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:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

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:

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s(X), incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
s/0

(4) Obligation:

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

nats → adx(zeros)
zeros → cons(0, zeros)
incr(cons(X, Y)) → cons(s, incr(Y))
adx(cons(X, Y)) → incr(cons(X, adx(Y)))
hd(cons(X, Y)) → X
tl(cons(X, Y)) → Y

S is empty.
Rewrite Strategy: FULL

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

The loop following loop proves infinite runtime complexity:
The rewrite sequence
zeros →⁺ cons(0, zeros)
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [ ].
The result substitution is [ ].

(6) BOUNDS(INF, INF)