The Runtime Complexity (innermost) 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:

g(Cons(x, xs), y) → Cons(x, xs)
h(Nil, y) → h(Nil, y)
h(Cons(x, xs), y) → f(Cons(x, xs), y)
g(Nil, y) → h(Nil, y)
f(Nil, y) → g(Nil, y)
f(Cons(x, xs), y) → h(Cons(x, xs), y)
sp1(x, y) → f(x, y)
r(x, y) → 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:

g(Cons(x, xs), y) → Cons(x, xs)
h(Nil, y) → h(Nil, y)
h(Cons(x, xs), y) → f(Cons(x, xs), y)
g(Nil, y) → h(Nil, y)
f(Nil, y) → g(Nil, y)
f(Cons(x, xs), y) → h(Cons(x, xs), y)
sp1(x, y) → f(x, y)
r(x, y) → x

S is empty.
Rewrite Strategy: INNERMOST

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

Sliced the following arguments:
g/1
Cons/0
Cons/1
h/1
f/1
sp1/1
r/1

(4) Obligation:

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

g(Cons) → Cons
h(Nil) → h(Nil)
h(Cons) → f(Cons)
g(Nil) → h(Nil)
f(Nil) → g(Nil)
f(Cons) → h(Cons)
sp1(x) → f(x)
r(x) → x

S is empty.
Rewrite Strategy: INNERMOST

(5) InfiniteLowerBoundProof (EQUIVALENT transformation)

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

(6) BOUNDS(INF, INF)