The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

fstsplit(0, x) → nil
fstsplit(s(n), nil) → nil
fstsplit(s(n), cons(h, t)) → cons(h, fstsplit(n, t))
sndsplit(0, x) → x
sndsplit(s(n), nil) → nil
sndsplit(s(n), cons(h, t)) → sndsplit(n, t)
empty(nil) → true
empty(cons(h, t)) → false
leq(0, m) → true
leq(s(n), 0) → false
leq(s(n), s(m)) → leq(n, m)
length(nil) → 0
length(cons(h, t)) → s(length(t))
app(nil, x) → x
app(cons(h, t), x) → cons(h, app(t, x))
map_f(pid, nil) → nil
map_f(pid, cons(h, t)) → app(f(pid, h), map_f(pid, t))
head(cons(h, t)) → h
tail(cons(h, t)) → t
ring(st_1, in_2, st_2, in_3, st_3, m) → if_1(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_1)))
if_1(st_1, in_2, st_2, in_3, st_3, m, false) → ring(sndsplit(m, st_1), cons(fstsplit(m, st_1), in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_2(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_2)))
if_2(st_1, in_2, st_2, in_3, st_3, m, true) → if_3(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_2)))
if_3(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, sndsplit(m, st_2), cons(fstsplit(m, st_2), in_3), st_3, m)
if_2(st_1, in_2, st_2, in_3, st_3, m, false) → if_4(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(two, head(in_2)), st_2))))
if_4(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, tail(in_2), sndsplit(m, app(map_f(two, head(in_2)), st_2)), cons(fstsplit(m, app(map_f(two, head(in_2)), st_2)), in_3), st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_5(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(two, head(in_2))))
if_5(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, tail(in_2), st_2, in_3, st_3, m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_6(st_1, in_2, st_2, in_3, st_3, m, leq(m, length(st_3)))
if_6(st_1, in_2, st_2, in_3, st_3, m, true) → if_7(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, st_3)))
if_7(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, in_3, sndsplit(m, st_3), m)
if_6(st_1, in_2, st_2, in_3, st_3, m, false) → if_8(st_1, in_2, st_2, in_3, st_3, m, empty(fstsplit(m, app(map_f(three, head(in_3)), st_3))))
if_8(st_1, in_2, st_2, in_3, st_3, m, false) → ring(st_1, in_2, st_2, tail(in_3), sndsplit(m, app(map_f(three, head(in_3)), st_3)), m)
ring(st_1, in_2, st_2, in_3, st_3, m) → if_9(st_1, in_2, st_2, in_3, st_3, m, empty(map_f(three, head(in_3))))
if_9(st_1, in_2, st_2, in_3, st_3, m, true) → ring(st_1, in_2, st_2, tail(in_3), st_3, m)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
fstsplit(s(n), cons(h, t)) →⁺ cons(h, fstsplit(n, t))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [n / s(n), t / cons(h, t)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)