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:

zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → 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:

zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
cons/0

(4) Obligation:

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

zeros → cons(n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X

S is empty.
Rewrite Strategy: FULL

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

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

(6) BOUNDS(INF, INF)