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__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(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(repItems(cons(X125049_3, X225050_3))) →⁺ cons(mark(mark(X125049_3)), cons(mark(X125049_3), repItems(X225050_3)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [X125049_3 / repItems(cons(X125049_3, X225050_3))].
The result substitution is [ ].

The rewrite sequence
mark(repItems(cons(X125049_3, X225050_3))) →⁺ cons(mark(mark(X125049_3)), cons(mark(X125049_3), repItems(X225050_3)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0].
The pumping substitution is [X125049_3 / repItems(cons(X125049_3, X225050_3))].
The result substitution is [ ].

(2) BOUNDS(2^n, INF)