Prove: $\{∃xp(x), ∃xp(x) → \forall x∃y[p(x)→q(y)]\} ⊢ \forall xq(x) ∨ ∃xr(x)$

Question

I don’t think I have understood how and when it’s legal to use universal generalization and/or existential generalization. I have done with this exercise this way: $\{∃xp(x), ∃xp(x) → \forall x∃y[p(x)→q(y)]\} ⊢ \forall xq(x) ∨ ∃xr(x)$

1. $\exists x\ p(x)$ Given
2. $\exists x\ p(x) \rightarrow \forall x\exists y[p(x)\rightarrow q(y)]$ Given
3. $p(a)$ EI 1
4. $p(a) \rightarrow [p(a) \rightarrow q(b)]$ EI & UI 2
5. $p(a) \rightarrow q(b)$ MP 3, 4
6. $q(b)$ MP 3, 5
7. $q(b) \vee r(a)$ Add 6

so, I don’t understand why and how should follow Ɐxq(x) ∨ ∃xr(x)

Answer 1

While at first I thought it was solvable, after looking at this problem for some time, I am convinced that it is a typo on behalf of the textbook. While it is possible to derive $\exists x q(x)$ from the premises, I do not think it possible to derive $\forall x q(x)$.

If the conclusion of the argument is false, so should be at least one of the premises. However, if you consider the case where q(1) is true, but q(2) is false, and $\forall x p(x)$ is true; then $\exists x p(x)$ is true, and $\exists y q(y)$ is true. This means that every premises is true, while the conclusion is false, and the argument is not valid.