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:

add0(x', Cons(x, xs)) → add0(Cons(Cons(Nil, Nil), x'), xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
add0(x, Nil) → x
goal(x, y) → add0(x, y)

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:

add0(x', Cons(x, xs)) → add0(Cons(Cons(Nil, Nil), x'), xs)
notEmpty(Cons(x, xs)) → True
notEmpty(Nil) → False
add0(x, Nil) → x
goal(x, y) → add0(x, y)

S is empty.
Rewrite Strategy: INNERMOST

(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:

add0(x', Cons(xs)) → add0(Cons(x'), xs)
notEmpty(Cons(xs)) → True
notEmpty(Nil) → False
add0(x, Nil) → x
goal(x, y) → add0(x, y)

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
add0(x', Cons(xs)) →⁺ add0(Cons(x'), 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 [x' / Cons(x')].

(6) BOUNDS(n^1, INF)