Here are a few hints:

- You should introduce names in
`intros`

and `induction`

. This makes the proofs easier to understand, because the names are more reasonable. (See the proof script skeleton below)
- In the second step, instead of
`simpl`

, you should use a rewriting lemma like `merge2`

. You should do a case analysis on `x <= y`

, where `x`

and `y`

are the first elements of the lists. In the first case, you can rewrite `merge (x :: l1) (y :: l2)`

with `merge2`

. For the second case, you need a symmetric lemma:

```
Lemma merge3 : forall (x1 x2:nat) r1 r2,
x2 < x1 ->
merge (x1::r1) (x2::r2) =
x2::merge (x1::r1) r2.
Proof.
intros.
simpl.
bdestruct (x1 <=? x2).
- omega.
- reflexivity.
Qed.
```

I also used the following two inversion lemmas:

```
Lemma sorted_cons_inv : forall x y l1, sorted (x :: y :: l1) -> x <= y /\ sorted (y :: l1).
Proof. intros. now inversion H. Qed.
Lemma sorted_cons_inv' : forall x l1, sorted (x :: l1) -> sorted l1.
Proof. intros. now inversion H. Qed.
```

Now, here’s the skeleton of the proof script. You should be able to fill in the gaps yourself.

```
intros l1.
induction l1 as [ | x l1 IHl1]; intros Hsorted1.
- intros [ | x l2']; simpl; tauto.
- induction l2 as [ | y l2 IHl2]; intros Hsorted2.
+ simpl. tauto.
+ bdestruct (x <=? y).
* (* ... *)
* (* ... *)
```

PS: Please surround code with ```, like this:

```
```
code
```
```