The Runtime Complexity (innermost) 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__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Rewrite Strategy: INNERMOST

(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__2nd(cons1(X, cons(Y, Z))) → mark(Y)
a__2nd(cons(X, X1)) → a__2nd(cons1(mark(X), mark(X1)))
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
mark(cons1(X1, X2)) → cons1(mark(X1), mark(X2))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

S is empty.
Rewrite Strategy: INNERMOST

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
a__2nd(cons(X, 2nd(cons(X11014_0, X21015_0)))) →⁺ a__2nd(cons1(mark(X), a__2nd(cons(mark(X11014_0), X21015_0))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1].
The pumping substitution is [X21015_0 / 2nd(cons(X11014_0, X21015_0))].
The result substitution is [X / mark(X11014_0)].

(4) BOUNDS(n^1, INF)