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:

list(Cons(x, xs)) → list(xs)
list(Nil) → True
list(Nil) → isEmpty[Match](Nil)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
goal(x) → list(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:

list(Cons(x, xs)) → list(xs)
list(Nil) → True
list(Nil) → isEmpty[Match](Nil)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
goal(x) → list(x)

S is empty.
Rewrite Strategy: INNERMOST

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

Sliced the following arguments:
Cons/0
isEmpty[Match]/0

(4) Obligation:

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

list(Cons(xs)) → list(xs)
list(Nil) → True
list(Nil) → isEmpty[Match]
notEmpty(Cons(xs)) → True
notEmpty(Nil) → False
goal(x) → list(x)

S is empty.
Rewrite Strategy: INNERMOST

(5) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
list(Cons(xs)) →⁺ list(xs)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [xs / Cons(xs)].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)