What is the principal difference between the two goals that make the second harder? (And what is the preferred way to solve it?)

```
Goal forall q,
(fix go a := match a with S a' => go a' | _ => 0 end)
q
=
fst ((fix go a := match a with S a' => go a' | _ => (0,0) end)
q).
Proof.
now induction 0.
Qed.
Goal forall q,
(fix go a r := match a with S a' => go a' (S r) | _ => 0 end)
q 0
=
fst ((fix go a r := match a with S a' => go a' (S r) | _ => (0,0) end)
q 0).
Proof.
induction 0.
- easy.
- give_up.
Abort.
```