The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(INF, INF).

(0) Obligation:

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

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(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:

2nd(cons1(X, cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X, X1))
from(X) → cons(X, from(s(X)))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
cons1/0
s/0

(4) Obligation:

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

2nd(cons1(cons(Y, Z))) → Y
2nd(cons(X, X1)) → 2nd(cons1(X1))
from(X) → cons(X, from(s))

S is empty.
Rewrite Strategy: FULL

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

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

(6) BOUNDS(INF, INF)