(0)
6: lastbit(s(s(x))) -> lastbit(x)
7: conv(0) -> cons(nil,0)
8: conv(s(x)) -> cons(conv(half(s(x))),lastbit(s(x)))

There are 5 dependency pairs:

9: HALF(s(s(x))) -> HALF(x)
10: LASTBIT(s(s(x))) -> LASTBIT(x)
11: CONV(s(x)) -> CONV(half(s(x)))
12: CONV(s(x)) -> HALF(s(x))
13: CONV(s(x)) -> LASTBIT(s(x))

The approximated dependency graph contains 3 SCCs:
{9},
{10}
and {11}.

- Consider the SCC {9}.
There are no usable rules.
By taking the polynomial interpretation
[HALF](x) = [s](x) = x + 1,
rule 9
is strictly decreasing.

- Consider the SCC {10}.
There are no usable rules.
By taking the polynomial interpretation
[LASTBIT](x) = [s](x) = x + 1,
rule 10
is strictly decreasing.

- Consider the SCC {11}.
The usable rules are {1-3}.
By taking the polynomial interpretation
[0] = 0,
[CONV](x) = x,
[s](x) = x + 1
and [half](x) = x - 1,
we obtain a quasi-model of the usable rules.
Furthermore, dependency pair 11
is strictly decreasing.

Hence the TRS is terminating.