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:

active(nats) → mark(cons(0, incr(nats)))
active(pairs) → mark(cons(0, incr(odds)))
active(odds) → mark(incr(pairs))
active(incr(cons(X, XS))) → mark(cons(s(X), incr(XS)))
active(head(cons(X, XS))) → mark(X)
active(tail(cons(X, XS))) → mark(XS)
active(cons(X1, X2)) → cons(active(X1), X2)
active(incr(X)) → incr(active(X))
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(tail(X)) → tail(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
incr(mark(X)) → mark(incr(X))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
tail(mark(X)) → mark(tail(X))
proper(nats) → ok(nats)
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(0) → ok(0)
proper(incr(X)) → incr(proper(X))
proper(pairs) → ok(pairs)
proper(odds) → ok(odds)
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(tail(X)) → tail(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
incr(ok(X)) → ok(incr(X))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
tail(ok(X)) → ok(tail(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
cons(mark(X1), X2) →⁺ mark(cons(X1, X2))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [X1 / mark(X1)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)