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:

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

a__nats → cons(0, incr(nats))
a__pairs → cons(0, incr(odds))
a__odds → a__incr(a__pairs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__head(cons(X, XS)) → mark(X)
a__tail(cons(X, XS)) → mark(XS)
mark(nats) → a__nats
mark(pairs) → a__pairs
mark(odds) → a__odds
mark(incr(X)) → a__incr(mark(X))
mark(head(X)) → a__head(mark(X))
mark(tail(X)) → a__tail(mark(X))
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__nats → nats
a__pairs → pairs
a__odds → odds
a__incr(X) → incr(X)
a__head(X) → head(X)
a__tail(X) → tail(X)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
a__tail(cons(X, tail(cons(X18666_3, X28667_3)))) →⁺ a__tail(cons(mark(X18666_3), X28667_3))
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [X28667_3 / tail(cons(X18666_3, X28667_3))].
The result substitution is [X / mark(X18666_3)].

(4) BOUNDS(n^1, INF)