@hammersklavier,

1. Ǝx(Gx -> Px)

2. ∀x(Px -> Ex)

ergo (Ǝx)(Gx -> Px)

First problem is that this argument begs the question. Assumption 1 is your conclusion, unless the parenthesis around your Ǝx makes the conclusion different from 1. But from my understanding, it shouldn't. All you really need to do is use a reiteration rule or a copy rule.

Do you mean to say?

1. Ǝx(Gx -> Px)

2. ∀x(Px -> Ex)

**ergo (Ǝx)(Gx -> Ex)**
---

1. Ǝx(Gx -> Px) | A

2. ∀x(Px -> Ex) | A

3. Negated conclusion | Provisional Assumption

4. Ax-(Gx -> Ex) | Quantifier Exchange, 3

5. Ga -> Pa | Existential Out, 1

6. -(Ga -> Ea) | Universal Out, 4

7. Pa -> Ea | Universal Out, 2

8. Ga . -Ea | Arrow Rule, 6

9. Pa | Modus Ponens, 5, 8 (one of the conjunctions)

10. Ea | Modus Ponens, 7, 9

11. -Ea | Conjunction out, 8

12. Ea . -Ea | Conjunction in, 10, 11

ergo (Ǝx)(Gx -> Ex) | Reductio ad absurdum or indirect proof

But even then, I thought existential statements were supposed to have conjunction symbols between the terms, not implication symbols.