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__eq(0, 0) → true
a__eq(s(X), s(Y)) → a__eq(X, Y)
a__eq(X, Y) → false
a__inf(X) → cons(X, inf(s(X)))
a__take(0, X) → nil
a__take(s(X), cons(Y, L)) → cons(Y, take(X, L))
a__length(nil) → 0
a__length(cons(X, L)) → s(length(L))
mark(eq(X1, X2)) → a__eq(X1, X2)
mark(inf(X)) → a__inf(mark(X))
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(length(X)) → a__length(mark(X))
mark(0) → 0
mark(true) → true
mark(s(X)) → s(X)
mark(false) → false
mark(cons(X1, X2)) → cons(X1, X2)
mark(nil) → nil
a__eq(X1, X2) → eq(X1, X2)
a__inf(X) → inf(X)
a__take(X1, X2) → take(X1, X2)
a__length(X) → length(X)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

The rewrite sequence
mark(inf(X)) →⁺ cons(mark(X), inf(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 / inf(X)].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)