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:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

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:

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf(X) → cons(X, inf(s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(Y, take(X, L))
length(nil) → 0
length(cons(X, L)) → s(length(L))

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
inf/0
cons/0

(4) Obligation:

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

eq(0, 0) → true
eq(s(X), s(Y)) → eq(X, Y)
eq(X, Y) → false
inf → cons(inf)
take(0, X) → nil
take(s(X), cons(L)) → cons(take(X, L))
length(nil) → 0
length(cons(L)) → s(length(L))

S is empty.
Rewrite Strategy: FULL

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

The loop following loop proves infinite runtime complexity:
The rewrite sequence
inf →⁺ cons(inf)
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)