The Runtime Complexity (full) 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:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

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:

sub(0, 0) → 0
sub(s(x), 0) → s(x)
sub(0, s(x)) → 0
sub(s(x), s(y)) → sub(x, y)
zero(nil) → zero2(0, nil)
zero(cons(x, xs)) → zero2(sub(x, x), cons(x, xs))
zero2(0, nil) → nil
zero2(0, cons(x, xs)) → cons(sub(x, x), zero(xs))
zero2(s(y), nil) → zero(nil)
zero2(s(y), cons(x, xs)) → zero(cons(x, xs))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

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

(4) BOUNDS(n^1, INF)