### (0) Obligation:

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

pairNscons(0, n__incr(n__oddNs))
oddNsincr(pairNs)
incr(cons(X, XS)) → cons(s(X), n__incr(activate(XS)))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
zip(nil, XS) → nil
zip(X, nil) → nil
zip(cons(X, XS), cons(Y, YS)) → cons(pair(X, Y), n__zip(activate(XS), activate(YS)))
tail(cons(X, XS)) → activate(XS)
repItems(nil) → nil
repItems(cons(X, XS)) → cons(X, n__cons(X, n__repItems(activate(XS))))
incr(X) → n__incr(X)
oddNsn__oddNs
take(X1, X2) → n__take(X1, X2)
zip(X1, X2) → n__zip(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
repItems(X) → n__repItems(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__oddNs) → oddNs
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__zip(X1, X2)) → zip(activate(X1), activate(X2))
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__repItems(X)) → repItems(activate(X))
activate(X) → X

Rewrite Strategy: FULL

### (1) InfiniteLowerBoundProof (EQUIVALENT transformation)

The loop following loop proves infinite runtime complexity:
The rewrite sequence
oddNs →+ cons(s(0), n__incr(incr(oddNs)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,0,0].
The pumping substitution is [ ].
The result substitution is [ ].