The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

a__from(X) → cons(mark(X), from(s(X)))
a__head(cons(X, XS)) → mark(X)
a__2nd(cons(X, XS)) → a__head(mark(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__sel(0, cons(X, XS)) → mark(X)
a__sel(s(N), cons(X, XS)) → a__sel(mark(N), mark(XS))
mark(from(X)) → a__from(mark(X))
mark(head(X)) → a__head(mark(X))
mark(2nd(X)) → a__2nd(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(sel(X1, X2)) → a__sel(mark(X1), mark(X2))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(0) → 0
mark(nil) → nil
a__from(X) → from(X)
a__head(X) → head(X)
a__2nd(X) → 2nd(X)
a__take(X1, X2) → take(X1, X2)
a__sel(X1, X2) → sel(X1, X2)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
mark(from(X)) →⁺ cons(mark(mark(X)), from(s(mark(X))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X / from(X)].
The result substitution is [ ].

The rewrite sequence
mark(from(X)) →⁺ cons(mark(mark(X)), from(s(mark(X))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0].
The pumping substitution is [X / from(X)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)