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:

mul0(Cons(x, xs), y) → add0(mul0(xs, y), y)
add0(Cons(x, xs), y) → add0(xs, Cons(S, y))
mul0(Nil, y) → Nil
add0(Nil, y) → y
goal(xs, ys) → mul0(xs, ys)

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:

mul0(Cons(x, xs), y) → add0(mul0(xs, y), y)
add0(Cons(x, xs), y) → add0(xs, Cons(S, y))
mul0(Nil, y) → Nil
add0(Nil, y) → y
goal(xs, ys) → mul0(xs, ys)

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:

mul0(Cons(xs), y) → add0(mul0(xs, y), y)
add0(Cons(xs), y) → add0(xs, Cons(y))
mul0(Nil, y) → Nil
add0(Nil, y) → y
goal(xs, ys) → mul0(xs, ys)

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
mul0(Cons(xs), y) →⁺ add0(mul0(xs, y), y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [xs / Cons(xs)].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)